Question

Find the absolute extrema, if they exist, as well as all values of x where they occur for the function:

\[ f(x) = (x² - 36)^(1/7) on the domain [-6, 7] ]

Answer 1

The absolute minimum of the function f(x) = (x² - 36)⁻¹/⁷ on the interval [-6, 7] is 0, occurring at x = -6, and the absolute maximum is ∙∙Root(1), occurring at x = 7.

Step-by-step explanation:

To find the absolute extrema of the function f(x) = (x² - 36)⁻¹/⁷ on the domain [-6, 7], we first need to find the critical points of the function within the given interval and then evaluate the function at the endpoints of the interval. Critical points occur where the derivative is zero or undefined, and for this function, we can find the derivative using the chain rule. However, the given function does not have a derivative at x = -6 and x = 6 because the function is not differentiable at places where the inside of the seventh root becomes zero (i.e., at x = ±6). Hence, the critical points are only the endpoints of the interval in this context.

Next, we evaluate f(x) at the endpoints of the interval: f(-6) and f(7). We find that f(-6) = 0 and f(7) = ∙∙Root(1). Since the function is continuous on the closed interval, the absolute minimum occurs at x = -6 and the absolute maximum occurs at x = 7.

Answer 2

The absolute minimum value of the function (f(x) = (x² - 36
`)^(1/7)`) on the domain ([-6, 7]) is (f(-6) = 0), occurring at (x = -6). The absolute maximum value of the function is (f(7) = 1), occurring at (x = 7).

Explanation:

The function (f(x) = (x² - 36
`)^(1/7)`) represents the seventh root of (x² - 36), which is essentially the seventh root of ((x - 6)(x + 6)). To find extrema on the given domain ([-6, 7]), we first evaluate the function at the endpoints and critical points within the domain.

By substituting (x = -6) into the function, we get f(-6) = (36 - 36
`)^(1/7)` = 0), establishing the absolute minimum at (x = -6). Evaluating x=7 yields (f(7) = (49 - 36
`)^(1/7)` = 1), indicating the absolute maximum at (x = 7).

To check for critical points within the domain, we compute the derivative of f(x) to find any potential points where extrema might occur. However, since the function is simple and lacks critical points beyond the endpoints, we evaluate the function at both endpoints to confirm the absolute minimum and maximum values.

In summary, the function f(x) reaches its absolute minimum of 0 at x = -6 and its absolute maximum of 1 at x = 7 on the given domain [-6, 7]. This is due to the nature of the function and the values of x within the specified range, making these the only extrema for this function in this domain.