Finding monotonicity and extreme values of $f$ with $e$

Question

I've done a lot of exercises related to finding monotonicity and extreme values using derivatives, but I am stuck at this one (stuck at three different things):

$f(x) = \sqrt{e^{2x}+e^{-x}}$

First, finding the domain of $f$.

$e^{2x}+e^{-x} \geq 0 \quad \quad \quad$ (stuck counter: 1) Let's ignore domain for now.

Finding the derivative and calculating solutions:

$f'(x)= \frac{2e^{2x}-e^{-x}}{2\sqrt{e^{2x}+e^{-x}}}$

$\frac{2e^{2x}-e^{-x}}{2\sqrt{e^{2x}+e^{-x}}} = 0 \quad/2\sqrt{e^{2x}+e^{-x}}$

$2e^{2x}-e^{-x} = 0$

$e^{-x}(2e^{3x}-1)=0$

First solution:

$e^{-x} = 0 \quad \quad \quad$ (stuck counter: 2)

Second solution:

$2e^{3x}-1=0$

$2e^{3x} = 1$

$e^{3x} = \frac{1}{2} \quad \quad \quad$ (stuck counter: 3)

I do know how to proceed further with the extremes/monotonicity, just not sure how to find the domain and how to approach these solutions. :/ I ignored finding the domain of $f'$ for now.

Thanks for help.

Answer 1

1. $\mathrm e^x > 0$ for all $x \in \mathbb R$, so you don't need to worry about the domain.

2. This equation has no solutions.

3. Take the logarithm.

Alternative way, consider the function $g(x ) = \mathrm e^{2x} + \mathrm e^{-x}$ instead, then since $\sqrt y$ is monotonically increasing on $[0,+\infty )$, so we could find the extreme values of $f$ via $g$. The latter one is easier for us to do the calculation.

Answer 2

By enforcing the substitution $e^x=z$ the problem boils down to studying the function $f(z) = z^2+\frac{1}{z}$ over $\mathbb{R}^+$. By the AM-GM inequality we have $$ f(z) = z^2+\frac{1}{2z}+\frac{1}{2z}\geq 3\sqrt[3]{z^2\cdot\frac{1}{2z}\cdot\frac{1}{2z}}=\frac{3}{2^{2/3}} $$ with equality attained only at $z=\frac{1}{2^{1/3}}$. Performing the inverse substitution we have that the unique stationary point of $\sqrt{e^{2x}+e^{-x}}$ is at $x=-\frac{1}{3}\log 2$ and the absolute minimum equals $\frac{3^{1/2}}{2^{1/3}}$. There are no local maxima since $f(z)$ is convex.