How Cheenta works to ensure student success?

Explore the Back-StoryThis is a Test of Mathematics Solution Subjective 124 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

** **Sketch on plain paper, the graph of $ y = \frac {x^2 + 1} {x^2 - 1} $

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

**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)

This is a Test of Mathematics Solution Subjective 124 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

** **Sketch on plain paper, the graph of $ y = \frac {x^2 + 1} {x^2 - 1} $

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

**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)

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?