Chapter 11.3, Problem 57E

To determine

To calculate : The derivative of the function, $f\left(x\right)=2x^3-8x$ and hence find the points on the graph of the function where the tangent line is horizontal. Also, verify your results using graphing utility.

Answer to Problem 57E

The derivative of the function, $f\left(x\right)=2x^3-8x$ is $f\text{'}\left(x\right)=6x^2-8$ and the points where the tangent line is horizontal are $\left(\frac{2\sqrt{3}}{3},-\frac{32\sqrt{3}}{9}\right)$ and $\left(-\frac{2\sqrt{3}}{3},\frac{32\sqrt{3}}{9}\right)$ .

Explanation of Solution

Given information : The function is $f\left(x\right)=2x^3-8x$ .

Formula used : The derivative of the function $f\left(x\right)$ is given by, $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ such that the limit exists.

The tangent line is horizontal at the points where the derivative of function is 0.

Calculation : Find the derivative of function, $f\left(x\right)=2x^3-8x$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\left[2{\left(x+h\right)}^3-8\left(x+h\right)\right]-\left(2x^3-8x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[2\left(x^3+h^3+3x^{2}h+3x{h}^2\right)-8\left(x+h\right)\right]-\left(2x^3-8x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{2x^3+2h^3+6x^{2}h+6x{h}^2-8x-8h-2x^3+8x}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(2h^2+6x^2+6xh-8\right)}{h}\\ =\underset{h\to 0}{\lim }2h^2+6x^2+6xh-8\\ =2{\left(0\right)}^2+6x^2+6x\left(0\right)-8\\ =6x^2-8\end{array}$

Thus, the derivative of the function is $f\text{'}\left(x\right)=6x^2-8$ .

Equate the derivative to 0.

  $\begin{array}{c}f\text{'}\left(x\right)=0\\ 6x^2-8=0\\ x^2=\frac{8}{6}\\ x^2=\frac{4}{3}\\ x=\pm \sqrt{\frac{4}{3}}\\ =\pm \frac{2\sqrt{3}}{3}\end{array}$

Substitute $x=\frac{2\sqrt{3}}{3}$ in the function, $f\left(x\right)=2x^3-8x$ .

  $\begin{array}{c}f\left(\frac{2\sqrt{3}}{3}\right)=2{\left(\frac{2\sqrt{3}}{3}\right)}^3-8\left(\frac{2\sqrt{3}}{3}\right)\\ =\frac{48\sqrt{3}}{27}-\frac{16\sqrt{3}}{3}\\ =\frac{48\sqrt{3}}{27}-\frac{144\sqrt{3}}{27}\\ =-\frac{96\sqrt{3}}{27}\\ =-\frac{32\sqrt{3}}{9}\end{array}$

Substitute $x=-\frac{2\sqrt{3}}{3}$ in the function, $f\left(x\right)=2x^3-8x$ .

  $\begin{array}{c}f\left(-\frac{2\sqrt{3}}{3}\right)=2{\left(-\frac{2\sqrt{3}}{3}\right)}^3-8\left(-\frac{2\sqrt{3}}{3}\right)\\ =-\frac{48\sqrt{3}}{27}+\frac{16\sqrt{3}}{3}\\ =-\frac{48\sqrt{3}}{27}+\frac{144\sqrt{3}}{27}\\ =\frac{96\sqrt{3}}{27}\\ =\frac{32\sqrt{3}}{9}\end{array}$

So, the points on the graph of the function where the tangent line is horizontal are $\left(\frac{2\sqrt{3}}{3},-\frac{32\sqrt{3}}{9}\right)$ and $\left(-\frac{2\sqrt{3}}{3},\frac{32\sqrt{3}}{9}\right)$ .

To verify the results,

Enter the equation of the function using $\text{Y}=$ key.

  ${\text{Y}}_1=2\text{X}^3-8\text{X}$

Then plot the function using GRAPH key.

  

From the graph it can be seen that tangent drawn at $x=1.15$ and $x=-1.15$ will be horizontal to x -axis.

As $\frac{2\sqrt{3}}{3}\approx 1.15$ and $-\frac{2\sqrt{3}}{3}\approx -1.15$ , the obtained results are verified.