Find the Absolute Max and Min over the Interval f(x)=5x^2-3x-1 , [-1,3]

Problem

ƒ(x)=5*x2−3*x−1,[−1,3]

Solution

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

d(ƒ(x))/d(x)=10*x−3

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

10*x−3=0

10*x=3

x=0.3

3. Verify the critical point is within the given interval [−1,3]

−1≤0.3≤3

4. Evaluate the function at the left endpoint x=−1

ƒ*(−1)=5*(−1)2−3*(−1)−1

ƒ*(−1)=5+3−1

ƒ*(−1)=7

5. Evaluate the function at the critical point x=0.3

ƒ(0.3)=5*(0.3)2−3*(0.3)−1

ƒ(0.3)=5*(0.09)−0.9−1

ƒ(0.3)=0.45−0.9−1

ƒ(0.3)=−1.45

6. Evaluate the function at the right endpoint x=3

ƒ(3)=5*(3)2−3*(3)−1

ƒ(3)=5*(9)−9−1

ƒ(3)=45−9−1

ƒ(3)=35

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

Values:7,−1.45,35

Final Answer

Absolute Max: *35* at *x=3, Absolute Min: −1.45* at *x=0.3

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