Final answer:
Absolute extreme values for the functions on their respective intervals were found. For f(x) = x^2 - 10 on [-2,3], the absolute maximum value is -1 and the absolute minimum is -6. For f(x) = (x+1)^4/3 on [-9,7], the absolute maximum value is 16 and the absolute minimum is 0.
Step-by-step explanation:
The task is to find the absolute extreme values of two functions on given intervals.
For the first function, f(x) = x2 - 10 on the interval [-2,3], we need to check the endpoints and any critical points inside the interval. The value of the function at x = -2 is f(-2) = (-2)2 - 10 = 4 - 10 = -6. At x = 3, the value is f(3) = (3)2 - 10 = 9 - 10 = -1. There are no critical points inside the interval since the derivative, f'(x) = 2x, is never zero in this case. Therefore, the absolute maximum value is f(3) = -1 and the absolute minimum value is f(-2) = -6.
For the second function, f(x) = (x+1)4/3 on the interval [-9,7], we find the values at the endpoints and check for critical points. At x = -9, f(-9) = ((-9)+1)4/3 = 04/3 = 0. At x = 7, f(7) = ((7)+1)4/3 = 84/3 = 16. The derivative of the function f'(x) = (4/3)(x+1)1/3 is undefined at x = -1, which is within the interval. However, this point corresponds to a corner or cusp, not an extreme value. Therefore, the absolute maximum value is f(7) = 16 and the absolute minimum value is f(-9) = 0.