Chapter 1, Problem 64RE

To determine

To graph:Thefunction and approximate any relative minima or maxima.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=x^3-4x^2-1$ .

Graph:

The first derivative of the function.

  $\begin{array}{c}f\left(x\right)=x^3-4x^2-1\\ f\text{'}\left(x\right)=3x^2-8x\end{array}$

Equate to zero.

  $\begin{array}{c}3x^2-8x=0\\ x\left(3x-8\right)=0\\ x=0,x=\frac{8}{3}\end{array}$

Second derivative of the function.

  $f\text{'}\text{'}\left(x\right)=6x-8$

The above equation, at $x=0$ .

  $\begin{array}{c}f\text{'}\text{'}\left(x\right)=6x-8\\ =-8\end{array}$

The given equation $f\left(x\right)=x^3-4x^2-1$ , at $x=0$ .

  $\begin{array}{c}f\left(x\right)=x^3-4x^2-1\\ f\left(0\right)=-1\end{array}$

  $f\text{'}\text{'}\left(x\right)<0$ for $x=0$ implies $\left(0,-1\right)$ is relative maxima.

The equation of second derivative, at $x=\frac{8}{3}$ ,

  $\begin{array}{c}f\text{'}\text{'}\left(x\right)=6x-8\\ =6\left(\frac{8}{3}\right)-8\\ =8\end{array}$

The given equation $f\left(x\right)=x^3-4x^2-1$ , at $x=\frac{8}{3}$ .

  $\begin{array}{c}f\left(x\right)=x^3-4x^2-1\\ f\left(\frac{8}{3}\right)={\left(\frac{8}{3}\right)}^3-4{\left(\frac{8}{3}\right)}^2-1\\ =-10.48\end{array}$

  $f\text{'}\text{'}\left(x\right)>0$ for $x=0$ implies $\left(\frac{8}{3},-10.48\right)$ is relative minimum.

To draw graph for the given function use $\text{T}1-83$ calculator and click on $"\text{Y}="$ and enter the above expression.

  

Click on window and change the window settings.

  

Click on the graph.

  

Interpretation:the function is relative maxima $\left(0,-1\right)$ and relative minimum at $\left(\frac{8}{3},-10.48\right)$ .