Chapter 13.3, Problem 25E

a.

To determine

To find : $f\text{'}\left(a\right)$if $f\left(x\right)=x^3-2x+4$.

Answer to Problem 25E

  $f\text{'}\left(a\right)=3a^2-2$

Explanation of Solution

Given information:

The given function is $f\left(x\right)=x^3-2x+4$.

Calculation :

The derivative of a function $f$ at a number $a$ , denoted by $f\text{'}\left(a\right)$ ,is

  $f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$

provided this limit exist.

According to definition, $f\text{'}\left(a\right)$ is

  $\begin{array}{l}f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{{\left(a+h\right)}^3-2\left(a+h\right)+4-\left(a^3-2a+4\right)}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{a^3+3a^{2}h+3a{h}^2+h^3-2a-2h+4-a^3+2a-4}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{3a^{2}h+3a{h}^2+h^3-2h}{h}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\cancel{h}\left(3a^2+3ah+h^2-2\right)}{\cancel{h}}\\ f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\left(3a^2+3ah+h^2-2\right)\\ f\text{'}\left(a\right)=3a^2-2\end{array}$

Hence,

  $f\text{'}\left(a\right)=3a^2-2$

b.

To determine

To find : Equation of the tangent lines to the graph f at the points whose x-coordinate are 0, 1, and 2.

Answer to Problem 25E

The equation of tangents whose x coordinate are 0, 1, and 2 are $y=-2x+4$ , $y=x+2$ and $y=10x-12$ respectively.

Explanation of Solution

Given information:

The function is $f\left(x\right)=x^3-2x+4$ and $f\text{'}\left(a\right)=3a^2-2$

Calculation :

The equation of a tangent line to the graph of $f$ at $x=a$ is the line passing through the point $\left(a,f\left(a\right)\right)$ , whose slope is $f\text{'}\left(a\right)$ . Therefore,

At $x=0$ , tangent line is

  $y-f\left(0\right)=f\text{'}\left(0\right)\left(x-0\right)$

Here,

  $\begin{array}{l}f\left(0\right)=0^3-2\cdot 0+4\\ f\left(0\right)=4\end{array}$ and

  $\begin{array}{l}f\text{'}\left(0\right)=3\cdot 0^2-2\\ f\text{'}\left(0\right)=-2\end{array}$

Thus the equation of the tangent at $x=0$ is

  $\begin{array}{l}y-4=-2\left(x-0\right)\\ y=-2x+4\end{array}$

Similarly, at $x=1$ , tangent line is

  $y-f\left(1\right)=f\text{'}\left(1\right)\left(x-1\right)$

Here,

  $\begin{array}{l}f\left(1\right)=1^3-2\cdot 1+4\\ f\left(1\right)=3\end{array}$ and

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

Thus the equation of the tangent at $x=1$ is

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

And, at $x=2$ , tangent line is

  $y-f\left(2\right)=f\text{'}\left(2\right)\left(x-2\right)$

Here,

  $\begin{array}{l}f\left(2\right)=2^3-2\cdot 2+4\\ f\left(2\right)=8\end{array}$ and

  $\begin{array}{l}f\text{'}\left(2\right)=3\cdot 2^2-2\\ f\text{'}\left(2\right)=10\end{array}$

Thus the equation of the tangent at $x=2$ is

  $\begin{array}{l}y-8=10\left(x-2\right)\\ y=10x-12\end{array}$

Hence,

The equation of tangents whose x coordinate are 0, 1, and 2 are $y=-2x+4$ , $y=x+2$ and $y=10x-12$ respectively.

c.

To determine

To graph : graph f and the three tangent lines.

Answer to Problem 25E

The graph is drawn below.

Explanation of Solution

Given information:

The function is $f\left(x\right)=x^3-2x+4$ and three tangent lines are $y=-2x+4$ , $y=x+2$ and $y=10x-12$.

Calculation :

The graph of the function f and the three tangent lines is drawn below: