How do you find the critical numbers of #x^4-2x^3-3x^2-5#?

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Answer 1 · 2018-06-04T05:58:12

Please see the explanation below

Let

#f(x)=x^4-2x^3-3x^2-5#

The first derivative is

#f'(x)=4x^3-6x^2-6x#

#=2x(2x^2-3x-3)#

The critical points are when #f'(x)=0#

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

#=>#, #{(x=0),(2x^2-3x-3=0):}#

#=>#, #{(x=0),(x=(3+sqrt(33))/(4)=2.186),(x=(3-sqrt(33))/4=-0.686):}#

The critical points are #(0,-5)# , #(2.186, -17.393)#, and #(-0.686,-5.545)#

graph{x^4-2x^3-3x^2-5 [-24.14, 27.18, -19.11, 6.56]}