Question

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

Answer 1

`x=(-(-12)+-sqrt((-12)^2-4(3)(11)))/(2(3))`

Explanation:

Rewrite the function as:

`(x^2-4x+5)^-1`

Then differentiate:

`-(x^2-4x+5)^-2(2x-4)`

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

`(-(x^2-4x+5)^-2)(2)+(2x-4)(-(-2)(x^2-4x+5)^-3(2x-4))`

Rewrite:

`2(2x+4)^2/(x^2-4x+5)^3 -2/(x^2-4x+5)^2`

Use common denominators and merge the fractions:

`(2(3x^2-12x+11))/(x^2-4x+5)^3`

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

`2(3x^2-12x+11)=0`

`3x^2-12x+11=0`

Use quadratic formula:

`x=(-(-12)+-sqrt((-12)^2-4(3)(11)))/(2(3))`