найдите разность между максимальным и минимальным значениями функции y=-(7x)/(x^2+1)

$y=-(7x)/(x^2+1)\\\\ y'=(-(7x)/(x^2+1))' = -((7x)/(x^2+1))' =\\ -((7x)'(x^2+1)-(x^2+1)'(7x))/(x^2+1)^2 =\\ -(7x^2+7-14x^2)/(x^2+1)^2 =\\ -(-7x^2+7)/(x^2+1)^2 = 7(x^2-1)/(x^2+1)^2\\\\ 7(x^2-1)/(x^2+1)^2 = 0\\\\ x^2-1=0, x_1 = 1, x_2 = -1\\$

$y(1) = -7/(2) = -3.5\\\\ y(-1) = 7/2 = 3.5\\\\ A = y_{max} - y_{min} = 7$