Question

What are the critical values, if any, of `f(x)=x+e^(-3^x)`?

Answer 1

First of all, `f(x)` is differentiable everywhere, with
`f'(x) = 1 + e^(-3^x)(-3^x)(ln3)`
`f'(x) = 1 - ((3^x)(ln3))/e^(3^x)`
The denominator of the expression dominates as x increases in absolute value, so that f'(x) has a horizontal asymptote at y = 1.

Now this function, f', is minimal when
`f''(x) = -(ln3)[(ln3)e^(-3^x)(-3^x)(3^x)+(ln3)e^(-3^x)(3^x)] = 0`
Divide by both exponentials (which are never zero), and by `(ln3)^2`:
`(-3^x)+1 = 0`
`3^x = 1`
`x = 0`
We know this is a minimum and not a maximum because `f'(0) < 1`, but we may also apply the first derivative test.

`f'(0) = 1 - ((1)(ln3))/e^(1)`
`f'(0) = 1 - (ln3)/e > 0`

Therefore f' is positive for all values of x.

Since f' is positive, f is strictly increasing on `(-oo,oo)`.
Therefore f has no critical points.