Answer
A. The domain is $(-\infty,0)\cup(0, \infty)$
B. There is no y-intercept.
There are no x-intercepts.
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \frac{e^x}{x^2} = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} \frac{e^x}{x^2} =\infty$
E. The function is decreasing on the interval $(0, 2)$
The function is increasing on the intervals $(-\infty, 0)\cup (2, \infty)$
F. $(2,1.85)$ is a local minimum.
G. The graph is concave up on the intervals $(-\infty, 0)\cup (0, \infty)$
There are no points of inflection.
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{e^x}{x^2}$
A. The function is defined for all real numbers except $x=0$.
The domain is $(-\infty,0)\cup(0, \infty)$
B. Since $x \neq 0,$ there is no y-intercept.
When $y = 0$:
$\frac{e^x}{x^2} = 0$
$e^x = 0$
There are no values of $x$ such that $e^x=0$
There are no x-intercepts.
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \frac{e^x}{x^2} = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} \frac{e^x}{x^2} =\frac{\infty}{\infty}$
$\lim\limits_{x \to \infty} \frac{e^x}{2x} =\frac{\infty}{\infty}$
$\lim\limits_{x \to \infty} \frac{e^x}{2} =\infty$
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{x^2~e^x-2x~e^x}{x^4} = \frac{(x-2)~e^x}{x^3} =0$
$(x-2)~e^x = 0$
$x = 2$
When $0 \lt x \lt 2$, then $y' \lt 0$
The function is decreasing on the interval $(0, 2)$
When $x \lt 0$ or $x \gt 2$, then $y' \gt 0$
The function is increasing on the intervals $(-\infty, 0)\cup (2, \infty)$
F. When $x = 2$, then $y = \frac{e^2}{2^2} = 1.85$
$(2,1.85)$ is a local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(e^x+xe^x-2e^x)(x^3)-(3x^2)(x-2)(e^x)}{x^6}$
$y'' = \frac{(x+x^2-2x)(e^x)-(3x-6)(e^x)}{x^4}$
$y'' = \frac{(x^2-4x+6)~e^x}{x^4} = 0$
$x^2-4x+6 = 0$
There are no values of $x$ such that $y'' = 0$
When $x \lt 0$ or $x \gt 0$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty, 0)\cup (0, \infty)$
There are no points of inflection.
H. We can see a sketch of the curve below.
