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### Section 3-9 : Chain Rule

29. Determine where $$V\left( z \right) = {z^4}{\left( {2z - 8} \right)^3}$$ is increasing and decreasing.

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We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

\begin{align*}V'\left( z \right) & = 4{z^3}{\left( {2z - 8} \right)^3} + {z^4}\left( 3 \right){\left( {2z - 8} \right)^2}\left( 2 \right)\\ & = 2{z^3}{\left( {2z - 8} \right)^2}\left[ {2\left( {2z - 8} \right) + 3z} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{2{z^3}{{\left( {2z - 8} \right)}^2}\left( {7z - 16} \right)}}\end{align*}

Note that we factored the derivative to help with the next step. In general, we don’t need to do this.

Show Step 2

Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. In this case it’s pretty easy to spot where the derivative will be zero.

$2{z^3}{\left( {2z - 8} \right)^2}\left( {7z - 16} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}z = 0,\,\,\,\,z = 4,\,\,\,\,z = \frac{{16}}{7} = 2.2857$ Show Step 3

To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals.

Here is the number line for this problem. From this we get the following increasing/decreasing information.

\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{Increasing :}} & \,\, - \infty < z < 0,\,\,\,\,\frac{{16}}{7} < z < 4,\,\,\,\,4 < z < \infty \\ {\mbox{Decreasing :}} & \,\,0 < z < \frac{{16}}{7}\end{align*}}