I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there. $$ f(x) = x^3+3x^2-mx+5 $$ I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?
Finding number of roots using Rolle's Theorem, and depending on parameter
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The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.
So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider $$ f'(x) = 3x^2 + 6x - m = 0.$$ As this is a quadratic, you can very quickly write down the two roots, $$ x = \frac{-6 \pm \sqrt{36 + 12m}}{6}.$$ When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f \to -\infty$ as $x \to -\infty$ and $f \to \infty$ as $x \to \infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.
More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.
You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is
$$ax^3+bx^2+cx+d=0$$
so in your case
$$a=1,\ b=3,\ c=-m,\ d=5$$
In general, the discriminant is
$$\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
You can substitute in your values and get a cubic expression in $m$. Then your equation has
To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.
Note that this did not need Rolle's theorem. Do you really need to use it?