Calculus A
Section 4.3 How Derivatives Affect the Shape of Graph Homework Solution
2.(a) f is increasing on (0, 1) and (3, 7).
(b) f is decresing on (1, 3).
(c) f is concave upward on (2, 4) and (5, 7).
(d) f is concave downward on (0, 2) and (4, 5).
(e) The points of reflection are (2,2) , (4,3) and (5,4).
7.(a)There is an IP at x = 3 beacuse the graph of f changes from CD to CU there. There is an IP at x = 5 because the graph of f changes from CU to CD there.
(b) There is an IP at x = 2 and x = 6 beacuse f0(x) has a maximum value there, and so f ”(x) changes from positive to negtive there. There is an IP at x = 4 because f0(x) has a minimum value there, and so f ”(x) changes from nega- tive to postive there.
(c) There is an IP at x = 1 beacuse f ”(x) changes from negative to positive there, and so the graph of f changes from concave downward to concave upward.
There is an IP at x = 7 beacuse f ”(x) changes from positive to negtive there, and so the graph of f changes from concave upward to concave downward.
8.(a) f is increasing on the intervals where f0(x) > 0, namely, (2, 4) and 6, 9).
(b) f has a local maximum where it changes from increasing to decreasing, that is, where f0 changes from positive to negtive ( at x = 4 ). Similarly, where f0 changes from negtive to positive, f has a local minimum ( at x = 2 and at x = 6 ).
(c)When f0 is increasing, its derivative f ” is positive and hence, f is concave upward. This happens on (1, 3), (5, 7), and (8, 9). Similarly, f is concave down- ward when f0 is decreasing– that is, (0, 1), (3, 5) , and (7, 8).
25. f0(0) = f0(2) = f0(4) = 0 ⇒ horizontal tangents at x = ±1.
f0(x) > 0 if x < 0 or 2 < x < 4 ⇒ f is increasing on (−∞, 0) and (2, 4) . f0(x) < 0 if 0 < x < 2 or x > 4 ⇒ f is decreasing on (0, 2) and (4, ∞) . f ”(x) > 0 if 1 < x < 3 ⇒ f is concave upward on (1, 3).
f ”(x) < 0 if x < 1 or x > 3 ⇒ f is concave downward on (−∞, 1) and (3, ∞) . The are reflection points when x = 1 and 3.
28. f0(x) > 0 if | x |< 2 ⇒ f is increasing on (−2, 2).
f0(x) < 0 if | x |> 2 ⇒ f is decreasing on (−∞, 2) and (2, ∞) .
f0(2) = 0, so f has a horizontal tangent (and local maximum) at x = 2.
limx→∞f (x) = 1 ⇒ y = 1 is a horizontal asynotote.
−f (x) = −f (x) ⇒ f is odd function (its graph is symmetric about the origin).
Finally, f ”(x) < 0 if 0 < x < 3 and f ”(x) > 0 if x > 3, so f is CD on (0,3) and CU on (3, ∞).
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32.(a) f is increasing where f0 is positive, on (0, 6) and (8, ∞), and decreasing where f0 is negtive on (0, 1) and (6, 8) .
(b) f has a local maximum where f0 chnages from positive or negative, at x = 6, and local minima where f0changes from negative to positive, at x = 1 and at x = 8.
(c) f is concave upward where f0 is increasing, that is , on (0, 2), (3, 5) , and (7, ∞), and concave downward where f0 is decreasing, that is, on (2,3) and (5,7).
(d) There are points of inflection where f changes its direction of concavity, at x = 2, x = 3, x = 5 and x = 7.
75.(a) Since f and g are positive, increasing, and CU on I with f ” and g” never equals to 0, we have f > 0, f0 ≥ 0, f ” > 0, g > 0, g0 ≥ 0, g” > 0 on I. Then (f g)0 = f0g + f g0 ⇒ (f g)” = f ”g + 2f0g0+ f g” ≥ f ”g + f g” > 0 on I ⇒ f + g is CU on I.
(b) In part (a)m if f and g are both decreasing instead of increasing, then f0 ≤ 0 and g0 ≤ 0 on I, so we still have 2f0g0 ≥ 0 on I. Thus, (f g)” = f ”g + 2f0g0+ f g” ≥ f ”g + f g” > 0 on I ⇒ f g is CU on I as in part (a).
78.(a) Let f (x) = ex − 1 − x. Now f (0) = e0 − 1 = 0, and for x ≥ 0, we have f0(x) = ex − 1 ≥ 0. Now, since f (0) = 0 and f is increasing on [0, ∞),f (x) ≥ 0f orx ≥ 0 ⇒ ex− 1 − x ≥ 0 ⇒ ex ≥ 1 + x.
(b) Let f (x) = ex − 1 − x − 12x2. Thus, f0(x) = ex − 1 − x, which is postive for x ≥ 0 by part (a). Thus, f (x) is increasing on (0, ∞), so on interval I, 0 = f (0) ≤ f (x) = ex− 1 − x −12x2 ⇒ ex ≥ 1 + x +12x2.
(c) By part (a), the result holds for n = 1. Suppose that ex ≥ 1 + x +x2!2 + ... +xk!k for x ≥ 0.
Let f (x) = ex− 1 − x −x2!2 − ... −xk!k −(k+1!)xk+1 . Then f0(x) = ex− 1 − x − ... −xk!k ≥ 0 by assumption. Hence, f (x) is increasing on (0, ∞). So 0 ≤ x implies that 0 = f (0) ≤ f (x) = ex−1−x−...−xk!k−(k+1)!xk+1 , and hence ex ≥ 1+x+...+xk!k+(k+1!)xk+1 for x ≥ 0. Therefore, for x ≥ 0, ex≥ 1 + x +x2!2 + ... +xn!n for every postive integer n , by mathmatical induction.
79. Let the cubic function be f (x) = ax3+bx2+cx+d ⇒ f0(x) = 3ax2+2bx+c ⇒ f ”(x) = 6ax + 2b.
So f is CU when 6ax + 2b > 0 ⇔ x > −b3a, CD when x < −b3a, and so the only pointg of reflection occurs when x = −b3a. If the graph has three x-intercepts x1, x2 and x3, then the expression for f (x) must factor as f (x) = a(x − x1)(x − x2)(x − x3).
Multiplying these factors together gives us f (x) = a[x3− (x1+ x2+ x3)x2+ (x1x2+ x1x3+ x2x3)x − x1x2x3].
Equating the coeficients of the x2−terms for the two fprms of f gives us b =
−a(x1+ x2+ x3).
Hence the x−coordinate of the point of inflection is −b3a = −a(x1+x3a2+x3) = x1+x32+x3.
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86. f (x) = cx + x21+3 ⇒ f0(x) = c − (x+3)2x 2. f0(x) > 0 ⇔ c > (x+3)2x 2 [call this g(x)].
Now f0 is positive (and hence f increasing) if c > g, so we’ll find the maximum value of g.
We find that g0(x) = 0 ⇔ x = ±1.
g0(x) > 0 on (0, 1) and g0(x) < 0 on (1, ∞), so g is increasing on (0, 1) and de- creasing on (1, ∞), and hence g has a maximum value on (0, ∞) of g(1) = 18. Also since g(x) ≤ 0 if x ≤ 0, the maximum value of g on (−∞, ∞) is 18. Thus, when c > 18, f is increasing. When c = 18, f0(x) > 0 on (−∞, 1) and (1, ∞), and hence f is increasing on these intervals. Since f is continuous, we may conclude that f is also increasing on (−∞, ∞) if c = 18. Therefore, f is increasing on (−∞, ∞) if c ≥ 18.
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