Final answer:
To find the absolute minimum value of f(x) = x³ – 7x + 9 on the interval [0, 3], evaluate the function at critical points and endpoints. There's only one critical point within the interval and the endpoints are x = 0 and x = 3. The absolute minimum value is 9 at x = 0.
Step-by-step explanation:
The question asks for the absolute maximum and absolute minimum values of the function f(x) = x³ – 7x + 9 on the interval [0, 3]. To find these values, we need to evaluate the function at the critical points and the endpoints of the interval.
First, we find the derivative of the function, f'(x) = 3x² - 7, to determine the critical points where f'(x) = 0 or where the derivative does not exist. Once the critical points are found, we will evaluate f(x) at these points and at the endpoints of the interval x = 0 and x = 3.
Solving f'(x) = 3x² - 7 = 0 gives us the critical points x = \(±\sqrt{\frac{7}{3}}\). However, since \(±\sqrt{\frac{7}{3}}\) is not in the interval [0, 3], we can ignore the negative value. Evaluating f(x) at the positive critical point and at the endpoints:
- f(0) = 0³ - 7(0) + 9 = 9
- f(\(\sqrt{\frac{7}{3}}\)) = \(\left(\sqrt{\frac{7}{3}}\right)\) ³ - 7(\(\sqrt{\frac{7}{3}}\)) + 9
- f(3) = 3³ - 7(3) + 9 = 27 - 21 + 9 = 15
After computing the values, the smallest value will be the absolute minimum. In this scenario, the absolute minimum value is 9 when x = 0.