Answer
a. $\frac{1}{e}$
b. (2, $\frac{2}{e^{2}}$)
Work Step by Step
a. Using the product rule, we get:
$f'(x)$ = $e^{-x}$ . $\frac{d}{dx}[x]$ + $x$ . $\frac{d}{dx}[e^{-x}]$
$f'(x)$ = $e^{-x}$ . $1$ + $x$ . $-e^{-x}$
$f'(x)$ = $e^{-x}-xe^{-x}$
First derivative test:
$f'(x) = 0$
$e^{-x}-xe^{-x} = 0$
$e^{-x} = xe^{-x}$
$1 = x$
The value of f'(x) for all x<1 is positive, and the value of f'(x) for all x>1 is negative. Hence we can conclude that we have an absolute maximum value at x = 1.
When x = 1, value of y:
$f(1) = (1)e^{-1}$
$f(1) = \frac{1}{e}$
b. Finding the second derivative, we have:
$f''(x) = \frac{d}{dx}[e^{-x}] - (e^{-x} . \frac{d}{dx}[x]$ + $x$ . $\frac{d}{dx}[e^{-x}])$
$f''(x) = -e^{-x} - (e^{-x} - xe^{-x})$
$f''(x) = -2e^{-x} + xe^{-x}$
Second Derivative test to find inflection point:
$f''(x) = 0$
$-2e^{-x} + xe^{-x} = 0$
$xe^{-x} = 2e^{-x}$
$x=2$
The value of $f''(x)$ for all $x<2$ is negative ($f'(x)$ is decreasing at the interval $(-\infty, 2)$, and the value of $f''(x)$ for all $x>2$ is positive ($f'(x)$ is increasing at the interval $(2, \infty)$. Hence we can conclude that $f(x)$ is concave down in the interval $(-\infty, 2)$ and concave up in the interval $(2, \infty)$.
This means that we have an inflection point at x =2.
When x = 2, value of y is:
$f(2) = (2)e^{-2}$
$f(2) = \frac{2}{e^{2}}$
Point of inflection = $(2, \frac{2}{e^{2}})$
