What are the critical values of #f(x)=x-xsqrt(e^x#?

1 Answer
May 12, 2018

Answer:

#x=0#

Explanation:

First of all, your function is well defined on #\mathbb{R}# and also its derivative. The critical values are the set of point where #f'(x) = 0#. In this case, you have

#f'(x) = 1- sqrt(e^x) - \frac{1}{2}xsqrt(e^x)#.

So, you have to solve

#1- sqrt(e^x) - \frac{1}{2}xsqrt(e^x)=0 \quad# which is equivalent to #\quad e^(x/2)(x+2)=2#. Now this is a nonlinear equation. But you could think in this way. You want to find #x# such that #\quad e^(x/2)(x+2) - 2=0#. But for which #x# we have

#\quad e^(x/2)(x+2) - 2>0#? An easy calculation gives us

#x > 0 \Rightarrow\quad e^(x/2)(x+2) - 2>0# and #x < 0 \Rightarrow\quad e^(x/2)(x+2) - 2<0#. So the only choice you have is #x=0#. We see that #f'(0) = 1-sqrt(e^0)-1/2 0sqrt(e^0)=0.# So the only critical point is #0#.