Chapter 13, Problem 9CRT

a.

To determine

To find: the derivative of the function.

Answer to Problem 9CRT

The derivative of the function is $g\text{'}\left(x\right)=3x^2$.

Explanation of Solution

Given information:

The function is $g\left(x\right)=x^3$.

Calculation: to find the derivative of the function, first differentiate it and simplify,

  $\begin{array}{l}g\left(x\right)=x^3\\ g\text{'}\left(x\right)=3x^2\text{ [differentiate]}\end{array}$

Thus, the derivative is $g\text{'}\left(x\right)=3x^2$.

b.

To determine

To find: the values.

Answer to Problem 9CRT

The values are $g\text{'}\left(-3\right)=27,g\text{'}\left(0\right)=0,g\text{'}\left(a\right)=3a^2$.

Explanation of Solution

Given information:

The function is $g\text{'}\left(x\right)=3x^2$.

Calculation: to find the values , substitute and then simplify,

  $\begin{array}{l}g\text{'}\left(x\right)=3x^2\\ g\text{'}\left(-3\right)=3{\left(-3\right)}^2\text{ [substitute]}\\ g\text{'}\left(-3\right)=27\text{ [simplify]}\\ g\text{'}\left(0\right)=3{\left(0\right)}^2\text{ [substitute]}\\ g\text{'}\left(0\right)=0\\ g\text{'}\left(a\right)=3{\left(a\right)}^2\text{ [substitute]}\\ g\text{'}\left(a\right)=3a^2\end{array}$

Thus, the values are $g\text{'}\left(-3\right)=27,g\text{'}\left(0\right)=0,g\text{'}\left(a\right)=3a^2$.

c.

To determine

To find: the equation of the line tangent at the point $\left(2,8\right)$.

Answer to Problem 9CRT

The equation of the line tangent is $y=12x-16$.

Explanation of Solution

Given information:

The function is $g\left(x\right)=x^3$.

Calculation: the slope of the line tangent is at the point $\left(2,8\right)$ ,

  $\begin{array}{l}g\text{'}\left(x\right)=3x^2\\ g\text{'}\left(2\right)=12\end{array}$

The equation of the line tangent,

  $y=12x+c$

Now, substitute and simplify,

  $\begin{array}{l}y=12x+c\\ 8=12\left(2\right)+c\text{ [susbtitute]}\\ c=-16\text{ [simplify]}\end{array}$

Thus, the equation of line tangent is $y=12x-16$.