Chapter 11.3, Problem 41E

a.

To determine

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

Answer to Problem 41E

The slope of the graph of $f\left(x\right)=x^2-1$ at the point $\left(2,3\right)$ is $m=4$

Explanation of Solution

Given:

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

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

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

  ${\left(a+b\right)}^2=a^2+b^2+2ab$

Calculation:

Given the function $f\left(x\right)=x^2-1$

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

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left({\left(x+h\right)}^2-1\right)-\left(x^2-1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{x^2+h^2+2xh-1-x^2+1}{h}{\left(a+b\right)}^2=a^2+b^2+2ab\\ =\underset{h\to 0}{\lim }\frac{h^2+2xh}{h}\end{array}$

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

At $\left(2,3\right)$ , the slope is

  $\begin{array}{c}m=2\left(2\right)\\ =4\end{array}$

Therefore, slope of the graph of $f\left(x\right)=x^2-1$ at the point $\left(2,3\right)$ is $m=4$

Conclusion:

Theslope of the graph of $f\left(x\right)=x^2-1$ at the point $\left(2,3\right)$ is $m=4$

b.

To determine

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

Answer to Problem 41E

The equation of the tangent line at the point $\left(2,3\right)$ is $y=4x-5$

Explanation of Solution

Given:

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

And from part a.) slope of the graph of $f\left(x\right)=x^2-1$ at the point $\left(2,3\right)$ is $m=4$

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 $f\left(x\right)=x^2-1$ at the point $\left(2,3\right)$ is $m=4$

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

  $y-b=m\left(x-a\right)$

i.e., $\begin{array}{c}y-3=4\left(x-2\right)\\ y-3=4x-8\\ y=4x-8+3\\ y=4x-5\end{array}$

Therefore, the equation of the tangent line at the point $\left(2,3\right)$ is $y=4x-5$

Conclusion:

The equation of the tangent line at the point $\left(2,3\right)$ is $y=4x-5$

c.

To determine

ToGraph: The function $f\left(x\right)=x^2-1$ and the tangent line $y=4x-5$

Answer to Problem 41E

The graph of the function $f\left(x\right)=x^2-1$ and the tangent line $y=4x-5$ is

  

Explanation of Solution

Given:

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

And from part b.) the equation of tangent line is $y=4x-5$

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 $f\left(x\right)=x^2-1$

And the tangent line $y=4x-5$

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

  

Conclusion:

The graph of the function $f\left(x\right)=x^2-1$ and the tangent line $y=4x-5$ is