Question

Show that the curve \(y = \left( {1 + x} \right)/\left( {1 + {x^2}} \right)\) has three inflection points and they lie on the one straight line.

Short answer

It is proved that the curve \(y = \left( {1 + x} \right)/\left( {1 + {x^2}} \right)\) has three inflection points and they lie on the one straight line.

Step-by-step solution

Given data

The given function is a curve that has three inflection points.

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Calculate the value of \({y^\prime }\) and \({y^{\prime \prime }}\)

Find \({y^\prime }\)

\(\begin{aligned}{c}y &= \frac{{1 + x}}{{1 + {x^2}}}\\{y^\prime } &= \frac{{(1)\left( {1 + {x^2}} \right) - (1 + x)(2x)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\{y^\prime } &= \frac{{1 + {x^2} - 2x - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\{y^\prime } &= \frac{{1 - {x^2} - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\end{aligned}\)

Calculate the value of \({y^{\prime \prime }}\):

\(\begin{array}{l}{y^{\prime \prime }} = \frac{{( - 2x - 2){{\left( {1 + {x^2}} \right)}^2} - \left( {1 - {x^2} - 2x} \right)(2)\left( {1 + {x^2}} \right)(2x)}}{{{{\left( {1 + {x^2}} \right)}^4}}}\\{y^{\prime \prime }} = \frac{{\left( {1 + {x^2}} \right)\left( {( - 2x - 2)\left( {1 + {x^2}} \right) - \left( {1 - {x^2} - 2x} \right)(4x)} \right)}}{{{{\left( {1 + {x^2}} \right)}^4}}}\\{y^{\prime \prime }} = \frac{{ - 2x - 2{x^3} - 2 - 2{x^2} - 4x + 4{x^3} + 8{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}}\\{y^{\prime \prime }} = \frac{{2{x^3} + 6{x^2} - 6x - 2}}{{{{\left( {1 + {x^2}} \right)}^3}}}\end{array}\)

Similarly, calculate further:

\(\begin{array}{l}{y^{\prime \prime }} = \frac{{2\left( {{x^3} + 3{x^2} - 3x - 1} \right)}}{{{{\left( {1 + {x^2}} \right)}^3}}}\\{y^{\prime \prime }} = \frac{{2(x - 1)\left( {{x^2} + 4x + 1} \right)}}{{{{\left( {1 + {x^2}} \right)}^3}}}\end{array}\)

Simplify the expression

Simplify the expression: \({y^{\prime \prime }} = 0\) when \(x = 1\) and

\(\begin{aligned}{c}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\x &= \frac{{ - (4) \pm \sqrt {{{(4)}^2} - 4(1)(1)} }}{{2(1)}}\\x &= \frac{{ - 4 \pm \sqrt {12} }}{2}\end{aligned}\)

Similarly, calculate further:

\(\begin{array}{c}x = \frac{{ - 4 \pm \sqrt {4 \cdot 3} }}{2}\\x = \frac{{ - 4 \pm 2\sqrt 3 }}{2}\\x = - 2 \pm \sqrt 3 \approx - 3.73, - 0.268\end{array}\)

Confirm that they're inflection points by testing the intervals between the zeros.

\(\begin{aligned}{c}( - \infty , - 3.73):{y^{\prime \prime }}( - 4) &\approx - 0.0002\\( - 3.73, - 0.268):{y^{\prime \prime }}( - 1) &= 1\\\left( { - 0.268,1} \right):{y^{\prime \prime }}(0) &= - 2\\(1,\infty ):{y^{\prime \prime }}(2) &= 0.208\end{aligned}\)

Adjacent intervals have different signs, so they are inflection points.

Calculate the value of \(y\)

Find the \(y\)- coordinates of the inflection point:

A:

\(\begin{array}{c}y( - 2 - \sqrt 3 ) = \frac{{1 + ( - 2 - \sqrt 3 )}}{{1 + {{( - 2 - \sqrt 3 )}^2}}}\\y( - 2 - \sqrt 3 ) = \frac{{ - 1 - \sqrt 3 }}{{8 + 4\sqrt 3 }} \cdot \frac{{8 - 4\sqrt 3 }}{{8 - 4\sqrt 3 }}\\y( - 2 - \sqrt 3 ) = \frac{{ - 8 - 4\sqrt 3 + 12}}{{64 - 48}}\\y( - 2 - \sqrt 3 ) = \frac{{4 - 4\sqrt 3 }}{{16}}\\y( - 2 - \sqrt 3 ) = \frac{{1 - \sqrt 3 }}{4}\end{array}\)

B:

\(\begin{array}{c}y( - 2 + \sqrt 3 ) = \frac{{1 + ( - 2 + \sqrt 3 )}}{{1 + {{( - 2 + \sqrt 3 )}^2}}}\\y( - 2 + \sqrt 3 ) = \frac{{ - 1 + \sqrt 3 }}{{8 - 4\sqrt 3 }} \cdot \frac{{8 + 4\sqrt 3 }}{{8 + 4\sqrt 3 }}\\y( - 2 + \sqrt 3 ) = \frac{{ - 8 + 4\sqrt 3 + 12}}{{64 - 48}}\\y( - 2 + \sqrt 3 ) = \frac{{4 + 4\sqrt 3 }}{{16}}\\y( - 2 + \sqrt 3 ) = \frac{{1 + \sqrt 3 }}{4}\end{array}\)

C:

\(\begin{array}{c}y(1) = \frac{{1 + (1)}}{{1 + {{(1)}^2}}}\\y(1) = 1\end{array}\)

Calculate the slope and sketch the graph

Find the slope from \(A\) to \(B\)

\(\begin{array}{c}\frac{{\frac{{1 + \sqrt 3 }}{4} - \frac{{1 - \sqrt 3 }}{4}}}{{ - 2 + \sqrt 3 - ( - 2 - \sqrt 3 )}} = \frac{{\frac{{2\sqrt 3 }}{4}}}{{2\sqrt 3 }}\\\frac{{\frac{{1 + \sqrt 3 }}{4} - \frac{{1 - \sqrt 3 }}{4}}}{{ - 2 + \sqrt 3 - ( - 2 - \sqrt 3 )}} = \frac{1}{4}\end{array}\)

Find the slope from \(B\) to \(C\)

\(\begin{array}{l}\frac{{\frac{{1 + \sqrt 3 }}{4} - 1}}{{ - 2 + \sqrt 3 - 1}} = \frac{{\frac{{ - 3 + \sqrt 3 }}{4}}}{{ - 3 + \sqrt 3 }}\\\frac{{\frac{{1 + \sqrt 3 }}{4} - 1}}{{ - 2 + \sqrt 3 - 1}} = \frac{1}{4}\end{array}\)

The slopes are equal, and since we used a common point for both slopes, the 3 points are on the same line ("collinear").

It is proved that the curve \(y = \left( {1 + x} \right)/\left( {1 + {x^2}} \right)\) has three inflection points and they lie on the one straight line.