Written Homework 7 Solutions

Section 4.3 20. Find the local maxima and minima using the First and Second Derivative tests:

x2 f (x) = x − 1

Solution: First start by finding the first derivative.

2x(x − 1) − x2 x2 − 2x f 0(x) = = (x − 1)2 (x − 1)2 It is easy to see that the numerator is zero when x ∈ 0, 2. Now to apply the First Derivative Test. For x = 0 values just to the right (0+) are negative while values just to the left (0−) are postitive. Zero is therefore a maximum. For x = 2 values just to the left (2−) are negative while values just the right (2+) are positive. Therefore 2 is a minimum. We can check this again using the second derivative which is:

(2x − 2)(x − 1)2 − (x2 − 2x)(2(x − 1)) (2x − 2)(x − 1) − 2x2 + 4x f 00(x) = = (x − 1)4 (x − 1)3

(2x − 2)(x − 1) − 2x2 + 4x 2x2 − 4x + 2 − 2x2 + 4x 2 = = (x − 1)3 (x − 1)3 (x − 1)3 Now to test the critical points: 2 f 00(0) = = −2 (−1)3 2 f 00(2) = = 2 (1)3 The second derivative test tells us that the critical point at 0 is a maximum and the critical point at 2 is a minimum.

64. Coffee is being poured at a constant rate into the mug shown in the text. Sketch a graph of the depth of the coffee as a function of time accounting for the shape of the graph in terms of concavity. What is the significance of the inflection point.

Solution: The graph should look roughly as follows. Homework 7 Solutions

Plot.jpg Note that up to the inflection point one has a concave up graph as the depth increases rapidly as the mug thins, but then turns to concave down as the mug widens after filling past the middle point of the mug.

2 2 66. Given the function f (x) = e−x /(2σ )

(a) Find the asymptote, maximum value, and inflection points of f . Solution: Note first that no vertical asymptote exists. Now to check for horizontal.

2 2 2 2 lim e−x /(2σ ) = e−(−∞) /(2σ ) = e−∞ = 0 x→−∞

2 2 2 2 lim e−x /(2σ ) = e−(∞) /(2σ ) = e−∞ = 0 x→∞ So there is a horizontal asymptote at 0. Now to obtain the first and second derivatives.

−x · e−x2/(2σ2) f 0(x) = σ2

−e−x2/(2σ2) x2 · e−x2/(2σ2) f 00(x) = + σ2 σ4 The only zero for the first derivative exists at x = 0. The second derivative at zero is negative so this critical point x = 0 is a maximum with value f (0) = 1. Setting the second derivative equal to zero gives. e−x2/(2σ2) x2 · e−x2/(2σ2) f 00(x) = 0 ⇔ = ⇔ σ2 = x2 ⇔ x = ±σ σ2 σ4 Therefore the inflection points are at x = ±σ. Page 2 of 11 Homework 7 Solutions

(b) What role does σ play in the shape of the curve? Solution: Given that it marks the inflection points and does not affect the asymptote nor the maximum, it widens the curve. (c) Plot four members of this family of curves on the same plot.

Function.jpg

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Section 4.4 24. Find the limit of eu/10 lim u→∞ u3 u/10 3 ∞ Solution: Note that as x → ∞ both e → ∞ and u → ∞. So this limit has indeterminate form “ ∞ ”. Use L’Hospital’s rule. u/10 u/10 e H 1 e lim = lim . u→∞ u3 u→∞ 10 3u2 ∞ The resulting limit still has indeterminate form “ ∞ ”. Use l’Hospital’s Rule again. u/10 u/10 1 e H 1 e lim = lim u→∞ 10 3u2 u→∞ 100 6u ∞ The resulting limit still has indeterminate form “ ∞ ”. Use l’Hospital’s Rule yet again. u/10 u/10 1 e H 1 e lim = lim = ∞. u→∞ 100 6u u→∞ 1000 6 So eu/10 lim = ∞. u→∞ u3

28. Find the limit of: x − sin(x) lim x→0 x − tan(x)

0 Solution: Plugging in x = 0 to the numerator and denominator gives the indeterminate form “ 0 ”. Use l’Hospital’s Rule. x − sin(x) H 1 − cos(x) lim = lim . x→0 x − tan(x) x→0 1 − sec2(x) 0 Plugging in x = 0 still gives the indeterminate form “ 0 ”. Use l’Hospital’s Rule again.

1 − cos(x) H sin(x) lim = lim x→0 1 − sec2(x) x→0 −2 sec2(x) tan(x) −1 sin(x) = lim cos3 x x→0 2 sin(x) 1 = − lim cos3 x 2 x→0 −1 = . 2 So the limit is -1/2.

34. Find the limit of: x lim x→0 tan−1(4x)

0 Solution: Plugging in x = 0 gives the indeterminate form “ 0 ”. Use l’Hospital’s Rule.

x H 1 lim = lim x→0 tan−1(4x) x→0 4/(1 + 16x2) 1 + 16x2 1 = lim = x→0 4 4 The limit is therefore 1/4. Page 4 of 11 Homework 7 Solutions

42. Find the limit of: √ lim xe−x/2 x→∞ √ Solution: This is an indeterminate product, i.e. as x → ∞, x → ∞ and e−x/2 → 0. In order to use L’Hospital’s rule, we must rewrite as an indeterminate fraction. √ √ x lim xe−x/2 = lim x→∞ x→∞ ex/2 ∞ Now, the limit has indeterminate form “ ∞ ” and we can use L’Hospital’s rule.

√ √1 x H 2 x lim = lim = 0 x→∞ ex/2 x→∞ 1 x/2 2 e where the last limit is zero because the numerator → 0 while the denominator → ∞.

56. Find the limit of: lim (tan(2x))x x→0+

Solution: Let L be the limit assuming it exists. Then:

ln(L) = lim ln(tan 2x)x = lim x ln tan 2x x→0+ x→0+ This limit is an indeterminate product form. Rewrite as a fraction and use l’Hospital’s Rule.

ln(L) = lim x ln tan 2x x→0+ ln tan 2x = lim x→0+ x−1 2 sec2 2x H = lim tan 2x x→0+ −x−2 −2x2 sec 2x = lim x→0+ sin 2x 2x = − lim lim sec 2x x→0+ sin 2x x→0 = 1 · 0

The limit is therefore: L = eln(L) = e0 = 1.

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Problem 4.5 √ 26. Use the guidelines of the section to sketch y = x 2 − x2

Solution:

(a) Domain √ √ y is defined for x2 ≤ 2 or for − 2 ≤ x ≤ 2. (b) Intercepts √ y-intercept at y = 0 2 − 0 = 0. √ √ √ x-intercept for x = 0 or 2 − x2. So then at x ∈ − 2, 0, 2. (c) Symmetry p y(−x) = (−x) 2 − (−x)2 = −y(x) The function is therefore odd. (d) Asymptotes Note that no asymptotes exist as y is not defined for x = ±∞ nor does the function go to infinity anywhere it is defined. Also no slant asymptotes exist as the degree between the numerator and denom- inator is more than 1. (e) Increasing or Decreasing The first derivative is: p x p x2 2 − 2x2 y0 = 2 − x2 + √ · −2x = 2 − x2 − √ = √ 2 2 − x2 2 − x2 2 − x2 This is a positive value for −1 < x < 1 and zero for x = ±1 and negative outside the range. (f) Local Maxima and Minima The second derivative is: √ 2 −2x −4x 2 − x2 − (2 − x ) √ 3 3 3 2 −8x + 4x + 2x − x 3x − 6x y00 = 2 2−x = = 2 − x2 (2 − x2)3/2 (2 − x2)3/2 Checking the critical points gives that y00(1) = −3 so it is a maximum y00(−1) = 3 and is a minimum. These have values of (x, y) = (1, 1) and (x, y) = (−1, −1). (g) Concavity and Points of Inflection Finding where the points of inflection are: √ √ 3x3 − 6x = 0 ⇔ x = 0, − 2, 2 √ √ Then for − 2 < x < 0, y is concave up (y(−1) > 0) and for 0 < x < 2, y is concave down (y(1) < 0). (h) Graph Here is the graph of the function.

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30. Use the guidelines of the section to sketch y = x5/3 − 5x2/3

Solution:

(a) Domain y is defined for all x. (b) Intercepts y-intercept at y = 05/3 − 5(0)2/3 = 0. x-intercept for: x5/3 = 5x2/3 ⇔ x = 5

(c) Symmetry y(−x) = (−x)5/3 − 5(−x)2/3 = −(x)5/3 − 5(x)2/3 The function is therefore neither even nor odd. (d) Asymptotes No vertical asymptotes exist as the function does not go to infinity. Also no slant asymptotes exist as the degree difference between numerator and denominator is larger than 1. Horizontal Asymptote: lim x5/3 − 5x2/3 = lim x5/3(1 − 5x−1) = ∞ x→∞ x→∞ lim x5/3 − 5x2/3 = lim x5/3(1 − 5x−1) = −∞ x→−∞ x→−∞ No horizontal asymptotes. (e) Increasing or Decreasing The first derivative is: 5x2/3 − 10x−1/3 5x−1/3(x − 2) y0 = = 3 3 So y0 = 0 when x = 2 and y0 does not exist at x = 0. There is a “cusp” at x = 0 because the term x−1/3 has right and left limits: lim x−1/3 = +∞ x→0+ lim x−1/3 = −∞. x→0− In particular, x−1/3 > 0 for x > 0 and x−1/3 < 0 for x < 0. So y is increasing for x > 2, decreasing for 0 < x < 2, and increasing for x < 0. (f) Local Maxima and Minima The second derivative is: 10x−1/3 + 10x−4/3 10x−4/3(x + 1) y00 = = 9 9 00 15 −1/3 00 Checking the critical points gives that y (2) = 9 ·2 so there is a minimum at x = 2. Note that y (0) DNE. But, since the function switches from increasing to decreasing at x = 0, it is a local maximum. (g) Concavity and Points of Inflection Looking at y00(x), we see that the concavity changes at x = −1. The concavity does not change at x = 0, because x−4/3 > 0 for all x , 0. Hence y is concave up for x > −1 (except at x = 0, where concavity isn’t defined) and concave down for x < −1.

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(h) Graph Here is the graph of the function.

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sin(x) 40. Use the guidelines of the section to sketch y = 2+cos(x)

Solution:

(a) Domain y is defined for all x. (b) Intercepts sin(0) y-intercept at y = 2+cos(0) = 0. x-intercept for whenever sin(x) = 0 or for x = n ∗ π where n is an integer. (c) Symmetry sin(−x) − sin(x) y(−x) = = 2 + cos(−x) 2 + cos(x) The function is therefore odd. The function also repeats with period 2π. (d) Asymptotes There are no vertical asymptotes as the denominator is never zero. No slant asymptote exists as the function oscillates forever. For the same reason there are no horizontal asymptotes. (e) Increasing or Decreasing The first derivative is: cos(x)(2 + cos(x)) − sin(x)(− sin(x)) (2 cos(x) + cos2(x)) + sin2(x) (2 cos(x) + 1) y0 = = = (2 + cos(x))2 (2 + cos(x))2 (2 + cos(x))2 This has a root at cos(x) = −1/2 or whenever x = 2π/3 + 2nπ or for −2π/3 + 2nπ where n is an integer. Note the denominator is always positive, so the the sign of y0 is the same as the sign of 2 cos x+1. Hence 2π 4π the intervals of increase are ( 3 + 2πn, 3 + 2πn), where n is any integer. The intervals of decrease are −2π 2π ( 3 + 2πn, 3 + 2πn), where again n is any integer. (f) Local Maxima and Minima The second derivative is: (−2 sin(x))(2 + cos(x))2 − 2(2 cos(x) + 1)(2 + cos(x))(− sin(x)) y00 = (2 + cos(x))4

(−2 sin(x))[(4 + 4 cos(x) + cos2(x)) − (2 cos(x) + 1)] y00 = (2 + cos(x))4 (−2 sin(x))(3 + 2 cos(x) + cos2(x)) y00 = (2 + cos(x))4 Checking the critical points gives that y00(2π/3 + 2nπ) = −0.77 so these are maxima. And y00(−2π/3 + 2nπ) = 0.77 so these are minima. (g) Concavity and Points of Inflection Finding where the points of inflection are when sin(x) = 0 (when x = nπ) or when: √ −2 ± 4 − 12 (3 + 2 cos(x) + cos2(x)) = 0 ⇔ (3 + 2u + u2) = 0 ⇔ u = 2 Hence 3 + 2 cos x + cos2 x is never zero, its always positive. Therefore the concavity changes exactly when sin x = 0. Note that y00(π/2) < 0.

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(h) Graph Here is the graph of the function.

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