Final answer:
The absolute maximum of the function f(x) = (x² - 25)⁰/₅ on the domain [-6,5] is approximately 1.615, and it occurs at x = -6. There is no absolute minimum within this domain.
Step-by-step explanation:
To find the absolute extrema of the function f(x) = (x² - 25)⁰/₅ on the domain [-6,5], we need to investigate both the endpoints of the domain and the critical points within the domain where the derivative is equal to zero or undefined. First, we evaluate the function at the endpoints:
- f(-6) = ((-6)² - 25)⁰/₅ = (36 - 25)⁰/₅ = (11)⁰/₅
- f(5) = (5² - 25)⁰/₅ = (25 - 25)⁰/₅ = 0⁰/₅
Next, we find the derivative of the function to determine the critical points:
f'(x) = ⅖/5 * (x² - 25)⁻⁴/₅ * 2x
The derivative is undefined at x = ±5, and equals zero when x² - 25 = 0, which yields x = ±5. Since the domain is restricted to [-6,5], we only consider x = -5 for critical points within the domain. Upon testing, f(-5) equals to 0, which is indeed a critical point but it does not contribute to an extremum since it's the same value as one of the endpoints. We now compare the function values:
- f(-6) = (11)⁰/₅ ≈ 1.615
- f(5) = 0
- f(-5) = 0
The absolute maximum value appears to be approximately 1.615, which occurs at x = -6. There is no absolute minimum within the domain because the function increases as x approaches -5 from the left and decreases as x approaches 5 from the right, with the function being zero at both -5 and 5.