Finding the local minimum of $e^{3x} + e^{-x}$ $${\begin{array}{l} f(x) \; = \; e^{3x} + e^{-x} \\ f'(x) \; = \; 3e^{3x} - e^{-x}\end{array}}$$
The interval on which $f$ is increasing is $\left( \frac{1}{4}\ln \left(\frac{1}{3}\right), \; \infty \right).$
The interval on which $f$ is decreasing is $\left(-\infty, \; \frac{1}{4}\ln \left(\frac{1}{3}\right) \right).$
When trying to find the local minimum, I used $\left( \frac{1}{4}\ln \left(\frac{1}{3}\right), \; \infty \right).$
The interval on which $f$ is decreasing is $\frac{1}{4}\ln \left(\frac{1}{3}\right)$ for $x$ in $f(x)$ and set it equal to zero. I ended up getting $.71$ for the answer, but it's not correct according to WebAssign. I checked my math to see if I did something wrong, but I keep getting the same answer.
 A: Another way, $f(x) = e^{3x}+\frac13 e^{-x}+\frac13 e^{-x}+\frac13 e^{-x} \ge \dfrac4{\sqrt[4]{27}}$ by AM-GM, with equality only when $3e^{4x}=1$.
A: $f(x)=\exp(3x)+\exp(-x)$, $f'(x)=3\exp(3x)-\exp(-x)$. Putting $f'(x)=0$ you get: $x=-\frac{1}{4}\ln(3)$. If you put this value in $f(x)$ you get: $f(-\frac{1}{4}\ln(3))=\frac{4}{3}3^{1/4}$
A: The optimums of a function is where the derivate (i.e. 'rate of change') is equal to zero. This is the points where the function is 'flat'. Thus you want to solve $f'(x^*)=0$, that is, you want to identify the points where the flatness conditions holds, in your case
$$ f'(x^*)=3e^{3x^*} - e^{-x}=0$$
Taking the log on both sides leads to
$$ \ln 3 + 3x^* = -x^* \implies x^* =-\frac{1}{4} \ln3 $$ 
Since there is only one solution this is the sole optimum of your function. (By constrast, you could for example end up with a quadratic equation for $x^*$, whichs means that there is 2 solutions)
The value of this optimum is $f(x^*)$. (You were off by a sign).
Extra: You can check if this is a minimum or maximum by computing the second derivative: $f''(x) = 9e^{3x} +e^{-x}\implies f''(x^*)=9\times3^{-3/4}+3^{1/4}$
which is positive: this is a minumum. 
