How do you find critical points for f(x) = x^4 - 8x^3 - 16x +5?

Sep 2, 2015

Use whatever tools you have for finding irrational solution to the cubic equation: ${x}^{3} - 6 {x}^{2} - 4 = 0$

Explanation:

For $f \left(x\right) = {x}^{4} - 8 {x}^{3} - 16 x + 5$,

we have $f ' \left(x\right) = 4 {x}^{3} - 24 {x}^{2} - 16 = 4 \left({x}^{3} - 6 {x}^{2} - 4\right)$.

So the critical numbers are the solutions to

${x}^{3} - 6 {x}^{2} - 4 = 0$

Descartes' Rule of Signs assures us that there is a positive root. But none of $1 , 2 , 4$ are roots, so the positive root is irrational.

We can also use the rule of signs to see that there are no negative roots. (So the remaining two roots are imaginary.)

Perhaps you are expected to know the general solution to the cubic, or you are using graphing technology, or you know Newton's or some other method for successive approximations to the root.