Section 2.8 The Derivative as a Function
EX.3
(a)- II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then negative again. The actual function values in graph II follow the same pattern.
(b)- IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c)- I, since the slopes of the tangents to graph (c) are negative for x > 0 and positive for x < 0, as are the function values of graph I.
(d)- III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then positive, then 0, then negative again, and the function values in graph III follow the same pattern.
EX.28
f (x) = x2x−32−1 = 14(2x + 3) + 542x−31 limh→0 (2(x+h)+3)−(2x+3)
h = 2
limh→0 h1(2(x+h)−31 − 2x−31 ) = (2x−3)−2 2
⇒ f0(x) = 12 −2(2x−3)5 2
Domains of f and f0 are both (−∞,32) ∪ (32, ∞).
EX.30
limh→0 (x+h)
32−x32
h = limh→0 h((x+h)(x+h)3/23+x−x(33/2)) = limh→0 (x+h)3x2+3xh+h3/2+x3/22 = 32√ x Domains of f and f0 are both [0, ∞).
EX.44
f - d ; f0 - c ; f00 - b ; f000 - a Reasons:
(1) Function d has horizontal tangents wherever function c is zero.
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(2) Function c decreases for x ≤ 0 and increases for x ≥ 0, and function b is negative for x ≤ 0 and positive for x ≥ 0.
(3) Function b is always increasing, and function a is always positive.
(Note that the derivative of an odd function will be an even function. This may also help.)
EX.52
(a) By definition g0(0) = limh→0 g(0+h)−g(0)
h = limh→0 1
h13 , which does not exist.
(b) By definition g0(a) = limx→ag(x)−g(a)x−a = limx→ax2/3x−a−a2/3 = limx→a(x1/3(x−a1/31/3−a)(x1/32/3)(x+x1/31/3+aa1/31/3+a) 2/3) =
2 3a1/3
(c) Since limx→0|g0(x)| = ∞ but g(x) is continuous at 0, g(x) has a vertical tangent line at (0, 0).
(d)
EX.55
(a)
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(b) It’s clear that f (x) is differentiable on both (−∞, 0) and (0, ∞), so we check the differentiability at x = 0:
limh→0 h|h|−0h = limh→0|h| = 0 Thus f is differentiable on <.
(c) Through differentiation, we know that
f0(x) =
2x, x > 0
−2x, x < 0 and from (b) we know that f0(0) = 0. Thus
f0(x) =
2x, x ≥ 0
−2x, x < 0 Or, f0(x) = 2 |x|.
EX.56
(a) f−0 (4) = limh→0− 5−(4+h)−1h = −1 f+0 (4) = limh→0+ 1/[5−(4+h)−1]
h = 1
(b)
(c) Since limx→4−f (x) = 4 = limx→4+f (x), while limx→0−f (x) = 0 6= 5 = limx→0+f (x) and f (5) is not defined, we conclude that f (x) is discontinuous at x = 0 and x = 5.
(d) Since differentiability implies continuity, a function is not differentiable wherever it is not contin- uous. In addition, from (a) we know that f is not differentiable at x = 4 since f−0 (4) 6= f+0 (4).
Therefore, we conclude that f is not differentiable at x = 0, 4 and 5.
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