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

Dec 31, 2015

First, let's note what each of these mean for a 2D graph on the Cartesian xy-plane.

DEFINITIONS

A critical point for some function $f \left(x\right)$ exists wherever some $f \left(c\right)$ exists such that one of the following is true:

• $\frac{\mathrm{df} \left(c\right)}{\mathrm{dx}} = 0$ (an extremum, or perhaps an inflection point)

• $\frac{\mathrm{df} \left(c\right)}{\mathrm{dx}}$ does not exist (a corner, cusp, etc)

...where $c$ takes on a chosen value within the domain of $f \left(x\right)$.

Concavity is essentially a descriptor noting that on either side of a critical point, the graph's behavior is the same, i.e. the graph increases on both sides, or decreases on both sides.

They exist where the first derivative is $0$ but the second derivative is not $0$, but either positive or negative.

An inflection point is almost like a point of concavity, but on either side of the critical point, the behavior is opposite. Thus, we call inflection points the point where the concavity changes when moving across the specified critical point.

They exist when the first derivative is $0$, with the condition that the second derivative is also $0$.

An example of inflection points? Try taking the second derivative of ${x}^{3}$; its critical point is at $x = 0$ since $\frac{d \left({x}^{3}\right)}{\mathrm{dx}} = 3 {x}^{2}$ and if $0 = 3 {x}^{2}$, $x = 0$.

THE GRAPH ITSELF

Before we start, yes, it kind of spoils where the critical point is, but here's the graph:

graph{1/(1+x^2) [-10, 10, -5, 5]}

CRITICAL POINT

You can find a critical point by taking the first derivative. All you know from the critical point, however, is that the derivative is $0$. You do not know yet whether it is a maximum, minimum, or inflection point.

For $f \left(x\right) = \frac{1}{1 + {x}^{2}}$, using the Power Rule and the Chain Rule, the derivative is:

$\textcolor{b l u e}{\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}} = - {\left(1 + {x}^{2}\right)}^{- 2} \cdot 2 x$

$= \textcolor{b l u e}{\frac{- 2 x}{1 + {x}^{2}} ^ 2}$

If we set this to equal $0$, we can search for the value(s) of $x$ that gives the critical point.

$\frac{\mathrm{df} \left(c\right)}{\mathrm{dx}} = 0 = \frac{- 2 x}{1 + {x}^{2}} ^ 2$

$= - 2 x$

$\textcolor{b l u e}{x = 0}$

Okay, so there is only one! Great. What is its concavity? Up or down?

CONCAVITY

To determine the concavity of the single inflection point, just take the second derivative of $f \left(x\right)$. Or, take the first derivative of the first derivative ($f ' \left(x\right)$, or $\frac{\mathrm{df}}{\mathrm{dx}}$); same thing.

$\setminus m a t h b f \left(\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{d \left[f \text{'} \left(x\right)\right]}{\mathrm{dx}}\right)$

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left[\frac{- 2 x}{1 + {x}^{2}} ^ 2\right]$

Using the Quotient Rule and Chain Rule, we get:

$= \frac{{\left(1 + {x}^{2}\right)}^{2} \left(- 2\right) - \left(- 2 x\right) \left(2 \left(1 + {x}^{2}\right) \cdot 2 x\right)}{1 + {x}^{2}} ^ 4$

$= \frac{- 2 {\left(1 + {x}^{2}\right)}^{{\cancel{2}}^{1}} + 8 {x}^{2} \cancel{\left(1 + {x}^{2}\right)}}{1 + {x}^{2}} ^ \left({\cancel{4}}^{3}\right)$

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

$= \textcolor{b l u e}{\frac{6 {x}^{2} - 2}{1 + {x}^{2}} ^ 3}$

Now, if we check the $x$ value where we found the critical point, we can look at the sign of this second derivative. If it's negative, the critical point is concave down, and vice versa.

$\frac{{d}^{2} f \left(c\right)}{{\mathrm{dx}}^{2}} :$

$\textcolor{b l u e}{\frac{{d}^{2} f \left(0\right)}{{\mathrm{dx}}^{2}}} = \frac{6 {\left(0\right)}^{2} - 2}{1 + {\left(0\right)}^{2}} ^ 3$

$= \textcolor{b l u e}{\frac{- 2}{1} ^ 3 < 0}$

Since $\frac{{d}^{2} f \left(c\right)}{{\mathrm{dx}}^{2}}$ is negative, the graph is concave down at $x = 0$.

INFLECTION POINTS

I'm going to say right upfront---this graph has no inflection points. It only has one extremum---a local maximum, which is concave down. It can't change concavity without becoming a new graph.

You can prove this by checking the second derivative at $x = c = 0$. If the second derivative had been $0$ instead of negative or positive, then an inflection point would be present. But it's negative.