Find the Local Maxima and Minima f(x)=x^3-6x^2

Problem

ƒ(x)=x3−6*x2

Solution

1. Find the first derivative of the function to determine the critical points.

d(ƒ(x))/d(x)=3*x2−12*x

2. Set the derivative to zero and solve for x to find the critical values.

3*x2−12*x=0

3*x*(x−4)=0

x=0,x=4

3. Find the second derivative to apply the Second Derivative Test for concavity.

d2(ƒ(x))/(d(x)2)=6*x−12

4. Evaluate the second derivative at x=0

ƒ(0)″=6*(0)−12=−12

Since ƒ(0)″<0 there is a local maximum at x=0

5. Evaluate the second derivative at x=4

ƒ(4)″=6*(4)−12=12

Since ƒ(4)″>0 there is a local minimum at x=4

6. Calculate the y-coordinates by substituting the critical values back into the original function ƒ(x)

ƒ(0)=(0)3−6*(0)2=0

ƒ(4)=(4)3−6*(4)2=64−96=−32

Final Answer

Local Maximum: *(0,0), Local Minimum: *(4,−32)

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