Chapter 11.3, Problem 43E

a.

To determine

ToEvaluate:The slope of the graph of $f\left(x\right)=x^3-2x$ at the point $\left(1,-1\right)$ .

Answer to Problem 43E

The slope of the graph of $f\left(x\right)=x^3-2x$ at the point $\left(1,-1\right)$ is $m=1$

Explanation of Solution

Given:

The function $f\left(x\right)=x^3-2x$ and the point $\left(1,-1\right)$

Concept Used:

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is equal to the slope of its tangent line at $\left(x,f\left(x\right)\right)$ , and is given by

  $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

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

Calculation:

Given the function $f\left(x\right)=x^3-2x$

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is given by

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

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

  $\begin{array}{c}=0^2+3x^2+3x\left(0\right)-2\text{ Substituting limit}\\ =3x^2-2\end{array}$

At $\left(1,-1\right)$ , the slope is

  $\begin{array}{c}m=3{\left(1\right)}^2-2\\ =3-2\\ =1\end{array}$

Therefore, slope of the graph of $f\left(x\right)=x^3-2x$ at the point $\left(1,-1\right)$ is $m=1$

Conclusion:

Theslope of the graph of $f\left(x\right)=x^3-2x$ at the point $\left(1,-1\right)$ is $m=1$

b.

To determine

ToEvaluate: The equation of the tangent line to the graph of $f\left(x\right)=x^3-2x$ at the point $\left(1,-1\right)$

Answer to Problem 43E

The equation of the tangent line at the point $\left(1,-1\right)$ is $y=x-2$

Explanation of Solution

Given:

The function $f\left(x\right)=x^3-2x$ and the point $\left(1,-1\right)$ .

And from part a.) slope of the graph of $f\left(x\right)=x^3-2x$ at the point $\left(1,-1\right)$ is $m=1$

Concept Used:

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

  $y-b=m\left(x-a\right)$

Calculation:

From part a.) the slope of the graph of $f\left(x\right)=x^3-2x$ at the point $\left(1,-1\right)$ is $m=1$

The equation of the tangent line at the point $\left(a,b\right)$ is given by

  $y-b=m\left(x-a\right)$

  $\begin{array}{c}\left(y-\left(-1\right)\right)=1\left(x-1\right)\\ y+1=x-1\\ y=x-1-1\\ y=x-2\end{array}$

Therefore, the equation of the tangent line at the point $\left(1,-1\right)$ is $y=x-2$

Conclusion:

The equation of the tangent line at the point $\left(1,-1\right)$ is $y=x-2$

c.

To determine

ToGraph:The function $f\left(x\right)=x^3-2x$ and the tangent line $y=x-2$

Answer to Problem 43E

The graph of the function $f\left(x\right)=x^3-2x$ and the tangent line $y=x-2$ is

  

Explanation of Solution

Given:

The function $f\left(x\right)=x^3-2x$ and the point $\left(1,-1\right)$ .

And from part b.) the equation of tangent line is $y=x-2$

Concept Used:

The graph of a function is the smooth curve, which is the collection of all the points that satisfy the particular function.

Calculation:

For the function $f\left(x\right)=x^3-2x$

And the tangent line $y=x-2$

Plotting both of them on the same graph using a graphing utility, we have

  

Conclusion:

The graph of the function $f\left(x\right)=x^3-2x$ and the tangent line $y=x-2$ is