Question

the maximum and minimum value of the function Y is equals to x cube minus 3 X square + 6 are​

Answer 1

Answer:

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Explanation:

y=x^3 - 3x^2 + 6

Differentiating with respect to x...

dy/dx=3x^2 - 6x

Equating to zero...

3x^2 - 6x = 0

x=2, x=0....

Again differentiating with respect to x...

d^2y/dx^2=6x - 6....

at x =2...

d^2y/dx^2=6x - 6....=6(2) - 6=6....that is positive... That is minimum...

y=x^3 - 3x^2 + 6=(2) ^3 - 3(2) ^2 + 6=2.....

at x =0...

d^2y/dx^2=6x - 6....=6(0) - 6=-6....that is negative... That is maximum...

y=x^3 - 3x^2 + 6=(0) ^3 - 3(0) ^2 + 6=6.....

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