Chapter 12.3, Problem 18E

(a)

To determine

To find: The slope of the graph of the function at the specified point.

Answer to Problem 18E

The slope of the graph of the function at the specified point is $3$ .

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=x^3$

The given point is $\left(1,1\right)$ .

Concept used:

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

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

The given function is,

  $f\left(x\right)=x^3$

The slope of the graph of the function at the specified point is calculated as,

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

Simplify further,

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

The slope at the point $\left(1,1\right)$ is calculated as,

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

Therefore, the slope of the graph of the function at the specified point is $3$ .

(b)

To determine

To find: The slope of the graph of the function at the specified point.

Answer to Problem 18E

The slope of the graph of the function at the specified point is $12$ .

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=x^3$

The given point is $\left(-2,-8\right)$ .

Concept used:

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

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

The given function is,

  $f\left(x\right)=x^3$

The slope of the graph of the function at the specified point is calculated as,

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

Simplify further,

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

The slope at the point $\left(-2,-8\right)$ is calculated as,

  $\begin{array}{c}m=3x^2\\ =3{\left(-2\right)}^2\\ =12\end{array}$

Therefore, the slope of the graph of the function at the specified point is $12$ .