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

Question

Aurora Alvarado · Accepted Answer

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.

Aurora Alvarado · Answer

undefined To find the critical values of $f(x) = x + e^{-3^x}$, we need to find where the derivative of $f(x)$ is zero or does not exist. The derivative of $f(x)$ is:

$$f'(x) = 1 - 3^x \cdot e^{-3^x} \cdot \ln(3)$$

Setting $f'(x)$ to zero and solving for $x$ gives:

$$1 - 3^x \cdot e^{-3^x} \cdot \ln(3) = 0$$

Since $e^{-3^x}$ is always positive, the only way for the expression to be zero is when $1 - 3^x = 0$. Solving this gives $x = \log_3(1)$, which is not a real number. Therefore, there are no critical values for $f(x)$.