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#.