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

May 22, 2017

They are $\left(1 , - 1\right)$, $\left(4 + \sqrt{3} , \frac{- 1 - \sqrt{3}}{4}\right)$ and $\left(4 - \sqrt{3} , \frac{- 1 + \sqrt{3}}{4}\right)$

#### Explanation:

Use the quotient rule to find $f ' \left(x\right)$ and $f ' ' \left(x\right)$

You should get

$f ' \left(x\right) = \frac{\left(1\right) \left({x}^{2} - 4 x + 5\right) - \left(\left(x - 3\right) \left(2 x - 4\right)\right)}{{x}^{2} - 4 x + 5}$

$= \frac{- {x}^{2} + 6 x - 7}{{x}^{2} - 4 x + 5} ^ 2$

And

$f ' ' \left(x\right) = \frac{\left(- 2 x + 6\right) {\left({x}^{2} - 4 x + 5\right)}^{2} - \left(- {x}^{2} + 6 x - 7\right) \left(2 \left({x}^{2} - 4 x + 5\right) \left(2 x - 4\right)\right)}{{x}^{2} - 4 x + 5} ^ 4$

$= \frac{\left(- 2 x + 6\right) \left({x}^{2} - 4 x + 5\right) - \left(- {x}^{2} + 6 x - 7\right) \left(2 \left(2 x - 4\right)\right)}{{x}^{2} - 4 x + 5} ^ 3$

$= \frac{2 \left({x}^{3} - 9 {x}^{2} + 21 x - 13\right)}{{x}^{2} - 4 x + 5} ^ 3$

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

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

$f ' ' \left(x\right) = 0$ where ${x}^{3} - 9 {x}^{2} + 21 x - 13 = 0$

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

${x}^{3} - 9 {x}^{2} + 21 x - 13 = \left(x - 1\right) \left({x}^{2} - 8 x + 13\right)$

Find the remaining zeros by the quadratic formula.

The zeros of the numerator are $1 , 4 \pm \sqrt{3}$.

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 $f \left(1\right)$, $f \left(4 + \sqrt{3}\right)$, and $f \left(4 - \sqrt{3}\right)$ to get inflection points:

$\left(1 , - 1\right)$, $\left(4 + \sqrt{3} , \frac{- 1 - \sqrt{3}}{4}\right)$ and $\left(4 - \sqrt{3} , \frac{- 1 + \sqrt{3}}{4}\right)$

Here is the graph of $f ' ' \left(x\right)$