f(x) = 2x² - 6x + 10 on (-$\infty$, $\infty$) has A) only an absolute maximum B) only an absolute minimum C) both an absolute maximum and an absolute minimum D) neither an absolute maximum nor an absolute minimum. E) cannot be determined

Question

Quizplus · Accepted Answer

First, note that the domain of the function is the interval (-F1F1F1S1F1F1F10, F1F1F1S1F1F1F10), which is a finite interval. Therefore, we only need to consider the values of the function at the endpoints and the critical points in the interior of the interval. To find the critical points, we need to find the derivative of the function and set it equal to 0: f'(x) = 4x² F1F1F1S1F1F1F10 6 Setting this equal to 0 and solving for x, we get: x = 0 Therefore, the only critical point is x = 0. Next, we evaluate the function at the endpoints and the critical point: f(-F1F1F1S1F1F1F10) = 2(-F1F1F1S1F1F1F10)² F1F1F1S1F1F1F10 6(-F1F1F1S1F1F1F10) + 10 = a finite negative number f(F1F1F1S1F1F1F10) = 2(F1F1F1S1F1F1F10)² F1F1F1S1F1F1F10 6(F1F1F1S1F1F1F10) + 10 = a finite positive number f(0) = 2(0)² F1F1F1S1F1F1F10 6(0) + 10 = 10 Therefore, the function has an absolute maximum at x = 0 (where the function attains its only critical point) and an absolute minimum at one of the endpoints. Since the function is decreasing as x approaches -F1F1F1S1F1F1F10 and increasing as x approaches F1F1F1S1F1F1F10, the absolute minimum must occur at x = -F1F1F1S1F1F1F10. Thus, the function has both an absolute maximum and an absolute minimum on the interval (-F1F1F1S1F1F1F10, F1F1F1S1F1F1F10), and the correct answer is (C) both an absolute maximum and an absolute minimum.

F(x) = 2x2 - 6x + 10 on (- ∞\infty∞ ∞\infty∞

F(x) = 2x² - 6x + 10 on (- $\infty$ $\infty$