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

##### 1 Answer

First of all,

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

Divide by both exponentials (which are never zero), and by

We know this is a minimum and not a maximum because

Therefore f' is positive for all values of x.

Since f' is positive, f is strictly increasing on

Therefore f has no critical points.