Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercept is $0$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (e^{2x}- e^x)=\lim\limits_{x \to -\infty} (e^x-1)~e^x = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} (e^{2x}- e^x) = \lim\limits_{x \to \infty} (e^x-1)~e^x =\infty$
E. The function is decreasing on the interval $(-\infty, -0.693)$
The function is increasing on the interval $(-0.693, \infty)$
F. $(-0.693,-0.25)$ is a local minimum.
G. The graph is concave down on the interval $(-\infty, -1.39)$
The graph is concave up on the interval $(-1.39, \infty)$
The point of inflection is $(-1.39, -0.19)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = e^{2x}- e^x$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = e^{2(0)}- e^0= 1-1 = 0$
The y-intercept is $0$
When $y = 0$:
$e^{2x}- e^x = 0$
$e^{2x}= e^x$
$e^x = 1$
$x = 0$
The x-intercept is $0$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (e^{2x}- e^x)=\lim\limits_{x \to -\infty} (e^x-1)~e^x = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} (e^{2x}- e^x) = \lim\limits_{x \to \infty} (e^x-1)~e^x =\infty$
E. We can find the values of $x$ such that $y' = 0$:
$y' = 2e^{2x}- e^x = 0$
$2e^{2x} = e^x$
$2e^x = 1$
$e^x = \frac{1}{2}$
$x = ln(\frac{1}{2})$
$x = -0.693$
The function is decreasing on the interval $(-\infty, -0.693)$
The function is increasing on the interval $(-0.693, \infty)$
F. When $x = -0.693$, then $y = e^{2(-0.693)}- e^{-0.693} = -0.25$
$(-0.693,-0.25)$ is a local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = 4e^{2x}-e^x = 0$
$4e^{2x} = e^x$
$4e^x = 1$
$e^x = \frac{1}{4}$
$x = ln(\frac{1}{4})$
$x = -1.39$
When $x \lt -1.39~~$, then $y'' \lt 0$
The graph is concave down on the interval $(-\infty, -1.39)$
When $x \gt -1.39$, then $y'' \gt 0$
The graph is concave up on the interval $(-1.39, \infty)$
When $x = -1.39$, then $y = e^{2(-1.39)}- e^{-1.39} = -0.19$
The point of inflection is $(-1.39, -0.19)$
H. We can see a sketch of the curve below.