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

1 Answer
Aug 14, 2017

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.