Find the Local Maxima and Minima x^3-12x

Problem

x3−12*x

Solution

1. Find the first derivative of the function ƒ(x)=x3−12*x to determine the slope of the tangent line.

(d(x3)−12*x)/d(x)=3*x2−12

2. Identify critical points by setting the first derivative equal to zero and solving for x

3*x2−12=0

3*x2=12

x2=4

x=±2

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

(d(3)*x2−12)/d(x)=6*x

4. 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.

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 up, indicating a local minimum.

6. Calculate the y-coordinates by substituting the x values back into the original function ƒ(x)=x3−12*x

ƒ*(−2)=(−2)3−12*(−2)=−8+24=16

ƒ(2)=(2)3−12*(2)=8−24=−16

Final Answer

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

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