Chapter 4.3, Problem 16E

To determine

To calculate : The points of inflection of the function $y=x^3\left(4-x\right)$ .

Answer to Problem 16E

The points of inflection of the function $y=x^3\left(4-x\right)$ are $\left(0,0\right)$ and $\left(2,16\right)$ .

Explanation of Solution

Given information : The function $y=x^3\left(4-x\right)$ .

Formula used : The point of inflection of the graph is the point where the function has a tangent line an where the concavity of the graph changes.

Use the power, product rule to differentiate.

Calculation :

Consider the function $y=x^3\left(4-x\right)$ .

Calculate the first derivative of the function using product and power rule.

  $\begin{array}{c}{y}^{\prime }=x^3\frac{d}{dx}\left(4-x\right)+\left(4-x\right)\frac{d}{dx}{x}^3\\ =x^3\left(-1\right)+\left(4-x\right)\left(3x^2\right)\\ =-x^3+12x^2-3x^3\\ =-4x^3+12x^2\end{array}$

Calculate the second derivative of the function using product and power rule.

  $\begin{array}{c}{y}^{″}=\frac{d}{dx}\left(-4x^3\right)+\frac{d}{dx}\left(12x^2\right)\\ =-12x^2+24x\end{array}$

Set $y^{″}=0$ and solve the equation.

  $\begin{array}{c}-12x^2+24x=0\\ -12x\left(x-2\right)=0\\ x=0;x=2\end{array}$

The inflection points are $\left(0,f\left(0\right)\right)$ and $\left(2,f\left(2\right)\right)$ .

Now,

  $\begin{array}{c}f\left(0\right)=y\left(0\right)\\ =0^3\left(4-0\right)\\ =0\end{array}$

And,

  $\begin{array}{c}f\left(2\right)=y\left(2\right)\\ =2^3\left(4-2\right)\\ =8\left(2\right)\\ =16\end{array}$

Draw the graph of the function $y=x^3\left(4-x\right)$ .

  

It can also be seen from the graph that the concavity of the graph changes at the point $\left(0,0\right)$ and $\left(2,16\right)$ .

Thus, the points of inflection of the function $y=x^3\left(4-x\right)$ are $\left(0,0\right)$ and $\left(2,16\right)$ .