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

May 12, 2018

$x = 0$

#### Explanation:

First of all, your function is well defined on $\setminus m a t h \boldsymbol{R}$ and also its derivative. The critical values are the set of point where $f ' \left(x\right) = 0$. In this case, you have

$f ' \left(x\right) = 1 - \sqrt{{e}^{x}} - \setminus \frac{1}{2} x \sqrt{{e}^{x}}$.

So, you have to solve

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

$\setminus \quad {e}^{\frac{x}{2}} \left(x + 2\right) - 2 > 0$? An easy calculation gives us

$x > 0 \setminus R i g h t a r r o w \setminus \quad {e}^{\frac{x}{2}} \left(x + 2\right) - 2 > 0$ and $x < 0 \setminus R i g h t a r r o w \setminus \quad {e}^{\frac{x}{2}} \left(x + 2\right) - 2 < 0$. So the only choice you have is $x = 0$. We see that $f ' \left(0\right) = 1 - \sqrt{{e}^{0}} - \frac{1}{2} 0 \sqrt{{e}^{0}} = 0.$ So the only critical point is $0$.