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y=x^3+16x^2-76x+96

Get the first derivative:

y'=3x^2+32x-76

Factor:

(3x+38)(x-2)

Get the value of x:

x=-38/3 ; x=2

Get the second derivative:

y''=6x+32

Substitute value of x:

x=-38/3

=6(-38/3)+32

=76+32

=108 > 0 Min

x=2

=6(2)+32

=12+32

=44 > 0 Min

Answer:

y=x^3+16x^2-76x+96 does not have relative extremum because y'' are both greater than 0.

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Q: Find the relative maximum relative minimum and zeros of fx-x3 16x2-76x 96?
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