Thank you once again to everyone for taking your time to help me! This has been extremely helpful. Jeff, your summary was immensely helpful, as it provided very clear reasoning. I also learned a bit about proper typesetting after reading your post, so thank you for that as well.

When I initially worked the problem, I did obtain cases for \(\displaystyle c>\frac{9}{32}, c=\frac{9}{32}\), and \(\displaystyle c=0\). I also built a table of my results and found that I could obtain 3 extrema for \(\displaystyle 0<c<\frac{9}{32}\). So, I was four-fifths of the way there, but I could not figure out how to reason through the case where \(\displaystyle c<0\). After working through your questions, it became much clearer to me how I might approach the problem. Let me address your questions to see if I'm on the right track.

Case 2

\(\displaystyle c = \dfrac{9}{32} \implies f'\left ( - \dfrac{3}{8} \right ) = 0 = f'(0).\)

But there cannot be exactly two distinct extrema for a polynomial of degree 4. Why not?

For a polynomial of degree 4, \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow\pm\infty\) and so there must be an odd number of extrema. If there were an even number, then either \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow\infty\) and \(\displaystyle f(x)\rightarrow-\infty\) as \(\displaystyle x\rightarrow-\infty\) OR \(\displaystyle f(x)\rightarrow-\infty\) as \(\displaystyle x\rightarrow\infty\) and \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow-\infty\), which would mean that the degree of the polynomial is odd.

Case 3a

\(\displaystyle -3 + \sqrt{9 - 32c} > 0 \implies 9 - 32c > 9 \implies c < 0 \implies \\

f''(0) < 0 \implies f(x) \text { has two local minima,}\\

\text {one on each side of } x = 0 \text { and a local maximum at } x = 0.\)

Do you see why?

Case 3c

\(\displaystyle 0 < c < \dfrac{9}{32}.\)

Can you do it?

I approached your final 2 questions in the same way. For the case where \(\displaystyle c<\frac{9}{32}\), I see that \(\displaystyle c=0\) is a "special case" of this inequality in that it is the only value of \(\displaystyle c<\frac{9}{32}\) for which f(x) has only one extremum, and therefore it breaks the inequality \(\displaystyle c<\frac{9}{32}\) into 2 parts: \(\displaystyle 0<c<\frac{9}{32}\) or \(\displaystyle c<0\). Starting with the first inequality, since \(\displaystyle 0<c<\frac{9}{32}\):

\(\displaystyle b^2>b^2-4ac>b^2-\frac{9a}{8}\). Since

*a* = 4 and

*b* = 3, \(\displaystyle 3>\sqrt{9-16c}>\frac{3}{\sqrt{2}}\), then \(\displaystyle 0<\frac{3-\sqrt{9-16c}}{8}<\frac{6-3\sqrt{2}}{16}\) and thus \(\displaystyle 0<x_1<\frac{6-3\sqrt{2}}{16}\) or \(\displaystyle \frac{3}{4}>\frac{3+\sqrt{9-16c}}{8}>\frac{6+3\sqrt{2}}{16}\) and thus \(\displaystyle \frac{3}{4}>x_2>\frac{6+3\sqrt{2}}{16}\).

So, there are 3 extrema when \(\displaystyle 0<c<\frac{9}{32}\): \(\displaystyle x_1, x_2\), and

*x* = 0 - two local minima and one local maximum.

When \(\displaystyle c<0\), \(\displaystyle b^2-4ac>b^2\). Since

*a* = 4 and

*b* = 3, then \(\displaystyle \sqrt{9-16c}>3\). Thus \(\displaystyle \frac{3-\sqrt{9-16c}}{8}<0\), or \(\displaystyle \frac{3+\sqrt{9-16c}}{8}<\frac{3}{4}\). Therefore \(\displaystyle x_3<0\) and \(\displaystyle x_4<\frac{3}{4}\).

Again, there are 3 extrema when \(\displaystyle c<0\): \(\displaystyle x_3, x_4\), and

*x* = 0 - two local minima and one local maximum.

So, with the exception of

*c *= 0, there are always 3 extrema when \(\displaystyle c<\frac{9}{32}\), and this covers the cases when \(\displaystyle c<0\).

If my reasoning is incorrect, please let me know. Thanks again!