Find the derivative of $f$.
Find all the points where f' does not exist or is equal to $0$. These are the only points where f' can "change signs".
Using the points from 2, determining on what intervals f'\gt 0 (that's where $f$ is increasing), and on what intervals f'\lt 0 (that's where $f$ is decreasing).
Local maxima of $f$ occur at point where $f$ is defined, and f' switches from positive (tangent lines that look like /
) to negative (tangent lines that look like \
).
Local minima of $f$ occur at points where $f$ is defined, and f' switches from negative (tangents \
) to positive (tangents /
).
Find the second derivative of $f$.
Repeat points 2 and 3 for f''. Intervals where f''\gt 0 are where $f$ is concave up; intervals where f''\lt 0 are where $f$ is concave down. (If you can't remember which is which, draw a concave up "cup", $\cup$, and notice that as you move left to right along the cup, the slope of the tangents increases; that means f' is increasing, that means f''\gt 0; draw the same with a concave down "dome", $\cap$, to notice the slopes of the tangents decrease as we move left to right along the dome).
Points of inflection are points where f'' changes from positive to negative, or from negative to positive.
For this particular function, you need to use the Chain Rule twice (well, you need to remember how to do derivatives, most of all).
\begin{align*} f'(x) &= \Bigl( e^{2x} + e^{-x}\Bigr)\\ &= \Bigl(e^{2x}\Bigr)' + \Bigl(e^{-x}\Bigr)'\\ &=e^{2x}(2x)' + e^{-x}(-x)'\\ &=e^{2x}(2) + e^{-x}(-1)\\ &=2e^{2x} - e^{-x}\\ &= 2e^{2x} - \frac{1}{e^x}\\ &= \frac{2e^{2x}e^x}{e^x} - \frac{1}{e^x}\\ &= \frac{2e^{3x} - 1}{e^x}\\ &= \frac{1}{e^x}(2e^{3x}-1). \end{align*}
This is always defined; and in order to be equal to $0$, you need $2e^{3x}-1=0$.