Section 3.2 : Interpretation of the Derivative
7. Answer the following questions about the function \(W\left( z \right) = 4{z^2} - 9z\).
- Is the function increasing or decreasing at \(z = - 1\)?
- Is the function increasing or decreasing at \(z = 2\)?
- Does the function ever stop changing? If yes, at what value(s) of \(z\) does the function stop changing?
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a Is the function increasing or decreasing at \(z = - 1\)? Show SolutionWe know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 5 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding.
So, from our previous work we know that the derivative is,
\[W'\left( z \right) = 8z - 9\]Now all that we need to do is to compute : \(W'\left( { - 1} \right) = - 17\). This is negative and so we know that the function must be decreasing at \(z = - 1\).
b Is the function increasing or decreasing at \(z = 2\)? Show Solution
Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.
The evaluation is : \(W'\left( 2 \right) = 7\). This is positive and so we know that the function must be increasing at \(z = 2\).
c Does the function ever stop changing? If yes, at what value(s) of \(z\) does the function stop changing? Show Solution
Here all that we’re really asking is if the derivative is ever zero. So we need to solve,
\[W'\left( z \right) = 0\hspace{0.25in} \to \hspace{0.25in}\,\,8z - 9 = 0\hspace{0.25in}\,\,\, \Rightarrow \,\hspace{0.25in}\,\,\,z = \frac{9}{8}\]So, the function will stop changing at \(z = \frac{9}{8}\).