Chapter 11, Problem 78CR

a)

To determine

To evaluate: the slope of the graph of f at the given point.

Answer to Problem 78CR

Slope at (2, 6) is 3.

Explanation of Solution

Given:

  $f\left(x\right)=x^2-x,\left(2,6\right)$

Calculation: -

It is given that,

  $\begin{array}{l}f\left(x\right)=x^2-x\\ f\text{'}\left(x\right)=2x-1\\ f\text{'}\left(2\right)=3\end{array}$

Hence, Slope at (2, 6) is 3.

b)

To determine

the equation of the tangent line to the graph at the point.

Answer to Problem 78CR

The equation of tangent is $11x-y-16=0$ .

Explanation of Solution

Given:

  $f\left(x\right)=x^2-x,\left(2,6\right)$

Calculation: -

It is given that,

  $\begin{array}{l}f\left(x\right)=x^3-x\\ f\text{'}\left(x\right)=3x^2-1\\ f\text{'}\left(2\right)=11\end{array}$

Hence, Slope at (2, 6) is 11.

Using one-point formula to find the equation of line,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-6=11\left(x-2\right)\\ y-6=11x-22\\ 11x-y-16=0\end{array}$

Hence, the equation of tangent is $11x-y-16=0$ .

c)

To determine

the graph of the function and the tangent line on it.

Answer to Problem 78CR

  

Explanation of Solution

Given:

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

Calculation: -It is given that,

  $\begin{array}{l}f\left(x\right)=x^3-x\\ f\text{'}\left(x\right)=3x^2-1\\ f\text{'}\left(2\right)=11\end{array}$

Hence, Slope at (2, 6) is 11.

Using one-point formula to find the equation of line,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-6=11\left(x-2\right)\\ y-6=11x-22\\ 11x-y-16=0\end{array}$

Hence, the equation of tangent is $11x-y-16=0$ .

Using graphing calculator to plot the graph,