# Thread: Inflection point

1. ## Inflection point

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

what are the inflection points?

I got f '(x) = - (x^2 + 2x -1) / (x^4 + 2x + 1)
I got f ''(x) = (2x^3 + 6x^2 - 6x - 2) / (x^6 + 3x^4 + 3x^2 + 1)

Now When I set f ''(x) = zero to find the inflection point(s), How can I solve for x....this is...amazingly hard.

2. Originally Posted by VkL
f(x) = (1+x) / (1 + x^2)

what are the inflection points?

I got f '(x) = - (x^2 + 2x -1) / (x^4 + 2x + 1)
I got f ''(x) = (2x^3 + 6x^2 - 6x - 2) / (x^6 + 3x^4 + 3x^2 + 1)

Now When I set f ''(x) = zero to find the inflection point(s), How can I solve for x....this is...amazingly hard.
Remember that a rational function can never have 0 in the denominator. To find the 0 of f''(x), all you need to do is set the numerator equal to zero. Once you,ve done this, find the roots of the cubic polynomial.

3. Originally Posted by VkL
f(x) = (1+x) / (1 + x^2)

what are the inflection points?

I got f '(x) = - (x^2 + 2x -1) / (x^4 + 2x + 1)
I got f ''(x) = (2x^3 + 6x^2 - 6x - 2) / (x^6 + 3x^4 + 3x^2 + 1)

Now When I set f ''(x) = zero to find the inflection point(s), How can I solve for x....this is...amazingly hard.
$f''(x) = \frac{2(x^3+3x^2-3x-1)}{(x^2+1)^3}
$

note the numerator ...

$2(x^3 + 3x^2 - 3x - 1)$

using the rational root theorem, $x = 1$ turns out to be a zero, resulting in the following factorization ...

$\frac{2(x - 1)(x^2+4x+1)}{(x^2+1)^3} = 0$

$x = 1$ , $x = -2 \pm \sqrt{3}$