# How do you find critical points of a logarithmic function y = e^x - 2e^(-x) - 3x?

Oct 29, 2015

Correction: That is not a logarithmic function, but here is how to find the critical numbers for the function.

#### Explanation:

$f \left(x\right) = {e}^{x} - 2 {e}^{- x} - 3 x$

Note that "Dom:(f) = (-oo,oo)

$f ' \left(x\right) = {e}^{x} - 2 {e}^{- x} \left(- 1\right) - 3$

$= {e}^{x} + 2 {e}^{- x} - 3$

$f ' \left(x\right)$ is never undefined, so we need only find its zeros.

We need to solve

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

As a first step, try getting rid of negative exponents:

${e}^{x} + \frac{2}{e} ^ x - 3 = 0$

Get a common denomiator, noting that ${e}^{x} \cdot {e}^{x} = {e}^{2 x}$, so we have

$\frac{{e}^{2 x} + 2 - 3 {e}^{x}}{e} ^ x = 0$.

So now we need to solve:

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

Again note that ${e}^{2 x} = {\left({e}^{x}\right)}^{2}$, so we can factor:

$\left({e}^{x} - 1\right) \left({e}^{x} - 2\right) = 0$

${e}^{x} = 1$ $\text{ }$ OR $\text{ }$ ${e}^{x} = 2$

$x = 0$ $\text{ }$ OR $\text{ }$ $x = \ln 2$

Both are in $\text{Dom} \left(f\right)$, so both are critical numbers.