## Calculus: Early Transcendentals 8th Edition

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.
$y = e^{2x}- e^x$ 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.