Find the second derivative = 0?

Question

f(x)=(1-x^2)/(1+x^2)
f'(x)= -4x/(1+x^2)^2

find where f''(x)=0

an explanation would be helpful!
Thanks

Answer 1

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

f'(x) = -2x/(1+x^2) - 2x(1-x^2)/(1+x^2)^2
f'(x) = [-2x(1+x^2) -2x(1-x^2)]/(1+x^2)^2
f'(x) = -4x /(1+x^2)^2

f"(x) = -4/(1+x^2)^2 - 2(2x)(-4x)/(1+x^2)^3
f"(x) = [-4(1+x^2) + 16x^2]/(1+x^2)^3
f"(x) = (12x^2 - 4) / (1+x^2)^3

f"(x) = 0
(12x^2 - 4) / (1+x^2)^3 = 0
12x^2 - 4 = 0
x^2 = 1/3
x = +/- 1/sqrt(3)