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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
118 SECTION 4.
CHAPTER 4
SECTION 4.
- f is differentiable on (0, 1), continuous on [0, 1]; and f (0) = f (1) = 0.
f ′(c) = 3c^2 − 1; 3 c^2 − 1 = 0 =⇒ c =
3 ∈/^ (0,^ 1)
- f is differentiable on (− 2 , 2), continuous on [− 2 , 2]; and f (−2) = f (2) = 0. f ′(c) = 4c^3 − 4 c; 4 c(c^2 − 1) = 0 =⇒ c = 0, ± 1
- f is differentiable on (0, 2 π), continuous on [0, 2 π]; and f (0) = f (2π) = 0. f ′(c) = 2 cos 2c; 2 cos 2c = 0 =⇒ 2 c = π 2 + nπ, and c = π 4 + nπ 2 , n = 0, ± 1 , ± 2... Thus, c = π 4 , 34 π , 54 π , 74 π
- f is differentiable on (0, 8), continuous on [0, 8]; and f (0) = f (8) = 0. f ′(c) = 23 c−^1 /^3 − 23 c−^2 /^3 =^23 c
c^2 /^3 f^
′(c) = 0 =⇒ c = 1.
- f ′(c) = 2c, f^ (b) b^ −−^ fa^ ( a)=^42 −−^11 = 3; 2 c = 3 =⇒ c = 3/ 2
- f ′(c) = 2 √^3 c − 4, f^ (b) b^ −−^ fa^ (a)= −^104 −−^ ( 1 − 1)= −3; 2 √^3 c − 4 = − 3 =⇒ c = 9/ 4
- f ′(c) = 3c^2 , f^ (b b)^ −−^ fa^ (a)=^273 −− 11 = 13; 3 c^2 = 13 =⇒ c =^13
39 is not in [a, b]
- f ′(c) = 23 c−^1 /^3 , f^ (b) b^ −−^ fa^ ( a)=^48 −−^11 =^37 ; 23 c−^1 /^3 = 37 =⇒ c = (14)
3 93
- f ′(c) = √−c 1 − c^2 , f^ (b b)^ −−^ fa^ (a)=^01 −−^10 = −1; √−c 1 − c^2 = − 1 =⇒ c =^12
2 is not in [a, b])
- f ′(c) = 3c^2 − 3, f^ (b) b^ −−^ fa^ (a)= 1 − −^2 (−−^2 1) = −2; 3 c^2 − 3 = − 2 =⇒ c = ±
- f is continuous on [− 1 , 1], differentiable on (− 1 , 1) and f (−1) = f (1) = 0. f ′(x) = −x(5^ −^ x
(3 + x^2 )^2
1 − x^2 , f ′(c) = 0 for c in (− 1 , 1) implies c = 0.
- (a) f ′(x) = 23 x−^1 /^3 = (^3) x^21 / 3 = 0 for all x ∈ (− 1 , 1). (b) f ′(0) does not exist. Therefore, f is not differentiable on (− 1 , 1).
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
SECTION 4.1 119
- No. By the mean-value theorem there exists at least one number c ∈ (0, 2) such that f ′(c) = f^ (2) 2 −−^ f 0 (0)=^32 > 1.
- No, by Rolle’s theorem: f (2) = f (3) = 1 but there is no value c ∈ (2, 3) such that f ′(c) = 0.
- By the mean-value theorem there is a number c ∈ (2, 6) such that f (6) − f (2) = f ′(c)(6 − 2) = f ′(c)4. Since 1 ≤ f ′(x) ≤ 3 for all x ∈ (2, 6), it follows that 4 ≤ f (6) − f (2) ≤ 12.
- f (x) = x^2 + x + 3, f ′(x) = 2x + 1. The slope of the line through (− 1 , 3) and (2, 9) is 2. Setting 2 x + 1 = 2, we get x = 12. The point on the graph of f where the tangent line is parallel to the line through (− 1 , 3) and (2, 9) is: (1/ 2 , 15 /4).
- f is everywhere continuous and everywhere differentiable except possibly at x = − 1. f is continuous at x = −1: as you can check,
x→−lim 1 −^ f^ (x) = 0,^ x→−lim 1 +^ f^ (x) = 0,^ and^ f^ (−1) = 0. f is differentiable at x = −1 and f ′(−1) = 2: as you can check,
hlim→ 0 −^ f^ (−1 +^ h h)^ −^ f^ (−1)= 2^ and^ hlim→ 0 +^ f^ (−1 +^ h h)^ −^ f^ (−1)= 2. Thus f satisfies the conditions of the mean-value theorem on every closed interval [ a, b ].
f ′(x) =
2 , x ≤ − 1 3 x^2 − 1 , x > − 1 f (2) − f (−3) 2 − (−3) =
f ′(c) = 2 with c ∈ (− 3 , 2) iff c = 1 or − 3 < c ≤ −1.
- f is continuous and differentiable everywhere; f ′(x) =
3 x^2 , x ≤ 1 3 , x > 1 f (2) − f (−1) 2 − (−1) =
For c ≤ 1 , f ′(c) = 3c^2 = 53 =⇒ c = ±
For c > 1 , f ′(c) = 3 = (^53)
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SECTION 4.1 121
- Let c and d be two consecutive roots of the equation P ′(x) = 0. The equation P (x) = 0 cannot have two or more roots between c and d for then, by Rolle’s theorem, P ′(x) would have to be zero somewhere between these two roots and thus between c and d. In this case c and d would no longer be consecutive roots of P ′(x) = 0.
- If f (x) = 0 at a 1 , a 2 ,... , an then by Rolle’s theorem, f ′(x) is zero at some number b 1 ∈ (a 1 , a 2 ), at some number b 2 ∈ (a 2 , a 3 ),... , at some number bn− 1 ∈ (an− 1 , an); f ′′(x), in turn, must be zero at some number c 1 ∈ (b 1 , b 2 ), at some number b 2 ∈ (b 2 , b 3 ),... , at some number cn− 2 ∈ (bn− 2 , bn− 1 ).
- Suppose that f has two fixed points a, b ∈ I, with a < b. Let g(x) = f (x) − x. Then g(a) = f (a) − a = 0 and g(b) = f (b) − b = 0. Since f is differentiable on I, we can conclude that g is differentiable on (a, b) and continuous on [a, b]. By Rolle’s theorem, there exists a number c ∈ (a, b) such that g′(c) = f ′(c) − 1 = 0 or f ′(c) = 1. This contradicts the assumption that f ′(x) < 1 on I.
- Set P (x) = x^3 + ax + b. It is obvious that for x sufficiently large, P (x) > 0 and for x sufficiently large negative, P (x) < 0. Thus, by the intermediate-value theorem, the equation P (x) = 0 has at least one real root. If a ≥ 0 , then P ′(x) = 3x^2 + a is positive, except possibly at 0, where it remains nonnegative. It follows that P is everywhere increasing and therefore it cannot take on the value 0 more than once. Suppose now that a < 0. Then − (^13)
3 |a| and (^13)
3 |a| are consecutive roots of the equation P ′(x) = 0 and thus, by Exercise 27, P cannot take on the value zero more than once between these two numbers.
- (a) f ′(x) = 3x^2 − 3 < 0 for all x in (− 1 , 1). Also, f is differentiable on (− 1 , 1) and continuous on [− 1 , 1]. Thus there cannot be a and b in (− 1 , 1) such that f (a) = f (b) = 0, or they would contradict Rolle’s theorem. (b) When f (x) = 0, b = 3x − x^3 = x(3 − x^2 ). When x is in (− 1 , 1), then |x(3 − x^2 )| < 2. Thus |b| < 2.
- f ′(x) = 3x^2 − 3 a^2 > 0 for all x in(−a, a). Also, f is differentiable on (−a, a) and continuous on [−a, a]. Thus there cannot be b and c in (−a, a) such that f (b) = f (c) = 0, or they would contradict Rolle’s theorem.
- For p(x) = xn^ + ax + b, p′(x) = nxn−^1 + a, which has at most one real zero for n even
x = − an^ n^1 −^1
If there were more than two distinct real roots of p(x), then by Rolle’s theorem there would be more than one zero of p′(x). Thus there are at most two distinct real roots of p(x).
- For p(x) = xn^ + ax + b, p′(x) = nxn−^1 + a, which has at most two real zeros for n odd. If there were more than three distinct real roots of p(x), then by Rolle’s theorem there would be more than two zeros of p′(x). Thus there are at most three distinct real roots of p(x).
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
122 SECTION 4.
- If x 1 = x 2 , then |f (x 1 ) − f (x 2 )| and |x 1 − x 2 | are both 0 and the inequality holds. If x 1 = x 2 , then by the mean-value theorem f (x 1 ) − f (x 2 ) x 1 − x 2 =^ f^
′(c)
for some number c between x 1 and x 2. Since |f ′(c)| ≤ 1: ∣∣ ∣∣^ f^ (x^1 )^ −^ f^ (x^2 ) x 1 − x 2
∣∣ ≤ 1 and thus |f (x 1 ) − f (x 2 )| ≤ |x 1 − x 2 |.
- See the proof of Theorem 4.2.2.
- Set, for instance, f (x) =
1 , a < x < b 0 , x = a, b
- (a) Let f (x) = cos x. Choose any numbers x and y, (assume x < y). By the mean-value theorem, there is a number c between x and y such that f (y) − f (x) y − x =^ f^
′(c) ⇒ |^ cos^ y^ −^ cos^ x| |y − x| =^ | −^ sin^ c| ≤^1 ⇒^ |^ cos^ x^ −^ cos^ y| ≤ |x^ −^ y| (b) Repeat the in part (a) with f (x) = sin x.
- (a) By the mean-value theorem, there exists a number c ∈ (a, b) such that f (b) − f (a) = f ′(c)(b − a). If f ′(x) ≤ M for all x ∈ (a, b), then it follows that f (b) ≤ f (a) + M (b − a) (b) If f ′(x) ≥ m for all x ∈ (a, b), then it follows that f (b) ≥ f (a) + m(b − a) (c) If |f ′(x)| ≤ K on (a, b), then −K ≤ f ′(x) ≤ K on (a, b) and the result follows from parts (a) and (b).
- Assume that g(x) = 0 for all x ∈ [a, b] and let h(x) = f g^ ((xx)). Then h is defined on [a, b] and h(a) = h(b) = 0. Therefore, by Rolle’s theorem, there exists a number c ∈ (a, b) such that h′(c) = g(c)f^
′(c) − f (c)g′(c) g^2 (c) = 0 Thus g(c)f ′(c) − f (c)g′(c) = 0 which contradicts the given condition f (x)g′(x) − g(x)f ′(x) = 0 for all x ∈ I. Thus, g has at least one zero in (a, b).
By reversing the roles of f and g, the same argument can be used to show that g cannot have two (or more) zeros on (a, b).
- We show first that the conditions on f and g imply that f and g cannot be simultaneously 0. Set h(x) = f 2 (x) + g(x). Then h′(x) = 2f (x)f ′(x) + 2g(x)g′(x) = 2f (x)[g(x)] + 2g(x)[−f (x)] = 0 =⇒ f 2 (x) + g^2 (x) = C constant.
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
124 SECTION 4.
- f (x) = 2x^3 + 3x^2 − 3 x − 2 is differentiable on (− 2 , 1), continuous on [− 2 , 1], and f (−2) = f (1) = 0. f ′(x) = 6x^2 + 6x − 3 f ′(c) = 0 at c 1 ∼= − 1. 366 , c 2 ∼= 0. 366
- f (x) = 1 − x^3 − cos(π x/2) is differentiable on (0, 1), continuous on [0, 1], and f (0) = f (1) = 0. f ′(x) = − 3 x^2 + π 2 sin(π x/2) f ′(c) = 0 at c ∼= 0. 676
- f (1) = 0; f satisfies Rolle’s theorem on [0, 1]; f ′(c) = 0 at c ∼= 0.6058.
- b ∼= 0. 5437 , c ∼= 0.3045, f (0.3045) ∼= − 0. 1749
0.2 0.4 0.6 0.8x
y
- x-intercepts: x = 0, x = 1; f ′(x) = 0 at x ∼= 0. 3874
- 0 c = 0
- x-intercepts: x = − 2 , x = 45 x = 2; f ′(x) = 0 at x ∼= − 1 .1005 and x ∼= 1. 5577
- f (x) = x^4 − 7 x^2 + 2; f ′(x) = 4x^3 − 14 x g(x) = 4x^3 − 14 x − f^ (3) 3 −−^ f 1 (1)= 4x^3 − 14 x − 12 g(c) = 0 at c ∼= 2. 205
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SECTION 4.2 125
- f (x) = x cos x + 4 sin x; f ′(x) = cos x − x sin x + 4 cos x = 5 cos x − x sin x and g(x) = 5 cos x − x sin x − f^ (π/2)^ − π^ f (−π/2)= 5 cos x − x sin x − (^) π^8 g(c) = 0 at c 1 ∼= − 0. 872 , c 2 ∼= 0. 872
- f (x) = x^3 − x^2 + x − 1; f ′(x) = 3x^2 − 2 x + 1; c = 8/3.
1 2 3 4 x
20
40
60
y
- f (x) = x^4 − 2 x^3 − x^2 − x + 1; f ′(x) = 4x^3 − 6 x^2 − 2 x − 1; c = 1/ 2 , − 0. 6180 , 1 .6180.
− 2 − 1 1 2 3 x 5
5
10
y
SECTION 4.
- f ′(x) = 3x^2 − 3 = 3
x^2 − 1
= 3(x + 1)(x − 1) f increases on (−∞, −1] and [1, ∞), decreases on [− 1 , 1]
- f ′(x) = 3x^2 − 6 x = 3x(x − 2) f increases on (−∞, 0] and [2, ∞), decreases on [0, 2]
- f ′(x) = 1 − (^) x^12 = x
x^2 =
(x + 1)(x − 1) x^2 f increases on (−∞, −1] and [1, ∞), decreases on [− 1 , 0) and (0, 1 ] (f is not defined at 0)
- f ′(x) = 3(x − 3)^2 ; f increases on (−∞, ∞)
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SECTION 4.2 127
- f (x) =
x^2 − x − 2 , x ≤ − 1 −x^2 + x + 2, − 1 < x < 2 x^2 − x − 2 , x ≥ 2
f ′(x) =
2 x − 1 , x ≤ − 1 − 2 x + 1, − 1 < x < 2 2 x − 1 , x ≥ 2 f increases on
[
]
and [2, ∞), decreases on (−∞, −1] and
[ 1
2 ,^2
]
- f ′(x) = 1 + sin x ≥ 0; f increases on [0, 2 π]
- f ′(x) = 1 + cos x ≥ 0; f increases on [0, 2 π]
- f ′(x) = −2 sin 2x − 2 sin x = −2 sin x (2 cos x + 1); f increases on
[ 2
3 π, π
]
, decreases on
[
0 , 23 π
]
- f ′(x) = −2 cos x sin x = −2 sin 2x; f increases on [π/ 2 , π], decreases on [0, π/2]
- f ′(x) =
3 + 2 sin 2x; f increases on
[
0 , 23 π
]
and
[ 5
6 π, π
]
, decreases on
[ 2
3 π,^56 π
]
- f ′(x) = 2 sin x cos x −
3 cos x = cos x
2 sin x −
f increases on [ 13 π, 12 π] and [ 23 π, π], decreases on [0, 13 π] and [ 12 π, 23 π]
- (^) dxd
( (^) x 3 3 −^ x
= f ′(x) =⇒ f (x) = x
3 3 −^ x^ +^ C f (1) = 2 =⇒ 2 = 13 − 1 + C, so C = 83. Thus, f (x) = 13 x^3 − x + 83.
- (^) dxd
x^2 − 5 x
= f ′(x) =⇒ f (x) = x^2 − 5 x + C f (2) = 4 =⇒ 4 = 4 − 10 + C, so C = 10. Thus, f (x) = x^2 − 5 x + 10.
- (^) dxd
x^5 + x^4 + x^3 + x^2 + x
= f ′(x) =⇒ f (x) = x^5 + x^4 + x^3 + x^2 + x + C f (0) = 5 =⇒ 5 = 0 + C, so C = 5. Thus, f (x) = x^5 + x^4 + x^3 + x^2 + x + 5.
- (^) dxd
− 2 x−^2
= f ′(x) =⇒ f (x) = − 2 x−^2 + C f (1) = 0 =⇒ 0 = −2 + C, so C = 2. Thus, f (x) = − 2 x−^2 + 2, x > 0.
- (^) dxd
4 x
3 x
3 / 2
= f ′(x) =⇒ f (x) =^34 x^4 /^3 − 23 x^3 /^2 + C
f (0) = 1 =⇒ 1 = 0 + C, so C = 1. Thus, f (x) = 34 x^4 /^3 − 23 x^3 /^2 + 1, x ≥ 0.
- (^) dxd
− 14 x−^4 − 254 x^4 /^5
= f ′(x) =⇒ f (x) = − 14 x−^4 − 254 x^4 /^5 + C f (1) = 0 =⇒ 0 = − 14 − 252 + C, so C = 132. Thus, f (x) = − 14 x−^4 − 254 x^4 /^5 + 132 , x > 0.
- (^) dxd (2x − cos x) = f ′(x) =⇒ f (x) = 2x − cos x + C f (0) = 3 =⇒ 3 = 0 − 1 + C, so C = 4. Thus, f (x) = 2x − cos x + 4.
- (^) dxd (2x^2 + sin x) = f ′(x) =⇒ f (x) = 2x^2 + sin x + C f (0) = 1 =⇒ 1 = 0 + C, so C = 1. Thus, f (x) = 2x^2 + sin x + 1.
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
128 SECTION 4.
- f ′(x) =
1 , x < − 3 − 1 , − 3 < x < − 1 1 , − 1 < x < 1 − 2 , 1 < x f increases on (−∞, −3) and [− 1 , 1]; decreases on [− 3 , −1] and [1, ∞)
- f ′(x) =
2(x − 1), x < 1 − 1 , 1 < x < 3 − 2 , x > 3 f decreases on (−∞, 1), [1, 3) and [3, ∞).
- f ′(x) =
− 2 x, x < 1 − 2 , 1 < x < 3 3 , 3 < x f increases on (−∞, 0] and [3, ∞); decreases on [0, 1) and [1, 3]
- f ′(x) =
1 , x < 0 2(x − 1), 0 < x < 3 − 1 , 3 < x < 7 2 , x > 7 f increases on (−∞, 0], [1, 3], (7, ∞); decreases on [0, 1] and [3, 7)
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
130 SECTION 4.
- Let f (x) = x − sin x. Then f ′(x) = 1 − cos x. (a) f ′(x) ≥ 0 for all x ∈ (−∞, ∞) and f ′(x) = 0 only at x = π 2 + nπ, n = 0, ± 1 , ± 2 ,... It follows from Theorem 4.2.3 that f is increasing on (−∞, ∞). (b) Since f is increasing on (−∞, ∞) and f (0) = 0 − sin 0 = 0, we have: f (x) > 0 for all x > 0 ⇒ x > sin x on (0, ∞); f (x) < 0 for all x < 0 ⇒ x < sin x on (−∞, 0).
- A proof is outlined just below the statement of the theorem.
- f ′(x) = 2 sec x(sec x tan x) = 2 sec^2 x tan x and g′(x) = 2 tan x sec^2 x. Therefore, f ′(x) = g′(x) for all x ∈ I.
- Evaluating sec^2 x − tan^2 x = C at x = 0 gives C = 1.
- Let f and g be functions such that f ′(x) = −g(x) and g′(x) = f (x). Then: (a) Differentiating f 2 (x) + g^2 (x) with respect to x, we have 2 f (x)f ′(x) + 2g(x)g′(x) = − 2 f (x)g(x) + 2g(x)f (x) = 0. Thus, f 2 (x) + g^2 (x) = C (constant). (b) f (0) = 0 and g(0) = 1 implies C = 1. (c) The functions f (x) = sin x, g(x) = cos x have these properties.
- (a) Let h(x) = f (x) − g(x). Then h′(x) = f ′(x) − g′(x) > 0 on (0, c), and h is increasing on (0, c). Since h(0) = f (0) − g(0) = 0, it follows that h(x) > 0 on (0, c). Thus, f (x) > g(x) on (0, c). (b) Again let h(x) = f (x) − g(x). Then h is increasing on (−c, 0) which implies that h(x) < 0 on this interval since h(0) = 0. Therefore, f (x) < g(x) on (−c, 0).
- Let f (x) = tan x and g(x) = x for x ∈ [0, π/2). Then f (0) = g(0) = 0 and f ′(x) = sec^2 x > g′(x) = 1 for x ∈ (0, π/2). Thus, tan x > x for x ∈ (0, π/2) by Exercise 56(a).
- Let f (x) = cos x −
1 − 12 x^2
for x ∈ [0, ∞). Then f (0) = 0 and f ′(x) = − sin x + x = x − sin x > 0 for x ∈ (0, ∞) by Exercise 51 (b). Thus, f (x) > 0 for x ∈ (0, ∞) which implies cos x > 1 − 12 x^2 on (0, ∞).
- Choose an integer n > 1. Let f (x) = (1 + x)n^ and g(x) = 1 + nx, x > 0. Then, f (0) = g(0) = 1 and f ′(x) = n(1 + x)n−^1 > g′(x) = n since (1 + x)n−^1 > 1 for x > 0. The result follows from Exercise 56(a).
- Let f (x) = sin x −
x − 16 x^3
. Then f (0) = 0 and f ′(x) = cos x −
1 − 12 x^2
0 by Exercise 58. Therefore, f (x) > f (0) = 0 for all x ∈ (0, ∞) which implies sin x > x − 16 x^3 on (0, ∞).
- 4 ◦^ ∼= 0.06981 radians. By Exercises 51 and 60,
- 6981 − (0.6981)
3 6 = 0.^06975 <^ sin 4
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
SECTION 4.2 131
- (a) Let f (x) = cos x − (1 − 12 x^2 + 241 x^4 ). Then f (0) = 0 and f ′(x) = − sin x + x − x
3 6 <^0 by Exercise 60. Therefore, f (x) < f (0) = 0 on all x ∈ (0, ∞), which implies cos x < 1 − 12 x^2 + 241 x^4 on (0, ∞). (b) 6◦^ = 30 π. Using this for x in 1 − 12 x^2 < cos x < 1 − 12 x^2 + 241 x^4 , =⇒ 0. 994517 < cos 6◦^ < 0. 994522.
- Let f (x) = 3x^4 − 10 x^3 − 4 x^2 + 10x + 9, x ∈ [− 2 , 5]. Then f ′(x) = 12x^3 − 30 x^2 − 8 x + 10. f ′(x) = 0 at x ∼= − 0. 633 , 0. 5 , 2. 633 f is decreasing on [− 2 , − 0 .633] and [0. 5 , 2 .633] f is increasing on [− 0. 633 , 0 .5] and [2. 633 , 5]
- Let f (x) = 2x^3 − x^2 − 13 x − 6 , x ∈ [− 3 , 4]. Then f ′(x) = 6x^2 − 2 x − 13. f ′(x) = 0 at x ∼= − 1. 315 , 1. 648 f is decreasing on [− 1. 315 , 1 .648] f is increasing on [− 3 , − 1 .315] and [1. 648 , 4]
- Let f (x) = x cos x − 3 sin 2x, x ∈ [0, 6]. Then f ′(x) = cos x − x sin x − 6 cos 2x. f ′(x) = 0 at x ∼= 0. 770 , 2. 155 , 3. 798 , 5. 812 f is decreasing on [0, 0 .770], [2. 155 , 3 .798] and [5. 812 , 6] f is increasing on [0. 770 , 2 .155] and [3. 798 , 5 .812]
- Let f (x) = x^4 + 3x^3 − 2 x^2 + 4x + 4, x ∈ [− 5 , 3]. Then f ′(x) = 4x^3 + 9x^2 − 4 x + 4. f ′(x) = 0 at x ∼= − 2. 747 f is decreasing on [− 5 , − 2 .747] f is increasing on [− 2. 747 , 3]
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
SECTION 4.3 133
- f ′(x) = − (^) x2(2 (^2) (xx + 1)+ 1) 2 ; critical pt − (^12) f
= −8 local max
- f (x) =
x^2 − 16 , x < − 4 16 − x^2 , − 4 ≤ x < 4 x^2 − 16 , x ≥ 4
f ′(x) =
2 x, x < − 4 − 2 x, − 4 < x < 4 2 x, x > 4 critical pts − 4 , 0 , 4; f (−4) = f (4) = 0 local minima, f (0) = 16 local max
- f ′(x) = x^2 (5x − 3)(x − 1); critical pts 0, 35 , 1
f
=^2
55 local max f (1) = 0 local min no local extreme at 0
- f ′(x) = 3
( (^) x − 2 x + 2
(x + 2)^2 ≥^ 0;^ critical pt^2 ,^ no local extreme values
- f ′(x) = (5 − 8 x)(x − 1)^2 ; critical pts 58 , 1
f
8
= 204827 local max no local extreme at 1
- f ′(x) = −(1 + x)^3 + (1 − x)(3)(1 + x)^2 = 2(1 + x)^2 (1 − 2 x); critical pts − 1 , (^12) f
2
= 2716 local max no local extreme at − 1
- f ′(x) = x(1 +(2 + x^ x) 2 ) ; critical pts − 2 , 0
f (−2) = −4 local max f (0) = 0 local min
- f ′(x) = (1 − x)^1 /^3 − 13 x(1 − x)−^2 /^3 = (^) 3(1^3 −− x^4 x) 2 / 3 ; critical pts 34 , 1 f (3/4) = (^443) / 3 local max no local extreme at 1
- f ′(x) = 13 x(7x + 12)(x + 2)−^2 /^3 ; critical pts − 2 , − 127 , 0
f
7
local max f (0) = 0 local min
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
134 SECTION 4.
- f ′(x) = (^) (x −+ 1)^12 + (^) (x −^1 2) 2 = (^) (x + 1)3(2x (^2) (−x^ 1)− 2) 2 ; critical pt (^12) f
2
= 43 local min
- f (x) =
2 − 3 x, x ≤ − (^12) x + 4, − 12 < x < 3 3 x − 2 , 3 ≤ x
f ′(x) =
− 3 , x < − (^12) 1 , − 12 < x < 3 3 , 3 < x critical pts − 12 , 3
f
= 72 local min no local extreme at 3
- f ′(x) = 73 x^4 /^3 − 73 x−^2 /^3 =^73 x
x^2 /^3 ;^ critical pts^ −^1 ,^0 ,^1 f (−1) = 6 local max, f (1) = −6 local min no local extreme at 0
- f ′(x) = 23 x−^4 /^3 (x − 1); critical pt 1 f (1) = 3 local min no local extreme at 0
- f ′(x) = (x^ + 1)3x
(^2) − x 3 (x + 1)^2 =^
x^2 (2x + 3) (x + 1)^2 ;^ critical pts^ −^
3 2 ,^0 f
= 274 local min no local extreme at 0
- f ′(x) = cos x − sin x; critical pts 14 π, 54 π f ′′(x) = − sin x − cos x, f ′′^
4 π
2 , f ′′^
4 π
f ( 14 π) =
2 local max, f ( 54 π) = −
2 local min
- f ′(x) = 1 − 2 sin 2x; critical pts 12 π , 512 π f ( 121 π) = 12 π+
2 local max f ( 125 π) =^512 π −
2 local min
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
136 SECTION 4.
- (^) P (x) = x^4 − 8 x^3 + 22x^2 − 24 x + 4 P ′(x) = 4x^3 − 24 x^2 + 44x − 24 P ′′(x) = 12x^2 − 48 x + 44 Since P ′(1) = 0, P ′(x) is divisible by x − 1. Division by x − 1 gives P ′(x) = (x − 1)
4 x^2 − 20 x + 24
= 4(x − 1)(x − 2)(x − 3). The critical pts are 1, 2 , 3. Since P ′′(1) > 0 , P ′′(2) < 0 , P ′′(3) > 0 , P (1) = − 5 is a local min, P (2) = − 4 is a local max, and P (3) = − 5 is a local min. Since P ′(x) < 0 for x < 0 , P decreases on (−∞, 0]. Since P (0) > 0 , P does not take on the value 0 on (−∞, 0]. Since P (0) > 0 and P (1) < 0, P takes on the value 0 at least once on (0, 1). Since P ′(x) < 0 on (0, 1), P decreases on [0, 1]. It follows that P takes on the value zero only once on [0, 1]. Since P ′(x) > 0 on (1, 2) and P ′(x) < 0 on (2, 3), P increases on [1, 2] and decreases on [2, 3]. Since P (1), P (2), P (3) are all negative, P cannot take on the value 0 between 1 and 3. Since P (3) < 0 and P (100) > 0 , P takes on the value 0 at least once on (3, 100). Since P ′(x) > 0 on (3, 100), P increases on [3, 100]. It follows that P takes on the value zero only once on [3, 100]. Since P ′(x) > 0 on (100, ∞), P increases on [100, ∞). Since P (100) > 0 , P does not take on the value 0 on [100, ∞).
f has a local maximum at x = 0; f has a local minimum at x = −1 and x = 2.
Let f (x) = Ax^2 + Bx + C. Then f ′(x) = 2Ax + B. f (−1) = 3 =⇒ A − B + C = 3; f (3) = − 1 =⇒ 9 A + 3B + C = − 1 Since f has a minimum at x = 2, f ′(2) = 4A + B = 0 Solving for A, B, C, we get A = 12 , B = − 2 , C = 12.
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:
SECTION 4.3 137
- Let f (x) = (^) x 2 ax (^) + b 2. Then f ′(x) = a
b^2 − x^2
(b^2 + x^2 )^2
. Now
f ′(0) = (^) ba 2 = 1 ⇒ a = b^2 and f ′(x) = b
2 (b (^2) − x 2 ) (b^2 + x^2 )^2
f ′(−2) = b^2
b^2 − 4
(b^2 + 4)^2 = 0 ⇒ b = ± 2 Thus, a = 4 and b = ± 2.
- (a) f (x) = xp(1 − x)q^ , p, q ≥ 2; f ′(x) = xp−^1 (1 − x)q−^1 [p − (p + q)x] f ′(x) = 0 =⇒ x = 0, x = 1, x = (^) p +p q (b) p even, p − 1 odd:
f has a local min at x = 0 (c) q even, q − 1 odd:
f has a local min at x = 1
(d) f ′′
( (^) p p + q
= − (p + q)
( (^) p p + q
)p− 1 ( (^) q p + q
)q− 1 < 0 ⇒ f has a local max at x = (^) p +p q.
- If p is a polynomial of degree n, then p′^ has degree n − 1. This implies that p′^ has at most n − 1 zeros, and it follows that p has at most n − 1 local extreme values.
- The function D(x) =
x^2 + [f (x)]^2 gives the distance from the origin to the point (x, f (x)) on the graph of f. Since the graph of f does not pass through the origin, D′(x) = x^ +^ f^ (x)f^
′(x) √ x^2 + [f (x)]^2 is defined for all x ∈ dom (f ). Suppose that D has a local extreme value at c. Then D′(c) = c^ +^ f^ (c)f^
′(c) √ c^2 + [f (c)]^2
= 0 ⇒ c + f (c)f ′(c) = 0 and f ′(c) = − (^) f (cc)
Suppose that c = 0. The slope of the line through (0, 0) and (c, f (c)) is given by m 1 = f^ ( cc )and the slope of the tangent line to the graph of f at x = c is given by m 2 = f ′(c) = − (^) f (cc). Since m 1 m 2 = − 1 , these two lines are perpendicular. If c = 0, then the tangent line to the graph of f is horizontal and the line through (0, 0) and (0, f (0)) is vertical.
- If f (x) = x^4 − 7 x^2 − 8 x − 3, then f ′(x) = 4x^3 − 14 x − 8 and f ′′(x) = 12x^2 − 14. Since f ′(2) = − 4 < 0 and f ′(3) = 58 > 0 , f ′^ has at least one zero in (2, 3). Since f ′′(x) > 0 for x ∈ (2, 3), f ′^ is increasing on this interval and so it has exactly one zero. Thus, f has exactly one critical point c in (2, 3).
