Find the Local Maxima and Minima f(x)=4x^3-12x

Problem

ƒ(x)=4*x3−12*x

Solution

1. Find the first derivative of the function to identify the slope of the tangent line.

(d(4)*x3−12*x)/d(x)=12*x2−12

2. Set the derivative to zero to find the critical points where the slope is horizontal.

12*x2−12=0

3. Solve for x by factoring or using algebraic manipulation.

12*(x2−1)=0

x2=1

x=1,x=−1

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

(d(12)*x2−12)/d(x)=24*x

5. Evaluate the second derivative at the critical points to determine if they are maxima or minima.

ƒ(1)″=24*(1)=24

ƒ″*(−1)=24*(−1)=−24

6. Interpret the results based on the sign of the second derivative. Since ƒ(1)″>0 there is a local minimum at x=1 Since ƒ″*(−1)<0 there is a local maximum at x=−1

7. Calculate the y-coordinates by plugging the critical points back into the original function ƒ(x)

ƒ(1)=4*(1)3−12*(1)=−8

ƒ*(−1)=4*(−1)3−12*(−1)=8

Final Answer

Local Maxima: *(−1,8), Local Minima: *(1,−8)

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