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 [MATH]c>\frac{9}{32}, c=\frac{9}{32}[/MATH], and [MATH]c=0[/MATH]. I also built a table of my results and found that I could obtain 3 extrema for [MATH]0<c<\frac{9}{32}[/MATH]. So, I was four-fifths of the way there, but I could not figure out how to reason through the case where [MATH]c<0[/MATH]. 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
[MATH]c = \dfrac{9}{32} \implies f'\left ( - \dfrac{3}{8} \right ) = 0 = f'(0).[/MATH]
But there cannot be exactly two distinct extrema for a polynomial of degree 4. Why not?
For a polynomial of degree 4, [MATH]f(x)\rightarrow\infty[/MATH] as [MATH]x\rightarrow\pm\infty[/MATH] and so there must be an odd number of extrema. If there were an even number, then either [MATH]f(x)\rightarrow\infty[/MATH] as [MATH]x\rightarrow\infty[/MATH] and [MATH]f(x)\rightarrow-\infty[/MATH] as [MATH]x\rightarrow-\infty[/MATH] OR [MATH]f(x)\rightarrow-\infty[/MATH] as [MATH]x\rightarrow\infty[/MATH] and [MATH]f(x)\rightarrow\infty[/MATH] as [MATH]x\rightarrow-\infty[/MATH], which would mean that the degree of the polynomial is odd.
Case 3a
[MATH]-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.[/MATH]Do you see why?
Case 3c
[MATH]0 < c < \dfrac{9}{32}.[/MATH]
Can you do it?
I approached your final 2 questions in the same way. For the case where [MATH]c<\frac{9}{32}[/MATH], I see that [MATH]c=0[/MATH] is a "special case" of this inequality in that it is the only value of [MATH]c<\frac{9}{32}[/MATH] for which f(x) has only one extremum, and therefore it breaks the inequality [MATH]c<\frac{9}{32}[/MATH] into 2 parts: [MATH]0<c<\frac{9}{32}[/MATH] or [MATH]c<0[/MATH]. Starting with the first inequality, since [MATH]0<c<\frac{9}{32}[/MATH]:
[MATH]b^2>b^2-4ac>b^2-\frac{9a}{8}[/MATH]. Since
a = 4 and
b = 3, [MATH]3>\sqrt{9-16c}>\frac{3}{\sqrt{2}}[/MATH], then [MATH]0<\frac{3-\sqrt{9-16c}}{8}<\frac{6-3\sqrt{2}}{16}[/MATH] and thus [MATH]0<x_1<\frac{6-3\sqrt{2}}{16}[/MATH] or [MATH]\frac{3}{4}>\frac{3+\sqrt{9-16c}}{8}>\frac{6+3\sqrt{2}}{16}[/MATH] and thus [MATH]\frac{3}{4}>x_2>\frac{6+3\sqrt{2}}{16}[/MATH].
So, there are 3 extrema when [MATH]0<c<\frac{9}{32}[/MATH]: [MATH]x_1, x_2[/MATH], and
x = 0 - two local minima and one local maximum.
When [MATH]c<0[/MATH], [MATH]b^2-4ac>b^2[/MATH]. Since
a = 4 and
b = 3, then [MATH]\sqrt{9-16c}>3[/MATH]. Thus [MATH]\frac{3-\sqrt{9-16c}}{8}<0[/MATH], or [MATH]\frac{3+\sqrt{9-16c}}{8}<\frac{3}{4}[/MATH]. Therefore [MATH]x_3<0[/MATH] and [MATH]x_4<\frac{3}{4}[/MATH].
Again, there are 3 extrema when [MATH]c<0[/MATH]: [MATH]x_3, x_4[/MATH], and
x = 0 - two local minima and one local maximum.
So, with the exception of
c = 0, there are always 3 extrema when [MATH]c<\frac{9}{32}[/MATH], and this covers the cases when [MATH]c<0[/MATH].
If my reasoning is incorrect, please let me know. Thanks again!