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Find the relative maximum relative minimum and zeros of fx-x3 16x2-76x 96?
Wiki User
∙ 14y ago
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.
Wiki User
∙ 14y ago
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