Let m be the absolute minimum value of the function f(x) = ( x¹² e¹⁹ )^ 1 x on the closed interval [e^2, e^4] . The value of e^4 m , where denotes the greatest integer function, is equal to

Step-by-step solution

Let g(x) = f(x) = 1 x (12 x - 19) = 12 x - 19 x . Differentiating with respect to x : g'(x) = x ( 12 x ) - (12 x - 19)(1) x^2 = 12 - 12 x + 19 x^2 = 31 - 12 x x^2 For critical points, g'(x) = 0 : 31 - 12 x = 0 x = 31 12 x = e^ 31/12 Since 31 12 2.58 , the critical point x = e^ 31/12 lies in the interval [e^2, e^4] . For x 0 and for x > e^ 31/12 , g'(x) The absolute minimum of f(x) (and hence g(x) ) on the closed interval must occur at one of the endpoints. Evaluating g(x) at the endpoints: At x = e^2 : g(e^2) = 12(