Answer
(a) $f$ is decreasing on the intervals $(-\infty, -\frac{ln(2)}{3})$
$f$ is increasing on the intervals $(-\frac{ln(2)}{3}, \infty)$
(b) There is no local maximum.
The local minimum is $f(-\frac{ln(2)}{3}) = 2^{-2/3}+2^{1/3}$
(c) The graph is concave up on the interval $(-\infty, \infty)$
There are no points of inflection.
Work Step by Step
(a) $f(x) = e^{2x}+e^{-x}$
We can find the points where $f'(x) = 0$:
$f'(x) = 2e^{2x}-e^{-x} = 0$
$2e^{2x} = e^{-x}$
$e^{3x} = \frac{1}{2}$
$3x = ln(\frac{1}{2})$
$3x = -ln(2)$
$x = -\frac{ln(2)}{3}$
When $x \lt -\frac{ln(2)}{3}$ then $f'(x) \lt 0$
$f$ is decreasing on the intervals $(-\infty, -\frac{ln(2)}{3})$
When $x \gt -\frac{ln(2)}{3}$ then $f'(x) \gt 0$
$f$ is increasing on the intervals $(-\frac{ln(2)}{3}, \infty)$
(b) $f(-\frac{ln(2)}{3}) = e^{(2)(-\frac{ln(2)}{3})}+e^{\frac{ln(2)}{3}}$
$f(-\frac{ln(2)}{3}) = [e^{ln(2)}]^{-2/3}+[e^{ln(2)}]^{1/3}$
$f(-\frac{ln(2)}{3}) = 2^{-2/3}+2^{1/3}$
There is no local maximum.
The local minimum is $f(-\frac{ln(2)}{3}) = 2^{-2/3}+2^{1/3}$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = 4e^{2x}+e^{-x} = 0$
Since $4e^{2x}+e^{-x} \gt 0$ for all values of $x$, there are no points where $f''(x) = 0$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on the interval $(-\infty, \infty)$
There are no points of inflection.
