Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 302: 50

Answer

(a) $y = 1$ is a horizontal asymptote. (b) $f$ is decreasing on the interval $(-\infty,0)$ $f$ is increasing on the interval $(0,\infty)$ (c) The local minimum is $f(0) = -1$ (d) The graph is concave down on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3},\infty)$ The graph is concave up on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$ The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{1}{2})$ and $(\frac{2\sqrt{3}}{3}, -\frac{1}{2})$ (e) We can see a sketch of the graph below.

Work Step by Step

(a) $f(x) = \frac{x^2-4}{x^2+4}$ $f(x)$ is defined for all $x$ so there are no vertical asymptotes. $\lim\limits_{x \to -\infty}f(x) = 1$ $\lim\limits_{x \to \infty}f(x) = 1$ $y = 1$ is a horizontal asymptote. (b) We can find the points where $f'(x) = 0$: $f'(x) = \frac{(2x)(x^2+4)- (x^2-4)(2x)}{(x^2+4)^2}$ $f'(x) = \frac{2x^3+8x-2x^3+8x}{(x^2+4)^2}$ $f'(x) = \frac{16x}{(x^2+4)^2} = 0$ $16x = 0$ $x = 0$ When $x \lt 0~~$ then $f'(x) \lt 0$ $f$ is decreasing on the interval $(-\infty,0)$ When $x \gt 0$ then $f'(x) \gt 0$ $f$ is increasing on the interval $(0,\infty)$ (c) $f(0) = \frac{0^2-4}{0^2+4} = -1$ The local minimum is $f(0) = -1$ There is no local maximum. (d) We can find the points where $f''(x) = 0$: $f''(x) = \frac{(16)(x^2+4)^2-(16x)(2)(x^2+4)(2x)}{(x^2+4)^4}$ $f''(x) = \frac{(16)(x^4+8x^2+16)-(64x^2)(x^2+4)}{(x^2+4)^4}$ $f''(x) = \frac{16x^4+128x^2+256-64x^4-256x^2}{(x^2+4)^4}$ $f''(x) = \frac{-48x^4-128x^2+256}{(x^2+4)^4}$ $f''(x) = \frac{(-48x^2+64)(x^2+4)}{(x^2+4)^4}$ $f''(x) = \frac{-48x^2+64}{(x^2+4)^3} = 0$ $48x^2= 64$ $x^2 = \frac{4}{3}$ $x = \pm \frac{2\sqrt{3}}{3}$ The graph is concave down when $f''(x) \lt 0$ The graph is concave down on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3},\infty)$ The graph is concave up when $f''(x) \gt 0$ The graph is concave up on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$ $f(-\frac{2\sqrt{3}}{3}) = \frac{(-\frac{2\sqrt{3}}{3})^2-4}{(-\frac{2\sqrt{3}}{3})^2+4} = -\frac{1}{2}$ $f(\frac{2\sqrt{3}}{3}) = \frac{(\frac{2\sqrt{3}}{3})^2-4}{(\frac{2\sqrt{3}}{3})^2+4} = -\frac{1}{2}$ The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{1}{2})$ and $(\frac{2\sqrt{3}}{3}, -\frac{1}{2})$ (e) We can see a sketch of the graph below.

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