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https://math.answers.com/algebra/Find_the_relative_maximum_relative_minimum_and_zeros_of_fx-x3_16x2-76x_96

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

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

https://math.answers.com/algebra/Find_the_relative_maximum_relative_minimum_and_zeros_of_fx-x3_16x2-76x_96

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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https://math.answers.com/algebra/Find_the_relative_maximum_relative_minimum_and_zeros_of_fx-x3_16x2-76x_96

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

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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      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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