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

Oct 19, 2017

Use the quotient rule and the definition of the derivative of the ${e}^{x}$ function (and/or the chain rule). x=4 is the only critical pt.

#### Explanation:

The critical points will occur where $f ' \left(x\right) = 0$. To find f'(x) we will need the quotient rule and the definition of the derivative of ${e}^{x}$. The former states that given $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} , f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$. Using that here...

$g \left(x\right) = x , g ' \left(x\right) = 1 , h \left(x\right) = {e}^{\sqrt{x}} , h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({e}^{\sqrt{x}}\right)$

The definition of $\frac{d}{\mathrm{dx}} {e}^{x} = \mathrm{dx} \cdot {e}^{x}$. THat in mind, we have ${e}^{\sqrt{x}} = {e}^{{x}^{\frac{1}{2}}}$, and $\frac{d}{\mathrm{dx}} {x}^{\frac{1}{2}} = \frac{1}{2 {x}^{\frac{1}{2}}} = \frac{1}{2 \sqrt{x}}$. Then $\frac{d}{\mathrm{dx}} {e}^{\sqrt{x}} = \frac{1}{2 \sqrt{x}} {e}^{\sqrt{x}}$, and thus...

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

Recall that ${\left({a}^{b}\right)}^{c} = {a}^{b \cdot c} \ldots$

-> f'(x) = ((e^sqrtx) - (sqrtx/2 e^sqrtx))/(e^(2sqrtx)

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

This can be further simplified if we wish, since ${a}^{b} / {a}^{c} = {a}^{b - c} \ldots$

$= \frac{1 - \frac{\sqrt{x}}{2}}{e} ^ \sqrt{x}$.

To find the critical points, we set this equal to 0...

$\frac{1 - \frac{\sqrt{x}}{2}}{e} ^ \sqrt{x} = 0 \to 1 - \frac{\sqrt{x}}{2} = 0 \to 1 = \frac{\sqrt{x}}{2} \to \sqrt{x} = 2 \to x = 4$

Thus, there is one real-valued critical point, and it occurs at $x = 4$

graph{x/(e^sqrtx) [-10, 10, -5, 5]}

The graph bears this out, as the maximum for the function appears to occur at $x = 4$.