Let f(x) be a polynomial of degree 5 , and have extrema at x = 1 and x = -1 . If _ x 0 ( f(x) x^3 ) = -5 , then f(2) - f(-2) is equal to:

Step-by-step solution

Let the polynomial of degree 5 be f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + k . Given _ x 0 ( f(x) x^3 ) = -5 , the terms of degree less than 3 must be zero, and the coefficient of x^3 must be -5 . Thus, k = 0 , e = 0 , d = 0 , and c = -5 . The polynomial becomes f(x) = ax^5 + bx^4 - 5x^3 . Differentiating with respect to x , we get: f'(x) = 5ax^4 + 4bx^3 - 15x^2 Since f(x) has extrema at x = 1 and x = -1 , we have f'(1) = 0 and f'(-1) = 0 . f'(1) = 5a + 4b - 15 = 0 f'(-1) = 5a - 4b - 15 = 0 Adding both equations, we