Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $1$
The x-intercept is $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (1-x)~e^x = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} (1-x)~e^x = -\infty$
E. The function is increasing on the interval $(-\infty, 0)$
The function is decreasing on the interval $(0, \infty)$
F. $(0,1)$ is a local maximum.
G. The graph is concave down on the interval $(-1, \infty)$
The graph is concave up on the interval $(-\infty, -1)$
The point of inflection is $(-1, 0.736)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = (1-x)~e^x$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = (1-0)~e^0 = 1$
The y-intercept is $1$
When $y = 0$:
$(1-x)~e^x = 0$
$1-x = 0$
$x = 1$
The x-intercept is $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (1-x)~e^x = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} (1-x)~e^x = -\infty$
E. We can find the values of $x$ such that $y' = 0$:
$y' = (-1)e^x+(1-x)~e^x = -x~e^x = 0$
$x=0$
The function is increasing on the interval $(-\infty, 0)$
The function is decreasing on the interval $(0, \infty)$
F. When $x = 0$, then $y = (1-0)~e^0= 1$
$(0,1)$ is a local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = -e^x-x~e^x = (-1-x)~e^x = 0$
$x = -1$
When $x \gt -1~~$, then $y'' \lt 0$
The graph is concave down on the interval $(-1, \infty)$
When $x \lt -1$, then $y'' \gt 0$
The graph is concave up on the interval $(-\infty, -1)$
When $x = -1$, then $y = [1-(-1)]~e^{-1}= \frac{2}{e} = 0.736$
The point of inflection is $(-1, 0.736)$
H. We can see a sketch of the curve below.