Question

Let g be the function defined by g(x)=(x²-x 1)eˣ. What is the absolute maximum value of g on the closed interval (-4,1)?

Answer 1

To find the absolute maximum value of the function g(x) = (x²-x+1)eˣ on the closed interval (-4,1), evaluate the function at the critical points and endpoints of the interval. The absolute maximum value is approximately 2.29.

Step-by-step explanation:

To find the absolute maximum value of the function g(x) = (x²-x+1)eˣ on the closed interval (-4,1), we need to evaluate the function at the critical points as well as the endpoints of the interval.

1. Find the critical points by taking the derivative of g(x) and setting it equal to 0:

g'(x) = (2x-1)(x²-x+1)eˣ

Solving for x:

2x - 1 = 0

x = 0.5

1. Evaluate g(x) at the critical point and the endpoints of the interval:

g(-4) = (-4²-(-4)+1)e^(-4) ≈ 0.07

g(0.5) = (0.5²-(0.5)+1)e^(0.5) ≈ 2.29

g(1) = (1²-1+1)e^1 = e

1. Compare the values obtained to find the absolute maximum:

The absolute maximum value of g(x) on the closed interval (-4,1) is approximately 2.29 and occurs at x = 0.5.