What are the points of inflection, if any, of #f(x)= (x^2+x)/(x^2+1) #?

Question

What are the points of inflection, if any, of #f(x)= (x^2+x)/(x^2+1)     #?

Charles Arocha · Accepted Answer

The #x#-coordinates of the three inflection points are #x=-1, 2 pm sqrt(3)# (explanation to come)

By the Quotient Rule, the derivative is #f'(x)=((x^2+1)(2x+1) - (x^2+x)(2x))/((x^2+1)^2)# #=(2x^3+x^2+2x+1-2x^3-2x^2)/((x^2+1)^2)# #=(-x^2+2x+1)/((x^2+1)^2)# The second derivative is (also using the Chain Rule), #f''(x)# #=((x^2+1)^2*(-2x+2)-(-x^2+2x+1)*2(x^2+1)*2x)/((x^2+1)^4)# #=((x^2+1)*(-2x+2)-(-x^2+2x+1)*4x)/((x^2+1)^3)# #=(-2x^3+2x^2-2x+2+4x^3-8x^2-4x)/((x^2+1)^3)# #=(2x^3-6x^2-6x+2)/((x^2+1)^3)#. Inflection points occur where #f''(x)# changes sign (also note that #f''(x)# is defined for all #x#). The #x#-values where #f''(x)=0# occur where #2x^3-6x^2-6x+2=0 \Leftrightarrow x^3-3x^2-3x+1=0#. You can either use your calculator or the Rational Root Theorem to help you find that #x=-1# is a root of this (once you know it's a root, it's easy to check: #(-1)^3-3(-1)^2-3(-1)+1=-1-3+3+1=0#.) Next, either use long division of polynomials or synthetic division to see that #x^3-3x^2-3x+1=(x+1)(x^2-4x+1)#. The quadratic formula gives the other two roots of #f''(x)# as #x=(4 pm sqrt{16-4})/(2)=2 pm 1/2 *sqrt{12}=2 pm sqrt{3}# You can most easily see that #f''(x)# changes sign at these three values by graphing it. But, since #f''(x)# is continuous for all #x#, you can just check its sign on various intervals by plugging in sample points. Therefore, the #x#-coordinates of the three inflection points of #f(x)# are #x=-1, 2 pm sqrt(3)#. By plugging these numbers into #f(x)# and doing various manipulations, you can find that the full rectangular coordinates of the three inflection points are: #(-1,0)#, #(2+sqrt{3},3/4+sqrt{3}/4)#, and #(2-sqrt{3},3/4-sqrt{3}/4)#.

Ezra Adams · Answer

undefined To find the points of inflection, we first need to find the second derivative of the function $ f(x) = \frac{x^2 + x}{x^2 + 1} $. After finding the second derivative, we'll set it equal to zero and solve for $ x $. These values of $ x $ will give us the potential points of inflection. Then we'll check the concavity of the function around these points to confirm if they are indeed points of inflection.

Given function: $ f(x) = \frac{x^2 + x}{x^2 + 1} $

First, find the first derivative $ f'(x) $ using the quotient rule:
$$ f'(x) = \frac{(x^2+1)(2x+1) - (x^2+x)(2x)}{(x^2+1)^2} $$

Now, simplify $ f'(x) $:
$$ f'(x) = \frac{2x^3 + 2x + 2x + 1 - 2x^3 - 2x^2}{(x^2+1)^2} $$
$$ f'(x) = \frac{2x + 1}{(x^2+1)^2} $$

Next, find the second derivative $ f''(x) $:
$$ f''(x) = \frac{d}{dx} \left( \frac{2x + 1}{(x^2+1)^2} \right) $$
$$ f''(x) = \frac{(x^2+1)^2(2) - (2x+1)(2)(2x)(x^2+1)}{(x^2+1)^4} $$
$$ f''(x) = \frac{2(x^4 + 2x^2 + 1) - 8x(x^2+1)(2x+1)}{(x^2+1)^3} $$
$$ f''(x) = \frac{2x^4 + 4x^2 + 2 - 16x^4 - 16x^3 - 8x}{(x^2+1)^3} $$
$$ f''(x) = \frac{-14x^4 - 16x^3 + 4x^2 - 8x + 2}{(x^2+1)^3} $$

Now, set $ f''(x) = 0 $ and solve for $ x $:
$$ -14x^4 - 16x^3 + 4x^2 - 8x + 2 = 0 $$

There isn't a simple algebraic solution to this quartic equation. You may use numerical methods such as graphing or a computational tool to find the approximate roots. Once you have the values of $ x $, plug them back into the second derivative to determine the concavity of the function around these points. If the concavity changes at these points, they are points of inflection.