Find the Absolute Max and Min over the Interval f(x)=x^3-3x^2-1 , [-3,4]

Problem

ƒ(x)=x3−3*x2−1,[−3,4]

Solution

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

(d(x3)−3*x2−1)/d(x)=3*x2−6*x

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

3*x2−6*x=0

3*x*(x−2)=0

x=0,x=2

3. Verify critical points are within the given interval [−3,4] Both 0 and 2 are inside the interval.

4. Evaluate the function at the critical points.

ƒ(0)=(0)3−3*(0)2−1=−1

ƒ(2)=(2)3−3*(2)2−1=8−12−1=−5

5. Evaluate the function at the endpoints of the interval.

ƒ*(−3)=(−3)3−3*(−3)2−1=−27−27−1=−55

ƒ(4)=(4)3−3*(4)2−1=64−48−1=15

6. Compare the values to determine the absolute maximum and minimum.

−55<−5<−1<15

Final Answer

Absolute Max: *15* at *x=4, Absolute Min: −55* at *x=−3

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