# What are the local extrema of f(x)= lnx/e^x?

Apr 14, 2017

$x = 1.763$

#### Explanation:

Take the derivative of $\ln \frac{x}{e} ^ x$ using quotient rule:

$f ' \left(x\right) = \frac{\left(\frac{1}{x}\right) {e}^{x} - \ln \left(x\right) \left({e}^{x}\right)}{e} ^ \left(2 x\right)$

Take out a ${e}^{x}$ from the top and move it down to the denominator:

$f ' \left(x\right) = \frac{\left(\frac{1}{x}\right) - \ln \left(x\right)}{e} ^ x$

Find when $f ' \left(x\right) = 0$ This only happens when the numerator is $0$:

$0 = \left(\frac{1}{x} - \ln \left(x\right)\right)$

You're going to need a graphing calculator for this one.

$x = 1.763$

Plugging in a number under $1.763$ would give you a positive outcome while plugging a number above $1.763$ would give you a negative outcome. So this is a local maximum.