Answer
A. The domain is $(-\frac{\pi}{2}, \frac{\pi}{2})$
B. The y-intercept is $0$
The x-intercept is $0$
C. The function is an even function.
D. $\lim\limits_{x \to (-\frac{\pi}{2})^+} x~tan~x = \infty$
$\lim\limits_{x \to (\frac{\pi}{2})^-} x~tan~x = \infty$
$x = -\frac{\pi}{2}$ and $x =\frac{\pi}{2}$ are vertical asymptotes.
E. The function is decreasing on the interval $(-\frac{\pi}{2}, 0)$
The function is increasing on the interval $(0, \frac{\pi}{2})$
F. The local minimum is $(0,0)$
G. The graph is concave up on the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$
There are no points of inflection.
H. We can see a sketch of the curve below.
Work Step by Step
$y = x~tan~x,~~~~-\frac{\pi}{2}\lt x \lt \frac{\pi}{2}$
A. The domain is given in the question as $(-\frac{\pi}{2}, \frac{\pi}{2})$
B. When $x=0$, then $y = (0)~tan~(0) = 0$
The y-intercept is $0$
When $y = 0$:
$x~tan~x = 0$
$x = 0$
The x-intercept is $0$
C. $(-x)~tan~(-x) = x~tan~x$
The function is an even function.
D. $\lim\limits_{x \to (-\frac{\pi}{2})^+} x~tan~x = \infty$
$\lim\limits_{x \to (\frac{\pi}{2})^-} x~tan~x = \infty$
$x = -\frac{\pi}{2}$ and $x =\frac{\pi}{2}$ are vertical asymptotes.
E. We can find values of $x$ such that $y' = 0$:
$y' = tan~x+x~sec^2~x = 0$
$x = 0$
When $-\frac{\pi}{2} \lt x \lt 0$, then $y' \lt 0$
The function is decreasing on the interval $(-\frac{\pi}{2}, 0)$
When $0 \lt x \lt \frac{\pi}{2}$, then $y' \gt 0$
The function is increasing on the interval $(0, \frac{\pi}{2})$
F. When $x=0$, then $y = (0)~tan~(0) = 0$
The local minimum is $(0,0)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = sec^2~x+sec^2~x+2x~sec^2~x~tan~x$
$y'' = 2~sec^2~x~(1+x~tan~x) = 0$
There are no values of $x$ in the domain such that $y'' = 0$
The graph is concave up on the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$
There are no points of inflection.
H. We can see a sketch of the curve below.
