Question

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

Answer 1

`c=(ln6)/7`

Explanation:

A critical point `c` is when `f'(c)=0` or does not exist.

To find `f'(x)`, know that according to the chain rule `d/dx[e^u]=e^u*(du)/dx`.

`f'(x)=e^x-6e^(-6x)`

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'(x)=0` when the numerator `=0` and will be undefined when the denominator `=0`.

`f'(x)=e^(-6x)(e^(7x)-6)`

`f'(x)=(e^(7x)-6)/e^(6x)`

`f'(x)=0` when `e^(7x)-6=0`. Solve this.

`e^(7x)=6`
`7x=ln6`
`x=(ln6)/7`

This is a critical value.

`f'(x)` isn't defined when `e^(6x)=0`. However, `e^(6x)` never can equal `0` so the function is never undefined.

Thus, the only critical point is `c=(ln6)/7~~0.256`.

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