Calculus I - The Shape of a Graph, Part II

Section 4.6 : The Shape of a Graph, Part II

11. For $h\left( z \right) = {z^4} - 2{z^3} - 12{z^2}$ answer each of the following questions.

Identify the critical points of the function.
Determine the intervals on which the function increases and decreases.
Classify the critical points as relative maximums, relative minimums or neither.
Determine the intervals on which the function is concave up and concave down.
Determine the inflection points of the function.
Use the information from steps (a) – (e) to sketch the graph of the function.

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a Identify the critical points of the function. Show Solution

The parts to this problem (with the exception of the last part) are just like the basic increasing/decreasing problems from the previous section and the basic concavity problems from earlier in this section. Because of that we will not be putting in quite as much detail here. If you are still unsure how to work the parts of this problem you should go back and work a few of the basic problems from the previous section and/or earlier in this section before proceeding.

We will need the 1^st derivative to start things off.

\[\underline {h'\left( z \right) = 4{z^3} - 6{z^2} - 24z = 2z\left( {2{z^2} - 3z - 12} \right)} \]

From the 1^st derivative we can see that the critical points of this function are then,

\[\underline {x = 0\,\,\,\,\,\,\,\& \,\,\,\,\,\,x = \frac{3 \pm \sqrt {105}}{4} = - 1.8117,\,\,3.3117} \]

b Determine the intervals on which the function increases and decreases. Show Solution

To answer this part all we need is the number line for the 1^st derivative.

$A number line for the first derivative of $h\left( z \right) = {z^4} - 2{z^3} - 12{z^2}$ with dashed vertical lines at $z = \frac{{3 - \sqrt {105} }}{4}$, $z = 0$ and $z = \frac{{3 + \sqrt {105} }}{4}$ dividing it into four regions. In the first region $h'\left( { - 2} \right) = - 8$ so $h'\left( z \right) < 0$, in the second $h'\left( { - 1} \right) = 14$ so $h'\left( z \right) > 0$, in the third $h'\left( 1 \right) = - 26$ so $h'\left( z \right) < 0$, and in the fourth $h'\left( 4 \right) = 64$ so $h'\left( z \right) > 0$.$

From this we get the following increasing/decreasing information for the function.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Increasing : }}\left( {\frac{3 - \sqrt {105}}{4},0} \right)\,\,\,\,\,\& \,\,\,\,\,\left( {\frac{3 + \sqrt {105}}{4},\infty } \right)\hspace{0.25in}{\mbox{Decreasing : }}\left( { - \infty ,\frac{3 - \sqrt {105}}{4}} \right)\,\,\,\,\& \,\,\,\,\left( {0,\frac{3 + \sqrt {105}}{4}} \right)}}\]

c Classify the critical points as relative maximums, relative minimums or neither. Show Solution

From the number line in the previous step we get the following classifications of the critical points.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{array}{c}z = \displaystyle \frac{3 \pm \sqrt {105}}{4}:\,\,\,{\mbox{Relative Minimum}}\hspace{0.5in}\hspace{0.5in}\hspace{0.5in}\,\,\,z = 0:\,\,\,{\mbox{Relative Maximum}}\,\end{array}}\]

d Determine the intervals on which the function is concave up and concave down. Show Solution

We’ll need the 2^nd derivative to find the list of possible inflection points.

\[\underline {h''\left( z \right) = 12{z^2} - 12z - 24 = 12\left( {z - 2} \right)\left( {z + 1} \right)} \]

The possible inflection points for this function are,

\[\underline {z = - 1\hspace{0.5in}\& \hspace{0.5in}\,z = 2} \]

To get the intervals of concavity we’ll need the number line for the 2^nd derivative.

$A number line for the second derivative of $h\left( z \right) = {z^4} - 2{z^3} - 12{z^2}$ with dashed vertical lines at $z = - 1$ and $z = 2$ dividing it into three regions. In the left region $h''\left( { - 2} \right) = 48$ so $h''\left( z \right) > 0$, in the middle region $h''\left( 0 \right) = - 24$ so $h''\left( z \right) < 0$, and in the right region $h''\left( 3 \right) = 48$ so $h''\left( z \right) > 0$.$

From this we get the following concavity information for the function.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Concave Up : }}\left( { - \infty , - 1} \right)\,\,\,\,\,\,\& \,\,\,\,\,\,\left( {2,\infty } \right)\hspace{0.5in}{\mbox{Concave Down : }}\left( { - 1,2} \right)}}\]

e Determine the inflection points of the function. Show Solution

From the concavity information in the previous step we can see that the inflection points for the function are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = - 1\hspace{0.5in}\& \,\,\,\,\,\,\,\,z = 2}}\]

f Use the information from steps (a) – (e) to sketch the graph of the function. Show Solution

Here is a sketch of the graph of this function using the information above. As we did in problems in this section we can start at the left and work our way to the right on the graph. As we do this we first pay attention to the increasing/decreasing information and then make sure that the curve has the correct concavity as we sketch it in.

$A sketch of the graph of $h\left( z \right) = {z^4} - 2{z^3} - 12{z^2}$. The graph comes down from the upper left to a valley at $\left( { - 1.8117, - 16.7209} \right)$, rises through the inflection point $\left( { - 1, - 9} \right)$ to a peak at the origin, then falls through the inflection point $\left( {2, - 48} \right)$ down to a valley at $\left( {3.3117, - 83.9666} \right)$ before rising steeply to the upper right. The labeled points are all marked on the graph.$

Note that because we used a computer to generate the sketch it is possible that your sketch won’t be quite the same. It should however, have the same points listed on the graph above, the same basic increasing/decreasing nature and the same basic concavity.