Question

Identify the critical numbers of the following functions and then use the first derivative test or second derivative test to determine which critical numbers are minimums, maximums or neither. Make certain to justify your answer.

f(x) = 3x^4 - 8x^3 - 24x^2 + 96x + 7

Answer 1

`\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`

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 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)`

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.