Question

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

Answer 1

`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`.