Chapter 12.3, Problem 61E

To determine

ToEvaluate:Thederivative of $f\left(x\right)=3x^3-9x$ . Use the derivative to determine any points on the graph of $f$ at which the tangent line is horizontal and use a graphing utility to verify the result.

Answer to Problem 61E

The derivative of $f\left(x\right)=3x^3-9x$ is $f\text{'}\left(x\right)=9x^2-9$

The point on the graph of $f$ at which the tangent line is horizontal are $\left(x,y\right)=\left(-1,6\right)$ and $\left(x,y\right)=\left(1,-6\right)$

From the graph of $f$ and the tangent line, it is verified that the point on the graph of $f$ at which the tangent line is horizontal are $\left(x,y\right)=\left(-1,6\right)$ and $\left(x,y\right)=\left(1,-6\right)$

Explanation of Solution

Given:

The function $f\left(x\right)=3x^3-9x$

Concept Used:

The derivative of $f$ at $x$ is given by

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ provided the limit exists.

The tangent line is horizontal when the slope of tangent line is zero.

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

  ${\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

Calculation:

Forthe function $f\left(x\right)=3x^3-9x$

The derivative of $f$ at $x$ is given by

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ provided the limit exists.

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\left(3{\left(x+h\right)}^3-9\left(x+h\right)\right)-\left(3x^3-9x\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{3x^3+3h^3+9x^{2}h+9x{h}^2-9x-9h-3x^3+9x}{h}\right){\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{3h^3+9x^{2}h+9x{h}^2-9h}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h\left(3h^2+9x^2+9xh-9\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(3h^2+9x^2+9xh-9\right)\\ =3{\left(0\right)}^2+9x^2+9x\left(0\right)-9\text{ Substituting limit}\end{array}$

  $=9x^2-9$

Therefore, the derivative of $f\left(x\right)=3x^3-9x$ is $f\text{'}\left(x\right)=9x^2-9$

The tangent line is horizontal when the slope of tangent line is zero.

Since the derivative of a function is the slope of its tangent line.

Substituting the derivative of $f$ or the slope of tangent equal to zero, we have

  $f\text{'}\left(x\right)=9x^2-9=0$

  $\begin{array}{c}9x^2-9=0\\ 9x^2=9\\ x^2=\frac{9}{9}\\ x^2=1\end{array}$

  $x=-1$

and

  $x=1$

At $x=-1$

Substituting in the function $y=f\left(x\right)=3x^3-9x$

  $\begin{array}{c}y=3{\left(-1\right)}^3-9\left(-1\right)\\ =-3+9\\ =6\end{array}$

At $x=1$

Substituting in the function $y=f\left(x\right)=3x^3-9x$

  $\begin{array}{c}y=3{\left(1\right)}^3-9\left(1\right)\\ =3-9\\ =-6\end{array}$

Therefore, the points on the graph of $f$ at which the tangent line horizontal are $\left(x,y\right)=\left(-1,6\right)$ and $\left(x,y\right)=\left(1,-6\right)$

For the equation of tangent line

Since, the tangent line is horizontal, so the slope of tangent line is $m=0$

For the point on the tangent line $\left(x,y\right)=\left(-1,6\right)$

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

  $\begin{array}{c}\left(y-6\right)=0\left(x+1\right)\\ y-6=0\\ y=6\end{array}$

For the point on the tangent line $\left(x,y\right)=\left(1,-6\right)$

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

  $\begin{array}{c}\left(y+6\right)=0\left(x-1\right)\\ y+6=0\\ y=-6\end{array}$

Plotting tangent line and the graph of $f$ , on the same graph using a graphing utility, we have

  

Therefore, from the above graph, it is verified that the points on the graph of $f$ at which the tangent line is horizontal are $\left(x,y\right)=\left(-1,6\right)$ and $\left(x,y\right)=\left(1,-6\right)$

Conclusion:

Thederivative of $f\left(x\right)=3x^3-9x$ is $f\text{'}\left(x\right)=9x^2-9$

The point on the graph of $f$ at which the tangent line is horizontal are $\left(x,y\right)=\left(-1,6\right)$ and $\left(x,y\right)=\left(1,-6\right)$

From the graph of $f$ and the tangent line, it is verified that the point on the graph of $f$ at which the tangent line is horizontal are $\left(x,y\right)=\left(-1,6\right)$ and $\left(x,y\right)=\left(1,-6\right)$