# If (x^2+1) / (x^2-2), what are the points of inflection, concavity and critical points?

Dec 3, 2015

There are no inflection points. The graph is concave up on$\left(- \infty , - \sqrt{2}\right)$ and $\left(\sqrt{2} , \infty\right)$, and concave down on $\left(- \sqrt{2} , \sqrt{2}\right)$. Crritical number is: $0$.

#### Explanation:

Let $f \left(x\right) = \frac{{x}^{2} + 1}{{x}^{2} - 2}$.

The domain of $f$ is $\left(- \infty , - \sqrt{2}\right) \cup \left(- \sqrt{2} , \sqrt{2}\right) \cup \left(\sqrt{2} , \infty\right)$.

To find critical numbers, we find $f ' \left(x\right) = \frac{- 6 x}{{x}^{2} - 2} ^ 2$ which is $0$ at $x = 0$ and is never undefined on the domain of $f$.

The only critical number is $0$.

To investigate concavity and inflection points, we find $f ' ' \left(x\right) = \frac{6 \left(3 {x}^{2} + 2\right)}{{x}^{2} - 2} ^ 3$ which is never $0$ and is undefined at $- \sqrt{2}$ and at $\sqrt{2}$

Sign of $f ' '$

{: (bb "Interval", bb"Sign of "f'',bb" Concavity"), ((-oo,-sqrt2)," "" " +" ", " "" Up"), ((-sqrt2,sqrt2), " "" " -, " " " Dn"), ((sqrt2 ,oo), " "" " +, " "" Up") :}

The graph of $f$ is concave up on the intervals $\left(- \infty , - \sqrt{2}\right)$ and $\left(\sqrt{2} , \infty\right)$. The graph is concave down on $\left(- \sqrt{2} , \sqrt{2}\right)$.

Although the concavioty changes at $x = \pm \sqrt{2}$, there is no point on the graph at $x = \pm \sqrt{2}$, so there are no inflection points.