Известно, чтоf (x) = x^2/x^2+1. Докажите, что: а) f (a) = f (-a) ; б) f (-a) + f (1/a) = 1.

Дано:

f (x) = x²: (x²+1)

а) f (a) = a²: (a²+1)

f (-a) = (-a) ²: ((-a) ²+1) = a²: (a²+1)

a²: (a²+1) = a²: (a²+1), значит f (a) = f (-a)

б) f (-a) + f (1/a) = 1.

f (-a) = a²: (a²+1)

f (1/a) = 1/a²: (1/a²+1)

a²: (a²+1) + 1/a²: (1/a²+1) = a²: (a²+1) + 1: (1+a²) = (a²+1) : (a²+1) = 1

f (x) = x^2 (/x^2+1)

f (a) = a ² / (a²+1)

f (-a) = (-a) ² / ((-a) ²+1) = a² / (a²+1)

f (a) = a (-a)

f (1/a) = 1/a²: (1/a²+1) = 1/a²*a² / (a²+1) = 1 / (a²+1)

f (-a) + f (1/a) = a² / (a²+1) + 1 / (a²+1) = (a²+1) / (a²+1) = 1