Let f: R R be a polynomial function of degree four having local extrema at x = -1 and x = 2 . If _ x 0 f(x) x^2 = -12 , and M and m are the absolute maximum and absolute minimum values of f(x) on the closed interval [-2, 3] respectively, then the value of M - m is equal to :

Step-by-step solution

Let the polynomial be f(x) = ax^4 + bx^3 + cx^2 + dx + e . Given _ x 0 f(x) x^2 = -12 , we have: _ x 0 ax^4 + bx^3 + cx^2 + dx + e x^2 = -12 For the limit to exist and be finite, d = 0 and e = 0 . Then _ x 0 (ax^2 + bx + c) = c = -12 . So, f(x) = ax^4 + bx^3 - 12x^2 . Differentiating with respect to x : f^ (x) = 4ax^3 + 3bx^2 - 24x = x(4ax^2 + 3bx - 24) Since f(x) has local extrema at x = -1 and x = 2 , f^ (-1) = 0 and f^ (2) = 0 . For x = -1 : 4a(-1)^2 + 3b(-1) - 24 = 0 4a - 3b = 24 For x = 2 : 4a(2)^2 + 3b(2) - 2