**Problem: **Sketch on plain paper, the graph of \(y = \frac {x^2 + 1} {x^2 – 1} \)

**Discussion:**

There are several steps to find graph of a function. We will use calculus to analyze the function. Here y=f(x)

**Domain:**The function is defined at all real numbers except x =1 and x = -1 which makes the denominator 0.**Even/Odd:**Clearly f(x) = f(-x). Hence it is sufficient to investigate the function for positive values of x and then reflect it about y axis.**Critical Points:**Next we investigate the critical points.are those values of x for which the first derivative of f(x) is either 0 or undefined. Since \(\displaystyle{y = \frac {x^2 + 1} {x^2 – 1}}\), then \(\displaystyle{f'(x) = \frac {\left(\frac{d}{dx}(x^2 + 1)\right)(x^2 -1) – \left(\frac{d}{dx}(x^2 – 1)\right)(x^2 + 1)} {(x^2 – 1)^2} }\).**Critical Points**

This implies \(\displaystyle{f'(x) = \frac{2x^3 – 2x – 2x^3 – 2x}{(x^2 – 1)^2 } = -\frac{4x}{(x^2 – 1)^2 }}\)

Hence critical points are x =0 , 1, -1**Monotonicity:**The first derivative is negative for all positive values of x (note that we are only investigating for positive x values, since we can then reflect the picture about y axis as previously found).*Hence the function is ‘decreasing’ for all positive value of x.***Second Derivative:**We compute the second derivative to understand a couple things:- convexity/concavity of the function
- examine whether the critical points are maxima, minima, inflection points.

\(\displaystyle{f”(x) = \frac {4(3x^2 +1)}{(x^2-1)^3}}\)

Clearly \(f”(0) = -4 \) implying at x = 0 we have local maxima. Since f(0) = – 1,**we have (0, -1) as a local maxima**.

Also the second derivative is negative from x = 0 to x = 1 and positive after x = 1**. Hence the curve is under-tangent (concave) from x = 0 to x = 1, and above-tangent (convex) from x =1 onward.**

**Vertical Asymptote:**We next examine what happens*near*x = 1. We want to know what happens when we approach x=1 from*left*and from*right*. To that end we compute the following limits:- \(\displaystyle {\lim_{x to 1^{-} } \frac{x^2 +1}{x^2-1} = -\infty}\) (since the denominator gets infinitesimally small with a negative sign, and numerator is about 2)
- \(\displaystyle {\lim_{x to 1^{+} } \frac{x^2 +1}{x^2-1} = +\infty}\) (since the denominator gets infinitesimally small with a positive sign, and numerator is about 2)

**Horizontal Asymptote:**Finally we examine what happens when x approaches \(+ \infty \). To that end we compute the following:

\(\displaystyle { \lim_{x to + \infty} \frac {x^2 +1}{x^2-1}= \lim_{\frac{1}{x} to 0} \frac {1+ \frac{1}{x^2}}{1-\frac{1}{x^2}} = 1} \)**Drawing the graph:**- Local Maxima at (0,-1)
- Even function hence we draw for positive x values and reflect about y axis
- Vertical asymptote at x =1
- From x = 0 to 1, the function decreasing to negative infinity, staying under tangent all the time.
- From x = 1 to positive infinity, the function decreases from positive infinity to 1 staying above tangent all the time.
- Horizontal Asymptote at y= 1

## Chatuspathi:

**What is this topic:**Graph Sketching using Calculus**What are some of the associated concept:**Maxima, Minima, Derivative, Convexity, Concavity, Asymptotes**Where can learn these topics:**Cheenta**Book Suggestions:**Calculus of One Variable by I.A. Maron, Play with Graph (Arihant Publication)