Chapter 11, Problem 77RE

(a)

To determine

To calculate:

The slope of the graph of following function:

  $f\left(x\right)=x^3+1$

Answer to Problem 77RE

The slope of the graph function $f$ at $\left(1,2\right)$ is $m=3$ .

Explanation of Solution

Given information:

  $f\left(x\right)=x^3+1$

Calculation:

Consider the function:

  $f\left(x\right)=x^3+1$

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

By using this formula:

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

The slope of graph of the function at $\left(1,2\right)$ is

  $\begin{array}{l}m=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ m=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ m=\underset{h\to 0}{\lim }\frac{\left({\left(1+h\right)}^3+1\right)-\left({\left(1\right)}^2+1\right)}{h}\\ m=\underset{h\to 0}{\lim }\frac{\left(\left(1+3h+3h^2+h^3\right)+1\right)-\left(2\right)}{h}\\ m=\underset{h\to 0}{\lim }\frac{\left(2+3h+3h^2+h^3\right)-2}{h}\\ m=\underset{h\to 0}{\lim }\frac{3h+3h^2+h^3}{h}\\ m=\underset{h\to 0}{\lim }\frac{3h+3h^2+h^3}{h}\end{array}$

Since, $h\to 0$ then, $h\ne 0$

Divide the denominator and numerator by $h$ and then compute the limit:

  $\begin{array}{l}m=\underset{h\to 0}{\lim }\frac{3h+3h^2+h^3}{h}=\underset{h\to 0}{\lim }\left(3+3h+h^2\right)\\ \underset{h\to 0}{\lim }\frac{3h+3h^2+h^3}{h}=3+3\left(0\right)+{\left(0\right)}^2\\ \underset{h\to 0}{\lim }\frac{3h+3h^2+h^3}{h}=3\end{array}$

Hence, the slope of the graph function $f$ at $\left(1,2\right)$ is $m=3$ .

(b)

To determine

To calculate:

The tangent line to graph the following equation:

  $f\left(x\right)=x^3+1$

Answer to Problem 77RE

The equation of tangent line at the point $\left(1,2\right)$ is $y=3x-1$ .

Explanation of Solution

Given information:

  $f\left(x\right)=x^3+1$

The equation of straight line is

  $y=mx+c$

Know that,

  $m=$ Slope of straight line

  $c=$ Constant

For the straight line, the slope at every point on it is similar.

By using the equation of the tangent:

  $\begin{array}{l}y=mx+c\\ y=3\cdot x+c\\ y=3x+c\end{array}$

Then, satisfy the equation:

  $\begin{array}{l}y=3x+c\\ 2=3\left(1\right)+c\\ c=-1\end{array}$

Hence, the equation of tangent line at the point $\left(3,17\right)$ is $y=3x-1$ .

(c)

To determine

To graph:

The tangent line and following function:

  $f\left(x\right)=x^3+1$

Answer to Problem 77RE

  

Graph of function:

Explanation of Solution

Given information:

  $f\left(x\right)=x^3+1$

  

Graph of function: