Question

What are the points of inflection, if any, of `f(x)=(x-3)/(x^2-4x+5)`?

Answer 1

They are `(1,-1)`, `(4+sqrt3,(-1-sqrt3)/4)` and `(4-sqrt3,(-1+sqrt3)/4)`

Explanation:

Use the quotient rule to find `f'(x)` and `f''(x)`

You should get

`f'(x) =((1)(x^2-4x+5)-((x-3)(2x-4)))/(x^2-4x+5)`

`= (-x^2+6x-7)/(x^2-4x+5)^2`

And

`f''(x) = ((-2x+6)(x^2-4x+5)^2-(-x^2+6x-7)(2(x^2-4x+5)(2x-4)))/(x^2-4x+5)^4`

`= ((-2x+6)(x^2-4x+5)-(-x^2+6x-7)(2(2x-4)))/(x^2-4x+5)^3`

`= (2(x^3-9x^2+21x-13))/(x^2-4x+5)^3`

To see where the concavity changes analyze the sign of `f''`

The denominator is never `0` for real values of `x`.

`f''(x) = 0` where `x^3-9x^2+21x-13=0`

Observe that `x=1` is a solution (Use the Rational Zeros Theorem), and

`x^3-9x^2+21x-13 = (x-1)(x^2-8x+13)`

Find the remaining zeros by the quadratic formula.

The zeros of the numerator are `1, 4+-sqrt3`.

None of these zeros have even multiplicity, so the sign of `f''` changes at each zero. Therefore each zero is the location (`x`-value) of an inflection point.

To find the corresponding `y` values, evaluate `f(1)`, `f(4+sqrt3)`, and `f(4-sqrt3)` to get inflection points:

`(1,-1)`, `(4+sqrt3,(-1-sqrt3)/4)` and `(4-sqrt3,(-1+sqrt3)/4)`

Here is the graph of `f''(x)`