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

Problem

ƒ(x)=x3−12*x

Solution

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

d(ƒ(x))/d(x)=3*x2−12

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

3*x2−12=0

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

3*(x2−4)=0

3*(x−2)*(x+2)=0

x=2,x=−2

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

d2(ƒ(x))/(d(x)2)=6*x

5. Test the critical point x=−2 by substituting it into the second derivative.

ƒ″*(−2)=6*(−2)=−12

Since ƒ″*(−2)<0 the function is concave down, indicating a local maximum.

6. Test the critical point x=2 by substituting it into the second derivative.

ƒ(2)″=6*(2)=12

Since ƒ(2)″>0 the function is concave up, indicating a local minimum.

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

ƒ*(−2)=(−2)3−12*(−2)=−8+24=16

ƒ(2)=(2)3−12*(2)=8−24=−16

Final Answer

Local Maximum at *(−2,16), Local Minimum at *(2,−16)

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