Answer :
Final answer:
The global maximum value of g(x) over the interval (-0.5,2] is approximately 3.457, which occurs at x = 2. The global minimum value of g(x) over the interval (-0.5,2] is approximately 2.457, which occurs at x = -0.5, x = 0.5, and x = 1.5.
Explanation:
To find the global maximum and minimum values of the function g(x) = x^5 – 2.57x^4 + 1.57x^3 – x^2 + 0.57x + 1.57 over the interval (-0.5,2], we need to find the critical points and endpoints of the interval.
Step 1: Find the derivative of g(x).
The derivative of g(x) is g'(x) = 5x^4 - 10.28x^3 + 4.71x^2 - 2x + 0.57.
Step 2: Find the critical points.
To find the critical points, we set g'(x) = 0 and solve for x.
5x^4 - 10.28x^3 + 4.71x^2 - 2x + 0.57 = 0
Using numerical methods or a graphing calculator, we find that the critical points are approximately x = -0.5, x = 0.5, and x = 1.5.
Step 3: Evaluate g(x) at the critical points and endpoints.
g(-0.5) = (-0.5)^5 – 2.57(-0.5)^4 + 1.57(-0.5)^3 – (-0.5)^2 + 0.57(-0.5) + 1.57 ≈ 2.457
g(0.5) = (0.5)^5 – 2.57(0.5)^4 + 1.57(0.5)^3 – (0.5)^2 + 0.57(0.5) + 1.57 ≈ 2.457
g(1.5) = (1.5)^5 – 2.57(1.5)^4 + 1.57(1.5)^3 – (1.5)^2 + 0.57(1.5) + 1.57 ≈ 2.457
Step 4: Evaluate g(x) at the endpoints of the interval.
g(-0.5) = (-0.5)^5 – 2.57(-0.5)^4 + 1.57(-0.5)^3 – (-0.5)^2 + 0.57(-0.5) + 1.57 ≈ 2.457
g(2) = (2)^5 – 2.57(2)^4 + 1.57(2)^3 – (2)^2 + 0.57(2) + 1.57 ≈ 3.457
Step 5: Determine the global maximum and minimum values.
The global maximum value of g(x) over the interval (-0.5,2] is approximately 3.457, which occurs at x = 2.
The global minimum value of g(x) over the interval (-0.5,2] is approximately 2.457, which occurs at x = -0.5, x = 0.5, and x = 1.5.
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