# What are the points of inflection, if any, of f(x)=1/(x^2-4x+5) ?

Mar 22, 2017

$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(3\right) \left(11\right)}}{2 \left(3\right)}$

#### Explanation:

Rewrite the function as:

${\left({x}^{2} - 4 x + 5\right)}^{-} 1$

Then differentiate:

$- {\left({x}^{2} - 4 x + 5\right)}^{-} 2 \left(2 x - 4\right)$

Then differentiate again. But now, we're going to have to use chain rule:

$\left(- {\left({x}^{2} - 4 x + 5\right)}^{-} 2\right) \left(2\right) + \left(2 x - 4\right) \left(- \left(- 2\right) {\left({x}^{2} - 4 x + 5\right)}^{-} 3 \left(2 x - 4\right)\right)$

Rewrite:

$2 {\left(2 x + 4\right)}^{2} / {\left({x}^{2} - 4 x + 5\right)}^{3} - \frac{2}{{x}^{2} - 4 x + 5} ^ 2$

Use common denominators and merge the fractions:

$\frac{2 \left(3 {x}^{2} - 12 x + 11\right)}{{x}^{2} - 4 x + 5} ^ 3$

Inflection points may happen when f"(x)=0 So only worry about the top part of the fraction:

$2 \left(3 {x}^{2} - 12 x + 11\right) = 0$

$3 {x}^{2} - 12 x + 11 = 0$

$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(3\right) \left(11\right)}}{2 \left(3\right)}$