Aplicaciones de la derivada calculo stewart

4

APPLICATIONS OF DIFFERENTIATION

4.1 Maximum and Minimum Values
1. A function f has an absolute minimum at x = c if f(c) is the smallest function value on the entire domain of f, whereas

f has a local minimum at c if f(c) is the smallest function value when x is near c.
3. Absolute maximum at s, absolute minimum at r, local maximum at c, local minima at b and r, neither a maximum nora

minimum at a and d.
5. Absolute maximum value is f(4) = 5; there is no absolute minimum value; local maximum values are f (4) = 5 and

f (6) = 4; local minimum values are f (2) = 2 and f(1) = f (5) = 3.
7. Absolute minimum at 2, absolute maximum at 3, 9. Absolute maximum at 5, absolute minimum at 2,

local minimum at 4

local maximum at 3, local minima at 2 and 4

11. (a)

(b)(c)

13. (a) Note: By the Extreme Value Theorem,

(b)

f must not be continuous; because if it were, it would attain an absolute minimum.

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APPLICATIONS OF DIFFERENTIATION

15. f (x) = 8 − 3x, x ≥ 1. Absolute maximum f (1) = 5; no

17. f (x) = x2 , 0 < x < 2. No absolute or local maximum or

local maximum. No absolute or local minimum.

minimumvalue.

19. f (x) = x2 , 0 ≤ x < 2. Absolute minimum f (0) = 0; no

local minimum. No absolute or local maximum.

21. f (x) = x2 , −3 ≤ x ≤ 2. Absolute maximum

f (−3) = 9. No local maximum. Absolute and local minimum f (0) = 0.

23. f (x) = ln x, 0 < x ≤ 2. Absolute maximum

25. f (x) = 1 −

√ x. Absolute maximum f (0) = 1; no local

f (2) = ln 2 ≈ 0.69; no local maximum. No absoluteor local minimum.

maximum. No absolute or local minimum.

27. f (x) =

1−x 2x − 4

if 0 ≤ x < 2 if 2 ≤ x ≤ 3

Absolute maximum f(3) = 2; no local maximum. No absolute or local minimum.

SECTION 4.1

MAXIMUM AND MINIMUM VALUES

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149

29. f (x) = 5x2 + 4x

⇒ f 0 (x) = 10x + 4. f 0 (x) = 0 ⇒ x = − 2 , so − 2 is the only critical number. 5 5 ⇒ f 0 (x) = 3x2 + 6x − 24 = 3(x2 +2x − 8).

31. f (x) = x3 + 3x2 − 24x

f 0 (x) = 0 ⇒ 3(x + 4)(x − 2) = 0 ⇒ x = −4, 2. These are the only critical numbers.
33. s(t) = 3t4 + 4t3 − 6t2

⇒ s0 (t) = 12t3 + 12t2 − 12t. s0 (t) = 0 ⇒ 12t(t2 + t − 1) ⇒

t = 0 or t2 + t − 1 = 0. Using the quadratic formula to solve the latter equation gives us √ √ −1 ± 12 − 4(1)(−1) −1 ± 5 −1 ± 5 = ≈ 0.618, −1.618. The three critical numbers are0, . t= 2(1) 2 2
35. g(y) =

y−1 y2 − y + 1

⇒

g 0 (y) =

(y 2 − y + 1)(1) − (y − 1)(2y − 1) y 2 − y + 1 − (2y 2 − 3y + 1) −y 2 + 2y y(2 − y) = = 2 = 2 . 2 − y + 1)2 2 − y + 1)2 (y (y (y − y + 1)2 (y − y + 1)2

g 0 (y) = 0 ⇒ y = 0, 2. The expression y 2 − y + 1 is never equal to 0, so g 0 (y) exists for all real numbers. The critical numbers are 0 and 2.
37. h(t) = t
3/4

− 2t1/4

⇒ h (t) = ⇒

0

3 −1/4 t 4 2 3

−

2 −3/4 t 4

=

1 −3/4 t (3t1/2 4

− 2) =

3

√ t−2 √ . 4 4 t3

h0 (t) = 0 ⇒ 3

√ t=2 ⇒

√ t=

⇒ t = 4 . h0 (t) does not exist at t = 0, so the critical numbers are 0 and 4 . 9 9

39. F (x) = x4/5 (x − 4)2

F 0 (x) = x4/5 · 2(x − 4) + (x − 4)2 · 4 x−1/5 = 1 x−1/5 (x − 4)[5 · x · 2 + (x − 4) · 4] 5 5 = (x − 4)(14x − 16) 2(x −4)(7x − 8) = 5x1/5 5x1/5

F 0 (x) = 0 ⇒ x = 4, 8 . F 0 (0) does not exist. Thus, the three critical numbers are 0, 8 , and 4. 7 7
41. f (θ) = 2 cos θ + sin2 θ

⇒ f 0 (θ) = −2 sin θ + 2 sin θ cos θ. f 0 (θ) = 0 ⇒ 2 sin θ (cos θ − 1) = 0 ⇒ sin θ = 0

or cos θ = 1 ⇒ θ = nπ [n an integer] or θ = 2nπ. The solutions θ = nπ include the solutions θ = 2nπ, so the critical numbers are θ = nπ.
43. f (x)= x2 e−3x

⇒ f 0 (x) = x2 (−3e−3x ) + e−3x (2x) = xe−3x (−3x + 2). f 0 (x) = 0 ⇒ x = 0,

2 3

[e−3x is never equal to 0]. f 0 (x) always exists, so the critical numbers are 0 and 2 . 3
45. The graph of f 0 (x) = 5e−0.1|x| sin x − 1 has 10 zeros and exists

everywhere, so f has 10 critical numbers.

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47. f (x) = 3x2 − 12x + 5,...