4.1Mathematical Applications

on a closed and bounded interval [a, b], then there exist two points, c1 and c2, in [a, b] such that f(c1) is the absolute minimum of f on [a, b] and f(c2) is the absolute maximum of f on [a, b].

Proof. Since [a, b] is a closed and bounded set and f is continuous on [a, b], Theorem 4.1.1 follows from Theorem 2.3.14.

Definition 4.1.4 A function f is said to be increasing on an open interval (a, b) if f(x1) < f(x2) for all x1 and x2 in (a, b) such that x1 < x2. The function f is said to be decreasing on (a, b) if f(x1) > f(x2) for all x1 and x2 in (a, b) such that x1 < x2. The function f is said to be non-decreasing on (a, b) if f(x1) ≤ f(x2) for all x1 and x2 in (a, b) such that x1 < x2. The function f is said to be non-increasing on (a, b) if f(x1) ≥ f(x2) for all x1 and x2 in (a, b) such that x1 < x2.

Theorem 4.1.2 Suppose that a function f is defined on some open interval (a, b) containing a number c such that f0(c) exists and f0(c) 6= 0. Then f(c) is not a local extremum of f.

Proof. Suppose that f0(c) 6= 0. Let = 21 |f0(c)|. Then > 0. Since > 0 and

for all x such that 0 < |x − c| < δ. Thus, if c − δ < x1 < c < x2 < c + δ, then either f(x1) < f(c) < f(x2) or f(x1) > f(c) > f(x2). It follows that f(c) is not a local extremum.

Theorem 4.1.3 If f is defined on an open interval (a, b) containing c, f(c) is a local extremum of f and f0(c) exists, then f0(c) = 0.

Proof. This theorem follows immediately from Theorem 4.1.2.

Theorem 4.1.4 (Rolle’s Theorem) Suppose that a function f is continuous on a closed and bounded interval [a, b], di erentiable on the open interval (a, b) and f(a) = f(b). Then there exists some c such that a < c < b and f0(c) = 0.

Proof. Since f is continuous on [a, b], there exist two numbers c1 and c2 on [a, b] such that f(c1) ≤ f(x) ≤ f(c2) for all x in [a, b]. (Extreme Value Theorem.) If f(c1) = f(c2), then the function f has a constant value on [a, b]

of Rolle’s Theorem.

Theorem 4.1.5 (The Mean Value Theorem) Suppose that a function f is continuous on a closed and bounded interval [a, b] and f is di erentiable on the open interval (a, b). Then there exists some number c such that a < c < b

and

f(b) − f(a) = f0(c). b − a

Proof. We define a function g(x) that is obtained by subtracting the line joining (a, f, (a)) and (b, f(b)) from the function f:

The g is continuous on [a, b] and di erentiable on (a, b). Furthermore, g(a) = g(b) = 0. By Rolle’s Theorem, there exists some number c such that a < c < b and

0= g0(c)

=f0(c) − f(b) − f(a) . b − a

Hence,

f(b) − f(a) = f0(c) b − a

as required.

Theorem 4.1.6 (Cauchy-Mean Value Theorem) Suppose that two functions f and g are continuous on a closed and bounded interval [a, b], di erentiable

Proof. We define a new function h on [a, b] as follows:

h(x) = f(x) − f(a) − f(b) − f(a) (g(x) − g(a)). g(b) − g(a)

Then h is continuous on [a, b] and di erentiable on (a, b). Furthermore,

h(a) = 0 and h(b) = 0.

By Rolle’s Theorem, there exist some c in (a, b) such that h0(c) = 0. Then

0 = h0(c) = f0(c) − f(b) − f(a) g0(c) g(b) − g(a)

as required. This completes the proof of Theorem 4.1.6.

Theorem 4.1.7 (L’Hospital’s Rule, 00 Form) Suppose f and g are di erentiable and g0(x) 6= 0 on an open interval (a, b) containing c (except possibly at c). Suppose that

where L is a real number, ∞, or −∞. Then

lim f(x) = lim f0(x) = L.

x→c g(x) x→c g0(x)

Proof. We define f(c) = 0 and g(c) = 0. Let x (c, b). Then f and g are continuous on [c, x], di erentiable on (c, x) and g0(y) 6= 0 on (c, x). By the Cauchy Mean Value Theorem, there exists some point y (c, x) such that

Similarly, we can prove that

lim f(x) = L.

x→c− g(x)

Therefore,

lim f(x) = lim f0(x) = L.

x→c g(x) x→c g0(x)

Remark 12 Theorem 4.1.7 is valid for one-sided limits as well as the twosided limit. This theorem is also true if c = ∞ or c = −∞.

Theorem 4.1.8 Theorem 4.1.7 is valid for the case when

Proof of Theorem 4.1.8 is omitted.

Example 4.1.1 Find each of the following limits using L’Hospital’s Rule.

Theorem 4.1.9 Suppose that two functions f and g are continuous on a closed and bounded interval [a, b] and are di erentiable on the open interval (a, b). Then the following statements are true:

(i) If f0(x) > 0 for each x in (a, b), then f is increasing on (a, b).

(ii)If f0(x) < 0 for each x in (a, b), then f is decreasing on (a, b).

(iii)If f0(x) ≥ 0 for each x in (a, b), then f is non-decreasing on (a, b).

(iv)If f0(x) ≤ 0 for each x in (a, b), then f is non-increasing on (a, b).

(v)If f0(x) = 0 for each x in (a, b), then f is constant on (a, b).

152CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

(vi)If f0(x) = g0(x) on (a, b), then f(x) = g(x)+C, for constant C, on (a, b).

Proof.

Part (i) Suppose a < x1 < x2 < b. Then f is continuous on [x1, x2] and di erentiable on (x1, x2). By the Mean Value Theorem, there exists some c such that a < x1 < c < x2 < b and

f(x2) − f(x1) = f0(c) > 0. x2 − x1

Since x2 − x1 > 0, it follows that f(x2) − f(x1) > 0 and f(x2) > f(x1). By definition, f is increasing on (a, b). The proof of Parts (ii)–(v) are similar and are left as an exercise.

Part (vi) Let F (x) = f(x) − g(x) for all x in [a, b]. Then F is continuous on [a, b] and di erentiable on (a, b). Furthermore, F 0(x) = 0 on (a, b). Hence, by Part (v), there exists some constant C such that for each x in (a, b),

F (x) = C, f(x) − g(x) = c, f(x) = g(x) + C.

This completes the proof of the theorem.

Theorem 4.1.10 (First Derivative Test for Extremum) Let f be continuous on an open interval (a, b) and a < c < b.

(i)If f0(x) > 0 on (a, c) and f0(x) < 0 on (c, b), then f(c) is a local maximum of f on (a, b).

(ii)If f0(x) < 0 on (a, c) and f0(x) > 0 on (c, b), then f(c) is a local minimum of f on (a, b).

Proof. This theorem follows immediately from Theorem 4.1.9 and its proof is left as an exercise.

Theorem 4.1.11 (Second Derivative Test for Extremum) Suppose that f, f0 and f00 exist on an open interval (a, b) and a < c < b. Then the following statements are true:

(i)If f0(c) = 0 and f00(c) > 0, then f(c) is a local minimum of f.

(ii)If f0(c) = 0 and f00(c) < 0, then f(c) is a local maximum of f.

(iii) If f0(c) = 0 and f00(c) = 0, then f(c) may or may not be a local extremum.

Proof.

Part (i) If f00(c) > 0, then by Theorem 4.1.2, there exists some δ > 0 such that for all x in (c − δ, c + δ),

Hence, f0(x) > 0 on (c, c + δ) and f0(x) < 0 on (c − δ, c). By the first derivative test, f(c) is a local minimum of f.

Part (ii) The proof of Part (ii) is similar to Part (i) and is left as an exercise.

Part (iii) Let f(x) = x3 and g(x) = x4. Then

f0(0) = g0(0) = f00(0) = g00(0).

However, f has no local extremum at 0 but g has a local maximum at 0. This completes the proof of this theorem.

Definition 4.1.5 (Concavity) Suppose that f is defined in some open interval (a, b) containing c and f0(c) exists. Let

y = g(x) = f0(c)(x − c) + f(c)

be the equation of the line tangent to the graph of f at c.

(i)If there exists δ > 0 such that f(x) > g(x) for all x in (c−δ, c+δ), x 6= c, then the graph of f is said to be concave upward at c. If the graph of f is concave upward at every c in (a, b), then it is said to be concave upward on (a, b).

(ii)If there exists δ > 0 such that f(x) < g(x) for all x in (c−δ, c+δ), x 6= c, then the graph of f is said to be concave downward at c. If the graph of f is concave downward at every c in (a, b), then it is said to be concave downward on (a, b).

(iii) The point (c, f(c)) is said to be a point of inflection if there exists some δ > 0 such that either

154CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

(i)the graph of f is concave upward on (c−δ, c) and concave downward on (c, c + δ), or

(ii)the graph of f is concave downward on (c−δ, c) and concave upward on (c, c + δ).

Remark 13 The first derivative test, second derivative test and concavity test are very useful in graphing functions.

Example 4.1.2 Let f(x) = x4 − 4x2, −3 ≤ x ≤ 3

(a)Locate the local extrema, and point extrema and points of inflections.

(b)Locate the intervals where the graph of f is increasing, decreasing, concave up and concave down.

(c)Sketch the graph of f. Determine the absolute maximum and the absolute minimum of the graph of f on [−3, 3].

Part (a)

(i) f(x) = x4 − 4x2 = x2(x2 − 4) = 0 → x = 0, x = −2, x = 2 are zeros of f.

√ √

(ii) f0(x) = 4x3 − 8x = 0 = 4x(x2 − 2) = 0 → x = 0, x = − 2 and x = 2 are the critical points of f.

(v) f00(x) changes sign at x = ± √ and hence ± √ , − are the points 3 3 9

of inflection of the graph of f.

(c)f(−3) = f(3) = 45 is the absolute maximum of f and is obtained at the

end points of the interval.

√√

Also, f(− 2) = f( 2) = −8 is the absolute minimum of f on [−3, 3]. We note that f(0) = 0 is a local maximum of f. The graph is sketched with the above information.

graph

Example 4.1.3 Consider g(x) = x2 −x2/3, −2 ≤ x ≤ 3. Sketch the graph of g, locating extrema, zeros, points of inflection, intervals where f is increasing or decreasing, and intervals where the graph of f is concave up or concave down.

Let us compute the zeros and critical points of g.

(i)g(x) = x2/3(x4/3 − 1) = 0 → x = 0, −1, 1.

g0(x) = 2x − 23 x−1/3 = 2x−1/3 x4/3 − 13 = 0 → x = ± 31 3/4.

(ii)g00(x) = 2 + 29 x−4/3 > 0 for all x, except x = 0, where g00(x) does not exist.

156CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

(iii)The point (0, 0) is not an inflection point, since the graph is concave up everywhere on (−∞, 0) (0, ∞).

Exercises 4.1 Verify that each of the following Exercises 1–2 satisfies the hypotheses and the conclusion of the Mean Value Theorem. Determine the value of the admissible c.

1.f(x) = x2 − 4x, −2 ≤ x ≤ 2

2.g(x) = x3 − x2 on [−2, 2]

3.Does the Mean Value Theorem apply to y = x2/3 on [−8, 8]? If not, why not?

4.Show that f(x) = x2 − x3 cannot have more than two zeros by using Rolle’s Theorem.

5.Show that f(x) = ln x is an increasing function. (Use Mean Value Theorem.)

6.Show that f(x) = e−x is a decreasing function.

7.How many real roots does f(x) = 12x4 − 14x2 + 2 have?

8.Show that if a polynomial has four zeros, then there exists some c such that f000(c) = 0.

A function f is said to satisfy a Lipschitz condition with constant M if

|f(x) − f(y)| ≤ M|x − y|

for all x and y. The number M is called a Lipschitz constant for f.

9.Show that f(x) = sin x satisfies a Lipschitz condition. Find a Lipschitz constant.

10.Show that g(x) = cos x satisfies a Lipschitz condition. Find a Lipschitz constant for g.

In each of the following exercises, sketch the graph of the given function over the given interval. Locate local extrema, absolute extrema, intervals where the function is increasing, decreasing, concave up or concave down. Locate the points of inflection and determine whether the points of inflection are oblique or not.