Show that the curve has 3 points of inflection and they all lie on 1 straight line: \[y=\frac{1+x}{1+x^{2}}\]

Y = (1 + x) / (1 + x^2) 

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

y'' 
= [(1 + x^2)^2 * (-2x - 2) - (-x^2 - 2x + 1)(2)(1 + x^2)(2x)] / (1 + x^2)^4 
= [(1 + x^2)(-2x - 2) - (4x)(-x^2 - 2x + 1)] / (1 + x^2)^3 
= [(-2x - 2x^3 - 2 - 2x^2) - (-4x^3 - 8x^2 + 4x)] / (1 + x^2)^3 
= [-2x - 2x^3 - 2 - 2x^2 + 4x^3 + 8x^2 - 4x] / (1 + x^2)^3 
= [2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3 

Setting y'' to zero, we have: 
y'' = 0 
[2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3 = 0 
(2x^3 + 6x^2 - 6x - 2) = 0 

Using trial and error, you will realise that x = 1 is a root. 
This means (x - 1) is a factor. 
Dividing 2x^3 + 6x^2 - 6x - 2 by x - 1 using long division, you will have 2x^2 + 8x + 2. 

2x^2 + 8x + 2 
= 2(x^2 + 4x) + 2 
= 2(x + 2)^2 - 2(2^2) + 2 
= 2(x + 2)^2 - 8 + 2 
= 2(x + 2)^2 - 6 

Setting 2x^2 + 8x + 2 to zero, we have: 
2(x + 2)^2 - 6 = 0 
2(x + 2)^2 = 6 
(x + 2)^2 = 3 
x + 2 = sqrt(3) or = -sqrt(3) 
x = -2 + sqrt(3) or x = -2 - sqrt(3) 

Note that -2 - sqrt(3) < -2 + sqrt(3) < 1 
We will choose random values belonging to each interval and test them out. 

-5 < -2 - sqrt(3) < -2 < -2 + sqrt(3) 
f''(-5) = [2(-5)^3 + 6(-5)^2 - 6(-5) - 2] / (1 + (-5)^2)^3 = -9/2197 < 0 
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0 
Note that one value is positive and the other is negative. 
Thus, x = -2 - sqrt(3) is an inflection point. 

-2 - sqrt(3) < -2 < -2 + sqrt(3) < 0 < 1 
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0 
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0 
Note that one value is positive and the other is negative. 
Thus, x = -2 + sqrt(3) is also an inflection point. 

-2 + sqrt(3) < 0 < 1 < 2 
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0 
f''(2) = [2(2)^3 + 6(2)^2 - 6(2) - 2] / (1 + (2)^2)^3 = 26/125 > 0 
Note that one value is positive and the other is negative. 
Thus, x = 1 is an inflection point. 

Hence, we have three inflection points in total. 

When x = -2 - sqrt(3), we have: 
y 
= (1 - 2 - sqrt(3)) / (1 + (-2 - sqrt(3))^2) 
= (-1 - sqrt(3)) / (1 + 4 + 4sqrt(3) + 3) 
= (-1 - sqrt(3)) / (8 + 4sqrt(3)) 

When x = -2 + sqrt(3), we have: 
y 
= (1 - 2 + sqrt(3)) / (1 + (-2 + sqrt(3))^2) 
= (-1 + sqrt(3)) / (1 + 4 - 4sqrt(3) + 3) 
= (-1 + sqrt(3)) / (8 - 4sqrt(3)) 

When x = 1, we have: 
y 
= (1 + 1) / (1 + 1^2) 
= 2 / 2 
= 1 

Using the slope formula, we have: 
(y - 1) / (x - 1) = [[(-1 + sqrt(3)) / (8 - 4sqrt(3))] - 1] / ( -2 + sqrt(3) - 1) 
(y - 1) / (x - 1) = 1/4, which is the equation of the line which the inflection points at x = 1 and x = -2 + sqrt(3) lies on. 

Note that I am skipping the intermediate steps for simplifying here, but the trick is to rationalise the denominator by multiplying a conjugate on both numerator and denominator. 

Now, we just need to check that the inflection point at x = -2 - sqrt(3) lies on the same line as well. 
L.H.S. 
= [[(-1 - sqrt(3)) / (8 + 4sqrt(3))] - 1] / (-2 - sqrt(3) - 1) 
= 1/4 
= R.H.S. 

Once again, I am skipping simplifying steps here. 

Anyway, this proves all three points of inflection lies on the same straight line.