3) if the derivative at some point is positive (negative), then in the
neighbourhood of this point the function increases (decreases); |
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4) if the functions f1 (x) and |
f2 (x) |
are differentiable on the interval (a; b) , |
f ′ (x) = |
f ′ (x) , at |
the points a |
and |
b functions are |
continuous, |
then the |
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difference between these functions is |
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f1 (x) − f2 (x) = const. |
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(Cauchy’s theorem). If the functions |
f (x) and |
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Theorem 3.14 |
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continuous on the segment [a; b] , differentiable at the |
interval |
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(a; b) , g (x) ≠ 0 , x (a; b) , then there’s at least one point c (a; b) , |
where the following statement takes place :
f (b) − f (a) |
= |
f ′(c) . |
g(b) − g(a) |
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g ′(c) |
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21.2. Taylor’s and Maclaurin’s formulas
In some cases Taylor’s formula is used for function investigation (calculation the value of a function, boundaries of a function).
Theorem 3.15 |
Let the function f (x) at the point |
x0 has the derivatives to |
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the (n + 1) -th order inclusively and |
x is arbitrary value from |
the given neighbourhood (x ≠ x0 ) . Then, there’s point c between points x0 and x , so that Taylor’s formula is true
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f |
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f |
′′(x0 ) |
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f (x) = f (x0 ) + |
(x0 ) |
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(x − x0 ) + |
(x − x0 )2 |
+ … |
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1! |
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2! |
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f |
(n) (x ) |
(x − x )n + |
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... + |
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R (x) , |
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n! |
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where R |
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f (n+1) (c) |
(x − x |
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is remainder term in the Lagrange’s form, |
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(n + 1)! |
0 |
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c = x0 + θ(x − x0 ) , 0 < θ < 1. |
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The expression |
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f (n) (x ) |
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f |
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(x − x0 )n |
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Pn (x) = f (x0 ) + |
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(x − x0 ) + …+ |
0 |
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1! |
n! |
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is called Taylor’s polynomial.
241
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If x0 = 0 then Taylor’s formula is called Maclaurin’s formula:
f (x) = f (0) + |
f ′(0) |
x + |
f ′′(0) |
x |
2 |
+ …+ |
f (n) (0) |
x |
n |
+ |
f (n+1) (c) |
x |
n+1 |
1! |
2! |
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n! |
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(n + 1)! |
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where c = θ x, 0 ≤ θ ≤ 1.
Let us consider some examples of expansions of elementary functions using Maclaurin’s formula:
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ex = 1+ |
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+ |
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x2 |
+ |
x3 |
+ …+ |
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xn |
+ Rn (x) ; |
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2! |
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1! |
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3! |
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n! |
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sin x = x − |
x3 |
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+ |
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− …+ (−1) |
n |
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x2n+1 |
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+ R2n+1 (x) ; |
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3! |
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5! |
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(2n + 1)! |
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cos x = 1− |
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+ |
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− …+ |
(−1) |
n |
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x2n |
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+ R2n (x) ; |
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2! |
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4! |
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(2n)! |
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ln(1+ x) = x − |
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+ |
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− …+ (−1) |
n−1 |
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xn |
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+ Rn (x) ; |
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(3.12) |
2 |
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arctg x = x − |
x3 |
+ |
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− …+ |
(−1) |
n |
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x2n+1 |
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+ R2n+1 (x) ; |
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3 |
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5 |
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2n + 1 |
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(1+ x)m = 1+ |
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x2 + |
m(m − 1)(m − 2) |
x3 |
+ …+ |
(3.13) |
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1! |
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2! |
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3! |
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+ |
m(m −1)(m − 2) |
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(m − (n −1)) |
xn + R |
(x) , |
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n! |
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If m = −1 then |
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n |
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+ Rn (x) , |
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(3.14) |
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= 1− x + x − x |
+…+ (−1) |
x |
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1+ x |
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1 |
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= 1+ x + x2 + x3 + …+ xn + Rn (x) . |
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(3.15) |
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1− x |
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21.3. L’Hospital’s rule
We use L’Hospital’s rule to disclosing indeterminate form of such kind as
242
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Theorem 3.16 (L’Hospital’s rule). Let functions ƒ(x) and g(x) be:
1) determined and differentiable in a neighbourhood of a point x0 , except,
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probably, the point x0 , and |
g′(x) ≠ 0 |
in this neighbourhood; |
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2) lim |
f (x) = lim |
g(x) = 0 |
or |
lim |
f (x) = lim g(x) = ∞ ; i.e. ƒ(x), g(x) |
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x→ x0 |
x→ x0 |
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x→ x0 |
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x→ x0 |
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are simultaneously infinitesimals or infinites as |
x → x0 ; |
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3) there is a finite lim |
f ′(x) |
. |
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x |
→ x0 |
g′(x) |
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f (x) |
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Then there is a limit of the quotient of functions lim |
and |
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g(x) |
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x→ x0 |
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lim |
f (x) |
= |
lim |
f ′(x) |
. |
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x→ x0 |
g(x) |
x→ x0 |
g′(x) |
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Let’s note, that the theorem is true and in that case, when x0 = ∞.
It should be noted, that sometimes students do a gross error, to search instead
of lim |
f ′(x) |
for |
lim |
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f (x) |
′ . |
g′(x) |
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x→ x |
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x→ x |
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g(x) |
0 |
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0 |
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L’Hospital’s rule is named after the mathematician which for the first time has published it. But this rule was proved for the first time by I. Bernulli, therefore L’Hospital’s rule is still named Bernulli-L’Hospital’s rule.
L’Hospital’s rule directly apply to disclosing indeterminate forms of a kind
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or |
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which are called the basic. Other indeterminate forms [0 ∞] , |
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[∞ − ∞] , |
1∞ |
, 00 |
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∞0 |
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are reduced to the basic. |
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10. Indeterminate form [0 ∞] ( lim |
f (x)g(x) , when lim f (x) = 0, |
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lim g(x) = ∞) |
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x→x0 |
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or ∞ |
x→x0 |
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x→x0 |
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is reduced to indeterminate forms 0 |
. Thus, |
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f (x)g(x) = |
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or |
f (x)g(x) = |
g(x) ∞ |
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20. Indeterminate form |
[ |
∞ − ∞ |
] |
( lim ( f (x) − g(x)) , when |
lim |
f (x) = ∞ |
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x→ x |
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x→ x |
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0 |
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and lim |
g(x) = ∞) |
is reduced to the indeterminate form 0 |
so: |
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243 |
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1 |
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f (x) − g(x) = |
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g(x) |
f (x) |
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f (x)g(x) |
30. Indeterminate forms |
1∞ , |
00 |
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∞0 |
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are reduced to the indeterminate |
form 0 ∞ with |
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the |
help |
of |
taking |
previous |
logarithm or representation of |
function f (x)g(x) |
as eg(x) ln f (x) |
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(using the basic logarithmic identity). |
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Micromodule 21
EXAMPLES OF PROBLEMS SOLUTION
Example 1. Expand functions using the Maclaurin’s formula:
a) |
f (x) = |
1 |
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b) f (x) = ln(2x |
2 |
+ 7x + 3). |
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4 + x 2 |
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Solution. а) we shall write down |
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= 4 |
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Then using the formula (3.14) we have |
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2n |
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f (x) = |
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1− |
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+ |
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− …+ (−1) |
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+ Rn |
(x) . |
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b) We use transformation |
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ln(2x2 + 7x + 3) = ln(1+ 2x)(x + 3) = ln(1+ 2x) + ln(x + 3) = |
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= ln(1+ 2x) + ln(1+ |
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Using the formula (3.12) twice, we shall get |
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ln(1+ 2x) = 2x − |
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+ Rn (x) , |
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n |
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ln(2x2 + 7x + 3) = ln 3 + |
∑ |
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+ |
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+ R (x) . |
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k |
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k =1 |
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244
http://vk.com/studentu_tk, http://studentu.tk/
Example 2. Calculate limit, using L’Hospital’s rule.
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Solution. We lim e3x − ex |
have indeterminate form 0 |
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, then lim |
e3x − ex |
= |
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x |
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3e3x − ex |
x→0 |
x |
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x→0 |
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= lim |
= 3 −1 = 2 . |
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1 |
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x→0 |
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Example 3. Calculate |
lim x2 ln x. |
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x → +0 |
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Solution. In this case we have indeterminate form 0 ∞ . We shall reduce to
indeterminate form ∞ |
and use L’Hospital’s rule: |
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∞ |
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lim x2 ln x = lim |
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ln x |
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lim |
(ln x)′ |
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= |
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lim |
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1/ x |
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= − |
1 |
lim x2 |
= 0 . |
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2 |
)′ |
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−2 / x3 |
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x→+0 |
x→+0 1/ x2 |
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x→+0 (1/ x |
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x→+0 |
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2 |
x→+0 |
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Example 4. Calculate lim |
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x + sin x |
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x→∞ |
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Solution. We have indeterminate form ∞ |
. We shall find |
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∞ |
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lim |
x + sin x |
= lim (1+ sin x) = 1+ 0 = 1. |
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x→∞ |
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x→∞ |
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Example 5. Calculate |
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lim (tgx − sec x) . |
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x→π |
2 |
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Solution. |
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[∞ − ∞] . |
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Here indeterminate form |
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We |
shall |
reduce |
it to |
indeterminate form 0 |
then use L’Hospital’s rule: |
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0 |
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sin x −1 |
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0 |
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lim |
(tg x − sec x) = |
lim (tg x − |
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) = |
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lim |
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= |
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= |
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cos x |
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cos x |
x |
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x→π / 2 |
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x |
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/ 2 |
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0 |
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= |
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lim |
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(sin x − 1)′ |
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= |
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lim |
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cos x |
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= 0 . |
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(cos x)′ |
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x→π / 2 |
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x→π / 2 − sin x |
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Example 6. Calculate |
lim (π − 2x)ños x . |
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x→π |
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value x = π |
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Solution. |
Substituting |
in |
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expression |
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the |
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, we |
shall |
receive |
indeterminate form 00. For |
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2 |
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convenience purposes |
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we |
execute |
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replacement |
π − 2x = t , then t → 0 . So, |
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245 |
http://vk.com/studentu_tk, http://studentu.tk/
lim (π − 2x)cos x |
= lim tsin(t / 2) = lim esin(t / 2) ln t = |
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lim sin(t / 2) ln t |
= eA , |
x→π / 2 |
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t→0 |
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t→0 |
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et→0 |
1 |
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ln t |
A = lim sin(t / 2) ln t = [0 ∞] = |
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1 |
lim t ln t = [0 ∞] = |
lim |
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= |
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t→0 |
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∞ |
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(ln t)′ |
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2 t→0 |
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2 t→0 1/ t |
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1 |
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1/ t |
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1 |
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= |
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= |
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lim |
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lim |
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= − |
lim t = 0 . |
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(1/ t)′ |
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−1/ t |
2 |
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∞ |
2 t→0 |
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2 t→0 |
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2 t→0 |
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Then
lim |
( |
π − 2x |
) |
cos x = e0 |
= 1. |
x→π / 2 |
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Micromodule 21
CLASS AND HOME ASSIGNMENTS
1.Expand a polynomial P4(x)=x4 – 5x3 + 5x2 + x + 2 in powers of binomial
x– 2.
Expand functions using Maclaurin’s formula:
1
2. 1− 2x .
5.x2 + 1 .
x− 2
8. x sin x .
3. |
1 |
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4. |
1 |
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2 + x |
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(x −1)(x + 2) |
6. |
x cos2 x . |
7. |
ln |
1− 2x |
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1+ x |
9. |
1 |
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10. ln(x2 − 3x + 2) . |
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1+ 4x |
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Calculate limits, using L’Hospital’s rule:
11. |
lim |
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7x − 3 |
. |
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12. lim (x2 |
+ 1)e− x . 13. |
lim(tg 2x)x . |
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x→∞ x2 + 2 |
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x→∞ |
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x cos 2x − sin x |
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16. lim x4 3 |
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14. |
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. 15. lim(ex + x) |
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x2 |
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lim |
x |
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x→0 |
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x3 |
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x→0 |
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x |
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17. |
lim ln(7 − 2x) ln(6 − 2x) . |
18. lim |
ln(cos 3x) |
. |
19. lim |
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ln2 x |
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x→3 |
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x→0 |
ln(cos 8x) |
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x→∞ |
x |
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arctg x |
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lim tg x ln |
2 |
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x . |
21. |
lim |
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lim |
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e |
x |
− 1 |
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x→0 |
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x→0 |
sin x |
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23. lim |
ex − esin x |
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24. |
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2 − (ex |
+ e− x ) cos x |
. 25. lim |
x + x2 |
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lim |
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x→0 |
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x − sin x |
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x→0 |
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x4 |
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x→∞ |
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x |
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246
http://vk.com/studentu_tk, http://studentu.tk/
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Answers |
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1. (x − 2)4 + 3(x − 2)3 − 7(x − 2) . 3. The instruction. Write |
down expression |
as |
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1 |
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1 |
. 5. The instruction. To reduce expression of a kind |
x + 2 − |
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5 |
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. |
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2 |
1+ x / 2 |
2(1 |
− x / 2) |
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11. ∞ . 12. 0. 13. 1. 14. –11/6. 15. e2 . 16. ∞ . 17. 0. 18. 3/8. 19. 0. 20. 0. 23. 1. 24. 1/3. 25. 0.
Micromodule 21
SELF-TEST ASSIGNMENTS
21.1. а) lim x ln2 x ; |
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x→0 |
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21.2. а) lim sin x ln2 x ; |
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x→0 |
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21.3. а) lim x2e− x ; |
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x→∞ |
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21.4. а) |
lim (x3 + 1)4− x ; |
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x→∞ |
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21.5. а) lim (x3 − x − 2) 3− x ; |
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x→∞ |
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21.6. а) lim |
1− cos x |
) |
ctg x |
; |
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x→0 |
( |
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21.7. а) lim |
ln x ln |
( |
x −1 |
; |
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x→1 |
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) |
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1 |
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21.8. а) |
lim x2 e |
x2 |
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x→0 |
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21.9. а) lim x4 2 x2 ;
x→0
− 1
21.10. а) lim x−2 2 x2 ;
x→0
− 1
21.11. а) lim x−4 5 x2 ;
x→0
b) |
lim |
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3x |
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c) lim(sin x)x . |
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x→∞ x2 − 5x + 2 |
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x→0 |
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b) lim |
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2x |
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c) lim(tg x)x . |
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x→∞ x2 + 3x + 1 |
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x→0 |
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b) lim |
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tg x − sin x |
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c) |
lim (tg x)2x − π . |
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x→0 |
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x − sin x |
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π |
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x→ 2 |
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b) lim |
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ln(x2 − 8) |
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; |
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c) |
lim |
( |
ctg x |
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sin x |
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x→3 2x2 − 5x − 3 |
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x→0 |
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1 |
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ex − e− x |
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b) lim |
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c) |
lim |
( |
ctg x |
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ln x . |
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x→0 sin x cos x |
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x→0 |
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π x |
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b) |
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; c) |
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cos |
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lim |
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− |
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lim |
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2 . |
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x→1 |
ln x |
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ln x |
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x→1 |
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b) lim |
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x cos x − sin x |
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x→0 |
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x3 |
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x→0 |
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1 |
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b) lim |
x − arctg x |
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c) |
lim |
(ex + x) |
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x |
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x→0 |
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x3 |
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x→0 |
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b) lim |
ex −cosx−xcosx |
; c) lim |
( |
sin x |
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tg x . |
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x→0 |
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x2 |
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x→0 |
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b) lim |
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ln cos x |
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c) lim |
( |
cos x |
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ctg2 x |
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x→0 ln cos 4x |
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x→0 |
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b) |
lim |
ln |
2 x |
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c) lim |
( |
cos 2x |
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x−2 |
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3 |
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x→∞ |
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x→0 |
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247
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1 |
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ln2 x |
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21.12. а) lim x−4 6− |
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b) lim |
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x4 |
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x→0 |
x→∞ 5 x4 |
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21.13. а) lim ln(5− 2x) ln(4 − 2x) ; |
b) lim |
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ln x |
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x→2 |
x→0 ctg x |
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21.14. а) |
lim ln(x − 3) ln(7 − 2x) ; |
b) lim |
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ln x |
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x→3 |
x→0 ctg x |
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21.15. а) |
lim ln(2x − 1) ln(2 − 2x) ; b) lim |
ln(sin 2x) |
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ln(sin 5x) |
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x→1 |
x→0 |
c) lim (arcsin x)x .
x→0
tg πx
c) lim(2 − x) 2 .
x→1
c) lim (tg x)tg 2x .
x→π / 4
1
; c) lim x1− x .
x→1
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21.16. а) lim |
1− cos x |
) |
ctg2 |
x ; |
b) lim |
ln(sin 3x) |
; |
c) |
lim ( |
2 |
arctg x)x . |
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x→0 |
( |
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x→0 ln(sin 7x) |
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x→+∞ |
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π |
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1 |
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ln(tg3 |
x) |
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21.17. а) lim(cos2 x − 3cos x + 2) e |
x2 |
; b) lim |
; c) |
lim |
xtg x . |
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ln(sin 4x) |
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x→0 |
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x→0 |
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x→+0 |
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1
21.18. а) lim(cos2 x − 1) e x2 ;
x→0
1
21.19. а) lim(1− cos2 x) e x2 ;
x→0
21.20. а) |
lim arcsin x ln2 (2x) ; |
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x→0 |
21.21. а) lim x3 ln2 x ; |
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x→0 |
21.22. а) |
lim x ln2 x ; |
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x→0 |
21.23. а) |
lim 4 x ln2 x ; |
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x→0 |
21.24. а) |
lim 3 x ln2 x ; |
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x→0 |
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1 |
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21.25. а) |
lim x8 2 |
x4 |
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x→0 |
b) lim |
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− |
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; c) |
lim |
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(tg x) |
2x−π |
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e |
x |
− 1 |
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x→0 |
x |
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x→π |
/ 2 |
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ln3 x |
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arctg x |
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b) lim |
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c) lim |
x2 |
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x→∞ |
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x→0 |
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1 |
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1 |
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sin x |
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b) lim |
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− |
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x2 |
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;c) |
lim |
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x |
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x→1 |
ln x |
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− 1 |
x→0 |
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b) lim |
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2x + 3x |
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c) |
lim xarcsin x . |
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x→∞ x3 − x + 1 |
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x→0 |
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b) |
lim |
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5x − 3x |
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; c) |
lim (− ln x)x . |
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x→∞ x2 − 6x + 4 |
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x→+0 |
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b) lim |
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3x2 |
+ 4x − 6 |
; c) lim(ctg x)sin x . |
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x→∞ |
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(1,1)x |
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x→0 |
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6x2 − x − 2 |
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1 |
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tg x |
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b) |
lim |
; c) |
lim |
x2 |
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x |
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x→∞ |
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(1, 5) |
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x→0 |
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x |
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2x2 +10x−20 |
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1 |
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b) |
lim |
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lim(x −1)2− x . |
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x→∞ |
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(2,5)x |
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x→2 |
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http://vk.com/studentu_tk, http://studentu.tk/
21.26. а) |
lim (x2 |
+ 4x + 5) 3− x ; |
b) lim |
ln(x2 + 3x + 1) |
; c) lim xarc tg 2x . |
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x→∞ |
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x→∞ |
ln(2x2 −1) |
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x→0 |
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21.27. а) lim (5x2 |
+ 3x + 2) 6− x ; |
b) lim |
ln(x2 + ex ) |
; |
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c) lim sin xarc tg x . |
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x→∞ |
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x→∞ |
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x |
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x→0 |
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21.28. а) |
lim x ln3 x ; |
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b) lim |
ln(3x2 + 2x ) |
;c) |
lim xx−sin x . |
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x |
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x→0 |
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x→∞ |
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x→0 |
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21.29. а) lim(x2 − 1) ln2 (x − 1) ; |
b) lim |
ln(4x2 + 3x ) |
;c) lim ln xln x . |
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x |
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x→1 |
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x→∞ |
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x→1 |
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− |
1 |
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1 |
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e− x |
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−6 |
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6 |
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− x |
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21.30. а) |
lim x |
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2 |
x |
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b) lim |
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− |
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; c) lim (e |
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) |
. |
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tg x |
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x→0 |
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x→0 |
sin x |
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e |
−1 |
x→∞ |
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Micromodule 22
BASIC THEORETICAL INFORMATION.
THE USAGE OF DERIVATIVE IN INVESTIGATION OF A FUNCTION
Monotony of a function. Extremum. Intervals of convexity and concavity, inflection points. Asymptotes. The greatest and the least meanings of the function. General scheme of investigation of a function and constructing of its graph.
Literature: [2, chapter 5], [4, part 5], [6, chapter 5, § 6], [7, chapter 6, §20, 21], [9], [10, chapter 4, §§ 27—31], [11, chapter 5, §1], [12, chapter 5, § 3, 4].
22.1. Increasing and decreasing functions. Relative extremum of a function
Definition 3.25. |
Function f (x) is |
called increasing |
(decreasing) on an |
interval |
(a; b) , if |
for any two points |
x1 |
and x2 from |
the given interval |
( x1 < x2 ), the following inequality is fulfilled |
f (x1 ) < f (x2 ) |
( f (x1 ) > f (x2 )) . |
Properties of increasing and decreasing functions: |
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1) |
if |
f |
′ |
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(x) > 0 for all x (a; b) , then function f (x) increases on (a; b); |
2) |
if |
f |
′ |
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x (a; b) , then function |
f (x) |
decreases on (a; b); |
(x) < 0 for all |
3) |
if |
f |
′ |
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x (a; b) ,then function |
f (x) |
is constant on (à; b) . |
(x) = 0 for all |
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249 |
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ff((xx0))
f(x) f(x0)
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О x0–δ x x0 |
x0+δ x |
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О x0–δ x x0 x0+δ x |
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x0 — point of maximum |
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x0 — is a point of minimum |
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Fig. 3.23 |
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Fig. 3.24 |
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Definition 3.26. Point x0 is |
called the point of relative maximum (or |
minimum) of function |
f (x) , if there exists such neighbourhood 0 < |
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x − x0 |
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< δ |
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of point x0 , which belongs to the domain of the function, and for all x |
from |
this neighbourhood the following inequality is fulfilled |
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f (x) < f (x0 ) ( f (x) > f (x0 )) . |
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Geometrical interpretation of the definition is clear from the Fig. 3.23 and 3.24. |
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Let’s define the conditions of relative extreme. |
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Theorem 3.17 |
(necessary condition |
of relative extreme). If |
a function |
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f (x) |
has at point x0 |
relative extremum and differentiable |
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at this point, then f ′(x0 ) = 0 . |
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Geometrical |
interpretation of |
the theorem 3.17. If function |
f (x) has in |
point x0 relative extremum and differentiable at this point, then in this point there exists the tangent line to the graph of function y = f (x) and this point
belongs to the x-axis.
Condition f ′(x0 ) = 0 is necessary, but is not enough for the function, which is differentiable at the point x0 , to have relative extremum. For example, the derivative of the function y = x3 at point x = 0 is equal to zero, but it does not have relative extremum at this point. So, function y =| x | has at point x = 0
minimum, but does not have the derivative at this point. Function y = 3 x is not
differentiable at point x = 0 and does not have extremum at this point.
Point, in which the derivative is equal to zero, is called stationary. Point, in which the derivative is equal to zero or does not exist are called critical. Critical point is the point of possible extremum.
250
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