![\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](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D3x%5E4-8x%5E3-24x%5E2%2B96x%2B7%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D12x%5E3-24x%5E2-48x%2B96%5Cimplies%200%3D12x%5E3-24x%5E2-48x%2B96%0A%5C%5C%5C%5C%5C%5C%0A0%3Dx%5E3-2x%5E2-4x%2B8)
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,
![\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)](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Br%7Crrrrrrr%7D%0A2%26%261%26-2%26-4%268%5C%5C%0A%26%26%262%260%26-8%5C%5C%0A%26%26--%26--%26--%26--%5C%5C%0A%26%261%260%26-4%260%26%5Cleftarrow%20remainder%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ax%5E2-0x-4%5Cimplies%20x%5E2-4%5Cimplies%20x%5E2-2%5E2%5Cimplies%20%28x-2%29%28x%2B2%29)
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.