Find the absolute minimum value on [0,\infty) for f(x) = \frac{2e^x}{x^5}.

Question:

Find the absolute minimum value on {eq}[0,\infty) {/eq} for {eq}f(x) = \frac{2e^x}{x^5}. {/eq}

Quotient Rule

The quotient rule is used to find the derivative of the quotient of two functions. Suppose that a differentiable function {eq}f(x) {/eq} can be written as

{eq}f(x)=\frac{g(x)}{h(x)} {/eq}

for some differentiable functions {eq}g(x) {/eq} and {eq}h(x) {/eq}.

Then the derivative of {eq}f(x) {/eq} is

{eq}f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^2} {/eq}

The function {eq}f(x) {/eq} can be written as

{eq}f(x)=\frac{g(x)}{h(x)} {/eq}

with

{eq}g(x)=2e^x {/eq} and {eq}h(x)=x^5 {/eq}.

The derivatives of these two functions are

{eq}g'(x)=2e^x {/eq} and {eq}h'(x)=5x^4 {/eq}

Using the quotient rule, the derivative of {eq}f(x) {/eq} is

{eq}\begin{align} f'(x)&=\frac{2e^xx^5-2e^x\cdot 5x^4}{(x^5)^2}\\ &=\frac{2e^xx^5-10e^xx^4}{x^10}\\ &=\frac{2e^x}{x^5}-\frac{10e^x}{x^6} \end{align} {/eq}

Set this derivative equal to zero and solve for {eq}x {/eq}:

{eq}\begin{align} \frac{2e^x}{x^5}-\frac{10e^x}{x^6}&=0\\ 2e^x(\frac{1}{x^5}-\frac{5}{x^6})&=0\\ \frac{1}{x^5}-\frac{5}{x^6}&=0\\ \frac{1}{x^5}(1-\frac{5}{x})&=0\\ 1-\frac{5}{x}&=0\\ 1&=\frac{5}{x}\\ x&=5 \end{align} {/eq}

The only critical point on this interval is at {eq}x=5 {/eq}. The function {eq}f(x) {/eq} is not defined at {eq}x=0 {/eq} and there is no other endpoint, so the critical point must be either an absolute minimum or absolute maximum.

Find the value of {eq}f(x) {/eq} at any other point in the interval:

{eq}\begin{align} f(1)&=\frac{2e^1}{1^5}\\ &=2e\\ &\approx 5.44 \end{align} {/eq}

Find {eq}f(5) {/eq}:

{eq}\begin{align} f(5)&=\frac{2e^5}{5^5}\\ &\approx 0.095 \end{align} {/eq}

Since {eq}f(5)<f(1) {/eq}, {eq}f(5) {/eq} cannot be the absolute maximum.

This means it must be a minimum. So {eq}f(5)=\frac{2e^5}{3125} {/eq} is the absolute minimum value.