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Show that the number of solutions of the equation $f(x) = 4x^5+x^3-2x-m$ has at least one solution and at most 3 solutions for any real values of $m$.

What I have done so far:

$f(x)=4x^5+x^3-2x-m\\f'(x)=20x^4+3x^2-2=(4x^2-1)(5x^2+2)\\f(m)=4m^5+m^3-3m=m(4m^4+m^2-3)=m(4m^2-3)(m^2+1)$

What I do next is to find test certain values of $m$

Case 1:

$-\frac 34<m\le0\\ f(-\frac 12)=\frac 34-m>0\\ f(-2)=-132-m<0\\ f(\frac 12)=-\frac34-m<0\\ f(2)=132-m>0$

Hence, by the Intermediate Value Theorem, I can conclude that there are at least 3 roots in Case 1. I will use Rolle's Theorem to prove that there are exactly 3 roots when $-\frac34<m\le0$

Case 2: $0<m<\frac34$

I will use similar argument to show that there are exactly 3 roots.

Case 3: $m\le-\frac{\sqrt3}{2}$

$f(-\frac 12)=\frac 34-m<0\\ f(m)<0\\ f(\frac 12)=-\frac34-m<0$

And yes, this is one of the problems I have encountered. I can sort of argue that $x=-\frac12$ is the local maximum and $x=\frac12$ is the local minimum and hence, there are no roots. I can even use Rolle's Theorem to show that there cannot be any roots. However, what value of $m$ can I substitute to show that in this particular case, there is at least one root by the Intermediate Value Theorem. (After doing this, I can use Rolle's Theorem to show that there is only 1 root). I will encounter the same problem for $m\ge\frac{\sqrt3}{2}$

But the greatest problem I have now is what values of $x$ do I substitute into $f(x)$ for $-\frac{\sqrt3}{2}<m\le-\frac34$ and $\frac34\le m<\frac{\sqrt3}{2}$ to show that there is at least 1 solution and at most 3 solutions?

Any help or any other solutions are appreciated! Thank you!

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  • $\begingroup$ At least one (real) solution is clear because the degree is odd. $\endgroup$ – Peter Mar 8 '17 at 15:47
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    $\begingroup$ Descartes's sign rule allows to prove that there are at most $4$ real solutions if $m\ne 0$. But exactly $4$ real solutions is impossible because the number of non-real solutions is always even. The case $m=0$ can be solved with the sign rule as well. $\endgroup$ – Peter Mar 8 '17 at 15:49
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Accidentally deleted comment. Leaving as answer:

For any $m$ you've found that there are exactly $2$ real solutions to $f'=0$, namely $x=\pm \frac{1}{2}$. Rolle's theorem says that between any two solutions to $f=0$, there is a solution to $f'=0$, so that shows there are at most $3$ solutions.

There is at least one because the polynomial is odd degree. More generally, you can see that $\lim_{x\to -\infty}f(x)=-\infty$ and $\lim_{x\to \infty}f(x)=\infty$ so by the intermediate value theorem there has to be at least one solution.

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  • $\begingroup$ Wow that's a new way to look at it!! Thanks!! Previously I've only learnt that "If f(x) = 0, then there exists a c such that f'(c) = 0" $\endgroup$ – Icycarus Mar 8 '17 at 16:01
  • $\begingroup$ I think this is the most intuitive way to interpret Rolle's theorem: If I start at $y$ and return to $y$ then at some point I turned around. $\endgroup$ – Callus Mar 8 '17 at 17:18

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