Find the Local Maxima and Minima f(x)=x^4-32x^2+256

Problem

ƒ(x)=x4−32*x2+256

Solution

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

d(ƒ(x))/d(x)=4*x3−64*x

2. Set the derivative to zero and solve for x to find the critical values.

4*x3−64*x=0

4*x*(x2−16)=0

4*x*(x−4)*(x+4)=0

x=−4,0,4

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

d2(ƒ(x))/(d(x)2)=12*x2−64

4. Evaluate the second derivative at each critical point to determine the nature of the extrema.

ƒ(0)″=12*(0)2−64=−64

ƒ″*(−4)=12*(−4)2−64=128

ƒ(4)″=12*(4)2−64=128

5. Interpret the results of the Second Derivative Test.
   Since ƒ(0)″<0 there is a local maximum at x=0
   Since ƒ″*(−4)>0 and ƒ(4)″>0 there are local minima at x=−4 and x=4

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

ƒ(0)=0−32*(0)2+256=256

ƒ*(−4)=(−4)4−32*(−4)2+256=256−512+256=0

ƒ(4)=4−32*(4)2+256=256−512+256=0

Final Answer

Local Maxima: *(0,256), Local Minima: *(−4,0),(4,0)

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