Paul's Online Notes
Home / Calculus I / Applications of Derivatives / The Shape of a Graph, Part II
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4-6 : The Shape of a Graph, Part II

16. Suppose that we know that $$f\left( x \right)$$ is a polynomial with critical points $$x = - 1$$, $$x = 2$$ and $$x = 6$$. If we also know that the 2nd derivative is $$f''\left( x \right) = - 3{x^2} + 14x - 4$$. If possible, classify each of the critical points as relative minimums, relative maximums. If it is not possible to classify the critical points clearly explain why they cannot be classified.

Hint : We do NOT need the 1st derivative to answer this question. We are in the 2nd derivative section and we did see a way in the notes on how to use the 2nd derivative (which we have nicely been given…) to classify most critical points.
Show Solution

This problem is not as difficult as many students originally make it out to be. We’ve been given the 2nd derivative and we saw how the 2nd derivative test can be used to classify most critical points so let’s use that.

First, we should note that because we have been told that $$f\left( x \right)$$ is a polynomial it should be fairly clear that, regardless of what the 1st derivative actually is, we should have,

$f'\left( { - 1} \right) = 0\hspace{0.5in}f'\left( 2 \right) = 0\hspace{0.5in}f'\left( 6 \right) = 0$

What this means is that we can use the 2nd derivative test as it only works for these kinds of critical points.

All we need to do then is plug the critical points into the 2nd derivative and use the 2nd derivative test to classify the critical points.

\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}f''\left( { - 1} \right) & = - 21 < 0\hspace{0.5in}{\mbox{Relative Maximum}}\\ f''\left( 2 \right) & = 12 > 0\hspace{0.5in}\hspace{0.25in}{\mbox{Relative Minimum}}\\ f''\left( 6 \right) & = - 28 < 0\hspace{0.5in}{\mbox{Relative Maximum}}\end{align*}}

So, in this case it was possible to classify all of the given critical points. Recall that if the 2nd derivative had been zero for any of them we would not have been able to classify that critical point without the 1st derivative which we don’t have for this case.