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 2 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−(1) =(2σ ) = e−∞ = 0 x!1 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 Page 3 of 11 Homework 7 Solutions Section 4.4 24. Find the limit of eu=10 lim u!1 u3 u=10 3 1 Solution: Note that as x ! 1 both e ! 1 and u ! 1. So this limit has indeterminate form “ 1 ”. Use L’Hospital’s rule. u=10 u=10 e H 1 e lim = lim : u!1 u3 u!1 10 3u2 1 The resulting limit still has indeterminate form “ 1 ”. Use l’Hospital’s Rule again. u=10 u=10 1 e H 1 e lim = lim u!1 10 3u2 u!1 100 6u 1 The resulting limit still has indeterminate form “ 1 ”. Use l’Hospital’s Rule yet again. u=10 u=10 1 e H 1 e lim = lim = 1: u!1 100 6u u!1 1000 6 So eu=10 lim = 1: u!1 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: p lim xe−x=2 x!1 p Solution: This is an indeterminate product, i.e. as x ! 1, x ! 1 and e−x=2 ! 0. In order to use L’Hospital’s rule, we must rewrite as an indeterminate fraction. p p x lim xe−x=2 = lim x!1 x!1 ex=2 1 Now, the limit has indeterminate form “ 1 ” and we can use L’Hospital’s rule. p p1 x H 2 x lim = lim = 0 x!1 ex=2 x!1 1 x=2 2 e where the last limit is zero because the numerator ! 0 while the denominator ! 1. 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: Page 5 of 11 Homework 7 Solutions Problem 4.5 p 26. Use the guidelines of the section to sketch y = x 2 − x2 Solution: (a) Domain p p y is defined for x2 ≤ 2 or for − 2 ≤ x ≤ 2. (b) Intercepts p y-intercept at y = 0 2 − 0 = 0. p p p x-intercept for x = 0 or 2 − x2. So then at x 2 − 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 + p · −2x = 2 − x2 − p = p 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: p 2 −2x −4x 2 − x2 − (2 − x ) p 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: p p 3x3 − 6x = 0 , x = 0; − 2; 2 p p 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. Page 6 of 11 Homework 7 Solutions Page 7 of 11 Homework 7 Solutions 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) = 1 x!1 x!1 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 = +1 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.