How do you find the critical numbers for $x^4-2x^3-3x^2-5$ to determine the maximum and minimum?

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Answer 1 · 2016-04-20T14:15:28

Please see the explanation section below,

$f(x) = x^4-2x^3-3x^2_5$ is a polynomial, its domain is $(-oo,oo)$

$f'(x) = 4x^3-6x^2-6x$ is never undefined and is $0$ at the solutions to

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

Factor out the common factor of $2x$ to get

$2x(2x^2-3x-3)=0$

This leads to two lower degree polynomial equations:

$2x=0$ $" "$ or $" "$ $2x^2-3x-3=0$

The first has solution $0$ and the solutions to the second may be found using the quadratic formula

$x= (-b+-sqrt(b^2-4ac))/(2a)$ .

We get $x=(3+-sqrt33)/4$ .

$f'(x)=0$ at

$0$ , $" "$ $(3+sqrt33)/4$ , $" "$ and $" "$ $(3-sqrt33)/4$

All of these solutions are in the domain of $f$ , so they are all critical numbers for $f$ .

How do you find the critical numbers for x^4-2x^3-3x^2-5x^4-2x^3-3x^2-5 to determine the maximum and minimum?