Final answer:
To find the absolute extrema of the function y = 5x² - 40 ln x on the interval [1,4], we find the derivative, set it equal to zero to find critical points, and evaluate the function at these points and the endpoints. The absolute maximum occurs at x = 4 and the minimum at x = 1.
Step-by-step explanation:
To locate the absolute extrema of the function y = 5x² - 40 ln x on the closed interval [1,4], we need to take the following steps:
- Find the derivative of the function to locate critical points.
- Check the function values at critical points and endpoints of the interval.
- Determine which of these values are the maximum and minimum.
The derivative of the function is y' = 10x - 40/x. Setting the derivative equal to zero gives the critical points: 10x - 40/x = 0, which simplifies to x² - 4 = 0. This gives us two critical points at x = 2 and x = -2. However, since x = -2 is not in our interval, we only consider x = 2.
Now we evaluate the function at the critical point and the endpoints of the interval:
- f(1) = 5(1)² - 40 ln(1) = 5
- f(2) = 5(2)² - 40 ln(2) = 20 - 40 ln(2)
- f(4) = 5(4)² - 40 ln(4) = 80 - 40 ln(4)
By comparing these values, we can find the absolute maximum and minimum on the interval. In this case, the maximum value occurs at x = 4 and the minimum value occurs at x = 1.
Therefore, the absolute maximum is (4, 80 - 40 ln(4)) and the absolute minimum is (1, 5).