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.

C. expression because any expression or equation may contain more than one variable