# What are the critical points of f (x) = e^x + e^-(6x)?

Dec 23, 2015

#### Answer:

$c = \frac{\ln 6}{7}$

#### Explanation:

A critical point $c$ is when $f ' \left(c\right) = 0$ or does not exist.

To find $f ' \left(x\right)$, know that according to the chain rule $\frac{d}{\mathrm{dx}} \left[{e}^{u}\right] = {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

$f ' \left(x\right) = {e}^{x} - 6 {e}^{- 6 x}$

Now, we have to know for what values of $x$ this equals $0$ or is undefined.

An easy way to do this is to turn it into a fraction—if a fraction, $f ' \left(x\right) = 0$ when the numerator $= 0$ and will be undefined when the denominator $= 0$.

$f ' \left(x\right) = {e}^{- 6 x} \left({e}^{7 x} - 6\right)$

$f ' \left(x\right) = \frac{{e}^{7 x} - 6}{e} ^ \left(6 x\right)$

$f ' \left(x\right) = 0$ when ${e}^{7 x} - 6 = 0$. Solve this.

${e}^{7 x} = 6$
$7 x = \ln 6$
$x = \frac{\ln 6}{7}$

This is a critical value.

$f ' \left(x\right)$ isn't defined when ${e}^{6 x} = 0$. However, ${e}^{6 x}$ never can equal $0$ so the function is never undefined.

Thus, the only critical point is $c = \frac{\ln 6}{7} \approx 0.256$.

graph{e^x+3^(-6x) [-10.04, 9.96, -5.04, 4.96]}