[tex]\bf f(x)=3x^4-8x^3-24x^2+96x+7
\\\\\\
\cfrac{dy}{dx}=12x^3-24x^2-48x+96\implies 0=12x^3-24x^2-48x+96
\\\\\\
0=x^3-2x^2-4x+8[/tex]
now, if we use the "rational root test" on the cubic, we have a few roots to work with, testing for x = 2 on a quick synthetic division will look like,
[tex]\bf \begin{array}{r|rrrrrrr}
2&&1&-2&-4&8\\
&&&2&0&-8\\
&&--&--&--&--\\
&&1&0&-4&0&\leftarrow remainder
\end{array}
\\\\\\
x^2-0x-4\implies x^2-4\implies x^2-2^2\implies (x-2)(x+2)[/tex]
now, the root we used for the synthetic division is x = 2, therefore, x - 2 = 0, thus the factor is then (x-2).
so we end up with 0 = (x-2)(x+2)(x-2), which of course will gives roots of -2 and 2, now, x-2 is there twice, so it has a multiplicity of 2, so the graph of the derivative, doesn't cross the x-axis, it simply bounces off of it, so it doesn't change signs there then.
so our critical points are ±2.
doing a first-derivative test, check the picture below on the left-hand-side, before the -2 the derivative is negative, so the original function is going down, between -2 and 2, it goes up, clearly that means at -2 there a minimum point, and after 2 it keeps on going up, recall it had a multiplicity of 2 for the derivative.
so at 2 there's no extrema, check the picture on the right-hand-side, is just a quick flattening and right back up.
