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

Problem

ƒ(x)=x3−3*x2−1,[−3,2]

Solution

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

d(ƒ(x))/d(x)=3*x2−6*x

2. Set the derivative to zero to find the critical values of x

3*x2−6*x=0

3*x*(x−2)=0

x=0,x=2

3. Identify critical points within the given interval [−3,2] Both x=0 and x=2 are within the interval.

4. Evaluate the function at the critical points and the endpoints of the interval.

ƒ*(−3)=(−3)3−3*(−3)2−1=−27−27−1=−55

ƒ(0)=(0)3−3*(0)2−1=−1

ƒ(2)=(2)3−3*(2)2−1=8−12−1=−5

5. Compare the values to determine the absolute maximum and minimum. The largest value is −1 and the smallest value is −55

Final Answer

Absolute Max: −1* at *x=0, Absolute Min: −55* at *x=−3

Want more problems? Check here!