Q. 1TB - Calculus

Question

Analyzing the behavior of a continuous function: Consider the function $f\left(x\right)=\frac{x^3}{e^x}$ on the interval $[0,\infty )$. Where is $f$ increasing and where is $f$ decreasing? Is the function bounded above and/or below? Does $f$ have a limit as $x\to \infty$? 

Step-by-Step Solution

Answer

The function $f\left(x\right)=\frac{x^3}{e^x}$ is increasing in the interval $[0,3)$ and decreasing $\left(3,\infty \right)$.

The function is bounded both above and below.

Yes, the function $f$ has a limit as $x\to \infty$. 

Step 1. Given Information

We are given a function $f\left(x\right)=\frac{x^3}{e^x}$.

The objective is to know where the function is decreasing and increasing in the interval $[0,\infty )$

Step 2. Find the derivative of the function

The derivative of the function $f\left(x\right)=\frac{x^3}{e^x}$ is given as:

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{3x^2·e^x-x^3·e^x}{e^{2x}}\\ & =& \frac{x^2{e}^x\left(3-x\right)}{e^{2x}}\\ & =& \frac{x^2\left(3-x\right)}{e^x}\end{array}$

Step 3. Use the derivative test.

The derivative of $f\left(x\right)=\frac{x^3}{e^x}$ is given as $f\text{'}\left(x\right)=\frac{x^2(3-x)}{e^x}$ and it is positive for $x<3$ and negative for $x>3$.

So using the derivative test, the function is increasing in the interval $[0,3)$ and decreasing in the interval $\left(3,\infty \right)$.

Step 4. Check the boundness

As the function is increasing from $[0,3)$ and decreasing from $\left(3,\infty \right)$. So the function has an upper bound at $x=3$. The function value $f\left(3\right)$ is the upper bound.

$\begin{array}{rcl}f\left(3\right)& =& \frac{3^3}{e^3}\\ & =& \frac{27}{e^3}\end{array}$

The function $f\left(x\right)=\frac{x^3}{e^x}$ is always non-negative, so zero is the lower bound.

Step 5. Does the function has a limit as x &#8594; &#8734; .

As the function decreases from $\left(3,\infty \right)$ and the lower bound of the function is zero. So as $x\to \infty$, the function value tends to zero.

So the function has a limit as $x\to \infty$.