Chapter 2.3, Problem 22E

Solution

To Sketch: The given function with local and minimum extrema and $x-$ intercepts. And explain its end behaviour with upper and lower limits.

Given: $f\left(x\right)=\left(2x+1\right){\left(x-4\right)}^3$

Graph:

Sketch the graph of $f\left(x\right)=\left(2x+1\right){\left(x-4\right)}^3$with all its extrema and zeros.

  

Find the extrema:

Find the first derivative of the function:

  $f\left(x\right)=8x^3-69x^2+168x-80$

Find the second derivative of the function:

  $f^{\prime \prime }\left(x\right)=24x^2-138x+168$

To find the local maximum and minimum values of the function, set the derivative equal to  $0$  and solve.

  $8x^3-69x^2+168x-80=0$

Factor the term:

  $\left(x-4\right)\left(8x^2-37x+20\right)=0$

Factor by grouping:

  $\left(x-4\right)\left(8x-5\right)\left(x-4\right)=0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

  $\begin{array}{c}x-4=0\\ 8x-5=0\\ x-4=0\end{array}$

Thus, the values of $x$ are:

  $\begin{array}{l}x=4\\ x=\frac{5}{8}\end{array}$

Evaluate the second derivative at $x=4$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

  $\begin{array}{l}=24{\left(4\right)}^2-138\cdot 4+168\\ =384-552+168\\ =-168+168\\ =0\end{array}$

Split $(-\infty ,\infty )$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

  $\left(-\infty ,\frac{5}{8}\right)\cup \left(\frac{5}{8},4\right)\cup (4,\infty )$

Substitute any number, such as $0$ , from the interval $\left(-\infty ,\frac{5}{8}\right)$ in the first derivative $8x^3-69x^2+168x-80$ to check if the result is negative or positive.

Replace the variable $x$ with $0$ in the expression.

  $\begin{array}{l}{f}^{\prime }\left(0\right)=8{\left(0\right)}^3-69{\left(0\right)}^2+168\left(0\right)-80\\ f^{\prime }\left(0\right)=-80\end{array}$

Substitute any number, such as $2$ , from the interval $\left(\frac{5}{8},4\right)$ in the first derivative $8x^3-69x^2+168x-80$ to check if the result is negative or positive.

Replace the variable $x$ with $2$ in the expression.

  $\begin{array}{l}{f}^{\prime }\left(2\right)=8{\left(2\right)}^3-69{\left(2\right)}^2+168\left(2\right)-80\\ f^{\prime }\left(2\right)=44\end{array}$

Since the first derivative changed signs from negative to positive around $x=\frac{5}{8}$ is a local minimum.

Since the first derivative did not change signs around $x=4$ , this is not a local maximum or minimum.

Find the value of $y:$

Substitute $x=\frac{5}{8}$ in $f\left(x\right)=\left(2x+1\right){\left(x-4\right)}^3$ to find $y$ value:

  $\begin{array}{l}y=\left(2\left(\frac{5}{8}\right)+1\right){\left(\frac{5}{8}-4\right)}^3\\ y=-86.497\end{array}$

The function $f$ has a local minimum of about $-86.497$ at $x=\frac{5}{8}$ .

Find the zeroes of the function $f\left(x\right)=\left(2x+1\right){\left(x-4\right)}^3$:

  $\begin{array}{c}2x+1=0\\ $ {(x-4)}^3=0$\end{array}$

The function has zeros at $x=-\frac{1}{2},4,4,4$

Thus, the function $f$ has $1$ extrema and $4$ zeros, the maximum number possible for a quadratic.

End behaviour:

Use the limits to describe its end behaviour.

  $\underset{x\to \infty }{\lim }f\left(x\right)=\infty \text{ and }\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$

The graph rises to the left and rises to the right