Chapter 4.2, Problem 16E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 16E

The local minimum value is $-\frac{49}{4}$ and there is no local maximum.

Explanation of Solution

Given:

  $g\left(x\right)=x^2-x-12$ .

Calculation:

Consider the given function,

  $g\left(x\right)=x^2-x-12$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{g}^{\prime }\left(x\right)=2x-1-0\\ =2x-1\end{array}$

Now to find the critical values, solve the first derivative by equating it to zero. That is,

  $\begin{array}{c}{g}^{\prime }\left(x\right)=0\\ 2x-1=0\\ 2x=1\\ x=\frac{1}{2}\end{array}$

Therefore $x=\frac{1}{2}$ is the critical value.

The graph of $g\left(x\right)=x^2-x-12$ falls to the left of $x=\frac{1}{2}$ and rises to the right of $x=\frac{1}{2}$ .

Therefore $x=\frac{1}{2}$ is the point of local minima.

Now the local minimum value is,

  $\begin{array}{c}g\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-\frac{1}{2}-12\\ =\frac{1}{4}-\frac{1}{2}-12\\ =-\frac{49}{4}\end{array}$

b.

To determine

To find: the intervals in which the given function is increasing.

Answer to Problem 16E

The given function increases in the interval $\left(\frac{1}{2},\infty \right)$ .

Explanation of Solution

Given:

  $g\left(x\right)=x^2-x-12$ .

Concept used:

The function increases in the interval at which the first derivative is positive.

Calculation:

Consider the given function,

  $g\left(x\right)=x^2-x-12$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{g}^{\prime }\left(x\right)=2x-1-0\\ =2x-1\end{array}$

Now the function increases in the interval for which,

  $\begin{array}{c}{g}^{\prime }\left(x\right)>0\\ 2x-1>0\\ 2x>1\\ x>\frac{1}{2}\end{array}$

Hence the given function increases in the interval $\left(\frac{1}{2},\infty \right)$ .

c.

To determine

To find: the intervals in which the given function is decreasing.

Answer to Problem 16E

The given function decreases in the interval $\left(-\infty ,\frac{1}{2}\right)$ .

Explanation of Solution

Given:

  $g\left(x\right)=x^2-x-12$ .

Concept used:

The function decreases in the interval at which the first derivative is negative.

Calculation:

Consider the given function,

  $g\left(x\right)=x^2-x-12$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{g}^{\prime }\left(x\right)=2x-1-0\\ =2x-1\end{array}$

Now the function decreases in the interval for which,

  $\begin{array}{c}{g}^{\prime }\left(x\right)<0\\ 2x-1<0\\ 2x<1\\ x<\frac{1}{2}\end{array}$

Hence the given function decreases in the interval $\left(-\infty ,\frac{1}{2}\right)$ .