Chapter 11.3, Problem 58E

To determine

ToEvaluate:Thederivative of $f\left(x\right)=x^3+3x$ . 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 58E

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

For no points on the graph of $f$ , the tangent line is horizontal.

From the above graph, it is verified that for no points on the graph of $f$ , the tangent line is horizontal.

Explanation of Solution

Given:

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

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 $f\left(x\right)=x^3+3x$

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.

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

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

  $=3x^2+3$

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

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

  $f\text{'}\left(x\right)=3x^2+3=0$

  $\begin{array}{c}3x^2+3=0\\ 3x^2=-3\\ x^2=\frac{-3}{3}\\ x^2=-1\end{array}$

No solution.

Since, there is no possible value of $x$ such that $x^2=-1$

Therefore, for no points on the graph of $f$ , the tangent line is horizontal.

Plotting the graph of $f$ , using a graphing utility, we have

  

Therefore, from the above graph, it is verified that for no points on the graph of $f$ , the tangent line is horizontal.

Conclusion:

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

For no points on the graph of $f$ , the tangent line is horizontal.

From the above graph, it is verified that for no points on the graph of $f$ , the tangent line is horizontal.