Find the Absolute Max and Min over the Interval f(x)=0.8x^3-4x^2-1 , [-30,4]

Problem

ƒ(x)=0.8*x3−4*x2−1,[−30,4]

Solution

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

d(ƒ(x))/d(x)=2.4*x2−8*x

2. Set the derivative to zero and solve for x to find the critical points within the interval.

2.4*x2−8*x=0

x*(2.4*x−8)=0

x=0

x=8/2.4=10/3≈3.33

3. Evaluate the function at the critical points x=0 and x=10/3 as both fall within the interval [−30,4]

ƒ(0)=0.8*(0)3−4*(0)2−1=−1

ƒ(10/3)=0.8*(10/3)3−4*(10/3)2−1=800/27−400/9−1=−427/27≈−15.81

4. Evaluate the function at the endpoints of the interval x=−30 and x=4

ƒ*(−30)=0.8*(−30)3−4*(−30)2−1=−21600−3600−1=−25201

ƒ(4)=0.8*(4)3−4*(4)2−1=51.2−64−1=−13.8

5. Compare all values to determine the absolute maximum and minimum.

−25201<−15.81<−13.8<−1

Final Answer

Absolute Max: *(0,−1), Absolute Min: *(−30,−25201)

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