Q. 83 - Calculus

Question

Determine the local extrema of the function,

$f\left(x\right)=x^3{e}^{-x}, I=[0,\infty ), J=(-\infty ,\infty )$.

Step-by-Step Solution

Answer

The value

Step 1. Given Information.

The function is,

$f\left(x\right)=x^3{e}^{-x}, I=[0,\infty ), J=(-\infty ,\infty )$.

Step 2. The critical point.

For critical points we consider, 

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \frac{d}{dx}\left(x^3{e}^{-x}\right)=0 \\ {x}^3(-e^{-x})+\left(e^{-x}\right)3x^2=0 \\ {x}^2{e}^{-x}(-x+3)=0 \end{gathered}$$

Dividing both sides by $e^{-x}$,

$x^2(-x+3)=0$

Either $x=0$

or, $$\begin{gathered} -x+3=0 \\ \Rightarrow x=3 \end{gathered}$$

Hence, $x=0,3$.

Step 3. Values of I .

Now,

$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }{x}^3{e}^{-x} \\ =0 \end{gathered}$$

style="width:30%" $$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{x}^3{e}^{-x} \\ =\underset{x\to \infty }{\lim }\frac{x^3}{e^x} \left[\frac{\infty }{\infty }\right] \\ =\underset{x\to \infty }{\lim }\frac{6x}{e^x} \\ =\underset{x\to \infty }{\lim }\frac{6}{e^x} \\ =\frac{6}{\infty } \\ =0 \end{gathered}$$

The graph of the function with limit $I=(0,\infty )$ is shown as,