Find the Absolute Max and Min over the Interval f(x)=x^4-4x^2 , [-3,4]

Problem

ƒ(x)=x4−4*x2,[−3,4]

Solution

1. Find the derivative of the function to locate critical points.

(d(x4)−4*x2)/d(x)=4*x3−8*x

2. Set the derivative to zero and solve for x to find the critical points within the interval [−3,4]

4*x3−8*x=0

4*x*(x2−2)=0

x=0,x=√(,2),x=−√(,2)

3. Evaluate the function at the critical points.

ƒ(0)=(0)4−4*(0)2=0

ƒ(√(,2))=(√(,2))4−4*(√(,2))2=4−8=−4

ƒ*(−√(,2))=(−√(,2))4−4*(−√(,2))2=4−8=−4

4. Evaluate the function at the boundaries of the interval [−3,4]

ƒ*(−3)=(−3)4−4*(−3)2=81−36=45

ƒ(4)=(4)4−4*(4)2=256−64=192

5. Compare all values to determine the absolute maximum and minimum.
   The values are 0, -4, -4, 45, 192$.

Final Answer

Absolute Max: *192* at *x=4, Absolute Min: −4* at *x=±√(,2)

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