Question

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

Answer 1

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

Explanation:

`f(x) = e^x - 2e^(-x) - 3x`

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

`f'(x) = e^x-2e^(-x)(-1)-3`

`= e^x+2e^(-x)-3`

`f'(x)` is never undefined, so we need only find its zeros.

We need to solve

`e^x+2e^(-x)-3 = 0`

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

`e^x+2/e^x-3 = 0`

Get a common denomiator, noting that `e^x*e^x = e^(2x)`, so we have

`(e^(2x)+2-3e^x)/e^x = 0`.

So now we need to solve:

`e^(2x)-3e^x+2 = 0`

Again note that `e^(2x)=(e^x)^2`, so we can factor:

`(e^x-1)(e^x-2) = 0`

`e^x=1` `" "` OR `" "` `e^x=2`

`x=0` `" "` OR `" "` `x=ln2`

Both are in `"Dom"(f)`, so both are critical numbers.