Find the Local Maxima and Minima f(x)=x^3-6x^2+1

Problem

ƒ(x)=x3−6*x2+1

Solution

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

(d(x3)−6*x2+1)/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.

(d(3)*x2−12*x)/d(x)=6*x−12

4. Evaluate the second derivative at the critical points.

ƒ(0)″=6*(0)−12=−12

ƒ(4)″=6*(4)−12=12

5. Determine the nature of the points. Since ƒ(0)″<0 there is a local maximum at x=0 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+1=1

ƒ(4)=(4)3−6*(4)2+1=64−96+1=−31

Final Answer

Local Maximum: *(0,1), Local Minimum: *(4,−31)

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