Question

The function f(x) = 2x^5 – 6x^4 + 2x^3 + 7x^2 - 6x has relative extrema at which of the following values of x?

A) -1.112
B) -0.747
C) 0.483
D) 1
E) 1.664
F) 2

Answer 1

To find the relative extrema of the function f(x) = 2x⁵ – 6x⁴ + 2x³ + 7x² - 6x, take the derivative of the function, set it equal to zero, and solve for x to find the critical points. The critical points are approximately x = -0.747, x = 0.483, and x = 2. So, the correct answer is F) 2.

Step-by-step explanation:

To find the relative extrema of the function f(x) = 2x⁵ – 6x⁴ + 2x³ + 7x² - 6x, we need to take the derivative of the function and find its critical points.

First, find the derivative of f(x): f'(x) = 10x⁴ - 24x³ + 6x² + 14x - 6.

Next, set f'(x) = 0 and solve for x to find the critical points: 10x⁴ - 24x³ + 6x² + 14x - 6 = 0.

Using a graphing calculator or factoring, we can find that the critical points are approximately x = -0.747, x = 0.483, and x = 2.

Therefore, the relative extrema of the function are found at x = -0.747, x = 0.483, and x = 2.

Thus, the correct answer is F) 2.