Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 54

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Show that all of the inflection points of $\displaystyle y = \frac{(1+x)}{(1+x^2)}$ lie on one straight line.

To get the inflection points, we set $f''(x) = 0$ and some for the solution, so...

if $\displaystyle y = \frac{(1+x)}{(1+x^2)}$, then

By using Quotient Rule,

$ \begin{equation} \begin{aligned} y' &= \frac{(1+x^2)(1) - (1+x)(2x)}{(1+x^2)^2}\\ \\ y' &= \frac{1+x^2 - 2x -2x^2}{(1+x^2)^2}\\ \\ y' &= \frac{-x^2 - 2x + 1}{(1+x^2)^2} \end{aligned} \end{equation} $

Again, by using Quotient Rule as well as Chain Rule

$ \begin{equation} \begin{aligned} y '' &= \frac{(1+x^2)^2 (-2x-2) - (-x^2-2x+1) \left( 2(1+x^2)(2x) \right)}{\left[ (1+x^2)^2\right]^2}\\ \\ y'' &= \frac{2(x-1)(x^2+4x+1)}{(1+x^2)^3} \end{aligned} \end{equation} $

when $y'' = 0$

$ \begin{equation} \begin{aligned} 0 & = \frac{2(x-1)(x^2+4x+1)}{(1+x^2)^3}\\ \\ 0 & = 2(x-1)(x^2+4x+1) \end{aligned} \end{equation} $

We have,

$x - 1 = 0$ and $x^2 + 4x + 1 = 0$ (by using Quadratic Formula)

$x = 1 $ and $x = -2 \pm \sqrt{3}$

Let's evaluate $f(x)$ with these inflection points. So,

$ \begin{equation} \begin{aligned} \text{when } x & =1 &&& \text{when } x &= -2 + \sqrt{3},\\ \\ y (1) &= \frac{1+1}{1+1^2} = 1 &&& y(-2+\sqrt{3}) &= \frac{1 + (-2+\sqrt{3})}{1 + (-2+\sqrt{3})^2} = \frac{1+\sqrt{3}}{4}\\ \\ \text{when } x &= 2 - \sqrt{3}\\ \\ y (2 - \sqrt{3}) &= \frac{1+ (-2 - \sqrt{3}) }{1 + (-2 - \sqrt{3})^2} = \frac{1-\sqrt{3}}{4} \end{aligned} \end{equation} $

If we get the equation of the line that pass through points (1,1) and $\displaystyle \left(-2+\sqrt{3},\frac{1+\sqrt{3}}{4} \right)$ by using point slope form..

$\displaystyle y- y_1 = \frac{y_2-y_1}{x_2-x_1} (x-x_1)$

$ \begin{equation} \begin{aligned} y -1 &= \frac{\frac{1+\sqrt{3}}{4}-1}{-2 + \sqrt{3}-1} (x-1)\\ \\ y &= \frac{1}{4} (x - 1) + 1\\ \\ y &= \frac{x}{4} + \frac{3}{4} \end{aligned} \end{equation} $

If we substitute the point $\displaystyle \left( 2-\sqrt{3}, 1- \frac{\sqrt{3}}{4}\right)$ to the equation of the line.

$ \begin{equation} \begin{aligned} 1 - \frac{\sqrt{3}}{4} &= \frac{2-\sqrt{3}}{4} + \frac{3}{4}\\ \\ 1 - \frac{\sqrt{3}}{4} &= 1 - \frac{\sqrt{3}}{4} \end{aligned} \end{equation} $

Notice that the points $\displaystyle \left( 2-\sqrt{3}, 1- \frac{\sqrt{3}}{4}\right)$ is also a solution of the tangent line. Therefore, we can say that all the inflection points lie on the same line