Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 23

Answer

A. The domain is $(-\infty, -2) \cup (1,\infty)$ B. There is no y-intercept. The x-intercepts are $-2$ and $1$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \sqrt{x^2+x-2} = \infty$ $\lim\limits_{x \to \infty} \sqrt{x^2+x-2} = \infty$ There is no horizontal asymptote. E. The function is decreasing on the interval $(-\infty, -2)$ The function is increasing on the interval $(1, \infty)$ F. There is no local minimum or local maximum. G. The graph is concave down on the intervals $(-\infty,-2)\cup (1, \infty)$ There are no points of inflection. H. We can see a sketch of the curve below.
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Work Step by Step

$y = \sqrt{x^2+x-2}$ A. The function is defined for all real numbers except when $x^2+x-2 \lt 0$: $(x+2)(x-1) \lt 0$ $-2 \lt x \lt 1$ The domain is $(-\infty, -2) \cup (1,\infty)$ B. Since $x \neq 0$, there is no y-intercept. When $y = 0$: $\sqrt{x^2+x-2} = 0$ $\sqrt{(x+2)(x-1)} = 0$ $x = -2$ or $x=1$ The x-intercepts are $-2$ and $1$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \sqrt{x^2+x-2} = \infty$ $\lim\limits_{x \to \infty} \sqrt{x^2+x-2} = \infty$ There is no horizontal asymptote. E. We can try to find the values of $x$ such that $y' = 0$: $y' = \frac{2x+1}{2~\sqrt{x^2+x-2}} = 0$ Note that $x = -\frac{1}{2}$ is not included in the domain. There are no values of $x$ in the domain such that $y' = 0$ When $x \lt -2$ then $y' \lt 0$ The function is decreasing on the interval $(-\infty, -2)$ When $x \gt 1$, then $y' \gt 0$ The function is increasing on the interval $(1, \infty)$ F. There is no local minimum or local maximum. G. We can find the values of $x$ such that $y'' = 0$: $y'' = \frac{(2)(2~\sqrt{x^2+x-2})-(2x+1)(\frac{2x+1}{\sqrt{x^2+x-2}})}{4~(x^2+x-2)}$ $y'' = \frac{(4)(x^2+x-2)-(2x+1)(2x+1)}{4~(x^2+x-2)^{3/2}}$ $y'' = \frac{4x^2+4x-8-4x^2-4x-1}{4~(x^2+x-2)^{3/2}}$ $y'' = \frac{-9}{4~(x^2+x-2)^{3/2}}$ There are no values of $x$ such that $y'' = 0$ When $x \lt -2$ or $x \gt 1$, then $y'' \lt 0$ The graph is concave down on the intervals $(-\infty,-2)\cup (1, \infty)$ There are no points of inflection. H. We can see a sketch of the curve below.
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