Find the Local Maxima and Minima f(x)=x^3-48x

Problem

ƒ(x)=x3−48*x

Solution

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

(d(x3)−48*x)/d(x)=3*x2−48

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

3*x2−48=0

3. Solve for x by isolating the variable.

3*x2=48

x2=16

x=±4

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

(d(3)*x2−48)/d(x)=6*x

5. Evaluate the second derivative at the critical points.

ƒ(4)″=6*(4)=24

ƒ″*(−4)=6*(−4)=−24

6. Determine the nature of the points based on the sign of the second derivative. Since ƒ(4)″>0 there is a local minimum at x=4 Since ƒ″*(−4)<0 there is a local maximum at x=−4

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

ƒ(4)=(4)3−48*(4)=64−192=−128

ƒ*(−4)=(−4)3−48*(−4)=−64+192=128

Final Answer

Local Maxima: *(−4,128), Local Minima: *(4,−128)

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