шамин с сдо
.pdff(x)
f′(x)
f(x) > 0 |
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f(x) |
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f(x) ≥ 0 |
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f(x) |
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f(x) < 0 |
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f(x) |
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f(x) ≤ 0 |
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f(x) |
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f(x) = |
x3 |
3x3 |
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3 − |
2 + 2x − |
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f′(x) = x2 − 3x + 2. |
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f′(x) = 0 |
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x1 = 1 x2 = 2 |
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2.0 |
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1.5 |
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1.0 |
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f(x) |
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0.5 |
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0.0 |
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−0.5 |
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0.0 |
0.5 |
1.0 |
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1.5 |
2.0 |
2.5 |
3.0 |
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x |
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f(x) |
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f′(x) |
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( |
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f′(x) > 0 |
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−∞ |
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f(x) |
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(1, 2) |
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f′(x) < 0 |
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f(x) |
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(2, |
∞ |
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f′(x) > 0 |
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f(x) |
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f(x)
f′(x) = 0, (a, b).
f(x) =
x3 |
[−1, 1] |
c = 0 |
f′(x) = 3x2 = 0
x = 0
f(x)
U(ε, c) |
c |
f(x)
f′(x) > 0, x < c, x U(ε, c)
f′(x) < 0, x > c, x U(ε, c),
c
f′(x) < 0, x < c, x U(ε, c)
f′(x) > 0, x > c, x |
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U(ε, c), |
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c |
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x U(ε, c) |
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x 6= c |
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f(x) |
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f(c) |
− |
f(x) = f′(ξ)(c |
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x), |
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ξ |
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c x |
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x < c |
f′(ξ) |
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x > c |
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f(c) − f(x) > 0.
c
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f(x) |
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c |
f(x) |
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f′′(c) < 0 |
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f′′(c) > 0. |
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c |
f′′(c) < 0 |
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f′(x) |
c |
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c |
f′(c) = 0 |
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c |
f′(x) |
c |
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c |
f(x) |
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c |
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f′′(x) > 0 |
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f(x) = cos x
x = 0
f′′(x)
f′′(x) = (cos x)′′ = − cos x,
f′′(0) = −1,
f(x) = x(1 − x)
f′(x) = 1 − 2x,
f′′(x) = −2.
f′(x) = 1 |
− |
2x = 0. |
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x0 |
= 1 |
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2 |
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x0 |
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f(x0) = 2 1 − 2 |
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4 . |
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1 |
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1 |
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1 |
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f(x)
M(x, f(x))
f(x)
(a, b)
(a, b)
f(x)
(a, b)
(a, b)
f(x)
(a, b)
(a, b)
f(x)
f′′(x0) < 0 f(x0)
(a, b)
y = f(x)
f′′(x) ≥
0 |
(a, b) |
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(a, b) |
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c |
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y |
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f(c) = f′(c)(x |
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c). |
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y = L(x) |
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f(x) L(x) |
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f(x) |
c |
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n = 1 |
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f(x) = f(c) + |
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f′(c) |
(x − c) + |
f′′(ξ) |
(x − c)2, |
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1! |
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2! |
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ξ
cx
f(x) − L(x) = f′′(ξ) (x − c)2 ≥ 0. 2
y = f(x)
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y = ln x |
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(0, ∞) |
1 |
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f′′(x) = − |
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< 0, x (0, ∞). |
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x2 |
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ln x |
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(c, f(c)) |
y = f(x) |
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c |
c |
y = f(x)
c
f′′(c) = 0,
f′′(x)
c |
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f(x) = x3 |
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f′′(x) = 6x |
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x = 0 |
x < 0 x > 0 |
x = 0 |
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y =
f(x) x → +∞
y = kx + b,
f(x)
f(x) = kx + b + α(x),
lim α(x) = 0
x→+∞
y = f(x) x → +∞ |
x = a |
lim f(x)
x→a+0
lim f(x)
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x→a−0 |
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+∞ |
−∞ |
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y = f(x) |
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x → +∞ |
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lim |
f(x) |
= k |
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x→+∞ |
x |
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x |
lim [f(x) |
− |
kx] = b. |
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+ |
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→ ∞ |
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y = f(x) |
x → |
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+∞ |
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lim |
f(x) |
= lim |
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kx + b + α(x) |
= k, |
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x |
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x |
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x→+∞ |
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x→+∞ |
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x |
lim [f(x) |
− |
kx] = |
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lim [b |
− |
α(x)] = b. |
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+ |
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x |
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→ ∞ |
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→ ∞ |
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α(x) = f(x) − kx − b
x→ +∞ x → −∞
y = |
3x3 + 2x2 + 1 |
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x2 + 1 |
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lim |
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3x3 + 2x2 + 1 |
= 3 = k. |
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x(x2 + 1) |
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x→±∞ |
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x→±∞ |
3x3 + 2x2 + 1 |
− |
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x2 + 1 |
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lim |
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3x |
= 2. |
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y = f(x)
y = L(x) = 3x + 2.
y = ln x
x = 0
lim ln x = −∞.
x→+0