# What are the critical points of f(x) = sqrt(e^(sqrtx)-sqrtx)?

Feb 25, 2017

$x = 0$

#### Explanation:

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

Through the chain rule:

$f ' \left(x\right) = \frac{1}{2} {\left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)$

Then:

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

Factoring from the final parentheses:

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

Rewriting:

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

$f ' \left(x\right) = \frac{{e}^{\sqrt{x}} - 1}{4 \sqrt{x} \sqrt{{e}^{\sqrt{x}} - \sqrt{x}}}$

If we want to find critical point, we need to find when $f ' \left(x\right) = 0$ or when $f '$ is undefined but $f$ is defined.

Setting $f ' \left(x\right) = 0$ gives ${e}^{\sqrt{x}} - 1 = 0 \implies {e}^{\sqrt{x}} = 1 \implies x = 0$.

This is also when $f '$ is undefined, since $\sqrt{x}$ is in the denominator. We also see that $\sqrt{{e}^{\sqrt{x}} - \sqrt{x}}$ is never undefined, as ${e}^{\sqrt{x}} > \sqrt{x}$ for all $x \ge 0$.

Thus the only critical point is $x = 0$. It's also one of the function's endpoints.