Find the Absolute Max and Min over the Interval f(x)=x^2-5 , [0,3]

Problem

ƒ(x)=x2−5,[0,3]

Solution

1. Find the derivative of the function to locate any critical points within the interval.

(d(x2)−5)/d(x)=2*x

2. Identify critical points by setting the derivative equal to zero and solving for x

2*x=0

x=0

3. Evaluate the function at the critical point x=0 and the endpoints of the interval x=0 and x=3

ƒ(0)=(0)2−5=−5

ƒ(3)=(3)2−5=4

4. Compare the values to determine the absolute maximum and minimum. The smallest value is the absolute minimum and the largest value is the absolute maximum.

−5<4

Final Answer

Absolute Max: *4, Absolute Min: −5

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