Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 36

A. The domain is $(-\frac{\pi}{2}, \frac{\pi}{2})$ B. The y-intercept is $0$ The x-intercepts are $-1.17, 0, 1.17$ C. The function is an odd function. D. $\lim\limits_{x \to (-\frac{\pi}{2})^+} (2x-tan~x) = \infty$ $\lim\limits_{x \to (\frac{\pi}{2})^-} (2x-tan~x) = -\infty$ $x = -\frac{\pi}{2}$ and $x =\frac{\pi}{2}$ are vertical asymptotes. E. The function is decreasing on the intervals $(-\frac{\pi}{2}, -\frac{\pi}{4})\cup (\frac{\pi}{4}, \frac{\pi}{2})$ The function is increasing on the interval $(-\frac{\pi}{4}, \frac{\pi}{4})$ F. The local minimum is $(-\frac{\pi}{4},-0.57)$ The local maximum is $(\frac{\pi}{4},0.57)$ G. The graph is concave up on the interval $(-\frac{\pi}{2}, 0)$ The graph is concave down on the interval $(0, \frac{\pi}{2})$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.

$y = 2x - tan~x,~~~~-\frac{\pi}{2}\lt x \lt \frac{\pi}{2}$ A. The domain is given in the question as $(-\frac{\pi}{2}, \frac{\pi}{2})$ B. When $x=0$, then $y = 2(0) - tan~(0) = 0$ The y-intercept is $0$ When $y = 0$: $2x-tan~x = 0$ $2x = tan~x$ $x = -1.17, 0, 1.17$ The x-intercepts are $-1.17, 0, 1.17$ C. $2(-x)-tan~(-x) = -(2x-tan~x)$ The function is an odd function. D. $\lim\limits_{x \to (-\frac{\pi}{2})^+} (2x-tan~x) = \infty$ $\lim\limits_{x \to (\frac{\pi}{2})^-} (2x-tan~x) = -\infty$ $x = -\frac{\pi}{2}$ and $x =\frac{\pi}{2}$ are vertical asymptotes. E. We can find values of $x$ such that $y' = 0$: $y' = 2-sec^2~x = 0$ $sec^2~x = 2$ $sec~x = \sqrt{2}$ $cos~x = \frac{1}{\sqrt{2}}$ $x = -\frac{\pi}{4}, \frac{\pi}{4}$ When $-\frac{\pi}{2} \lt x \lt -\frac{\pi}{4}$ or $\frac{\pi}{4} \lt x \lt \frac{\pi}{2}$, then $y' \lt 0$ The function is decreasing on the intervals $(-\frac{\pi}{2}, -\frac{\pi}{4})\cup (\frac{\pi}{4}, \frac{\pi}{2})$ When $-\frac{\pi}{4} \lt x \lt \frac{\pi}{4}$, then $y' \gt 0$ The function is increasing on the interval $(-\frac{\pi}{4}, \frac{\pi}{4})$ F. When $x=-\frac{\pi}{4}$, then $y = 2(-\frac{\pi}{4}) - tan~(-\frac{\pi}{4}) = -0.57$ The local minimum is $(-\frac{\pi}{4},-0.57)$ When $x=\frac{\pi}{4}$, then $y = 2(\frac{\pi}{4}) - tan~(\frac{\pi}{4}) = 0.57$ The local maximum is $(\frac{\pi}{4},0.57)$ G. We can find the values of $x$ such that $y'' = 0$: $y'' = -2~sec^2~x~tan~x = 0$ $-\frac{2~sin~x}{cos^3~x} = 0$ $sin~x = 0$ $x = 0$ When $-\frac{\pi}{2}\lt x \lt 0$, then $y'' \gt 0$ The graph is concave up on the interval $(-\frac{\pi}{2}, 0)$ When $0 \lt x \lt \frac{\pi}{2}$, then $y'' \lt 0$ The graph is concave down on the interval $(0, \frac{\pi}{2})$ When $x=0$, then $y = 2(0) - tan~(0) = 0$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.