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Chapter 3: Problem 13

Find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=\frac{x}{x^{2}+1}\)

Short Answer

Expert verified
The points of inflection are at \(x=0\), \(x=-\frac{1}{\sqrt{3}}\), and \(x=\frac{1}{\sqrt{3}}\). The function is concave up on the intervals \(-\infty < x < -\frac{1}{\sqrt{3}}\) and \(0 < x < \frac{1}{\sqrt{3}}\), and concave down on \(-\frac{1}{\sqrt{3}} < x < 0\) and \(\frac{1}{\sqrt{3}} < x < \infty\).

Step by step solution

01

Find the First Derivative

Firstly, differentiate the function \(f(x)=\frac{x}{x^{2}+1}\) using the quotient rule. The quotient rule states that the derivative of \(\frac{u}{v}\) is \(\frac{v\cdot u' - u\cdot v'}{v^2}\), where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\) respectively. So, \(f'(x)=\frac{(x^{2}+1)\cdot 1 - x\cdot 2x}{(x^{2}+1)^{2}}=\frac{1-x^{2}}{(x^{2}+1)^{2}}\).
02

Find the Second Derivative

Next, find the second derivative \(f''(x)\) by differentiating \(f'(x)=\frac{1-x^{2}}{(x^{2}+1)^{2}}\) again using the quotient rule. The result is \(f''(x)=\frac{2x(3x^{2}-1)}{(1+x^{2})^{3}}\).
03

Solve \(f''(x) = 0\)

Set the second derivative equal to zero and solve for \(x\). This gives the values of \(x\) where possible points of inflection occur. Solving \(\frac{2x(3x^{2}-1)}{(1+x^{2})^{3}}=0\), we find \(x= 0, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\).
04

Test the intervals

Take a number in each of the intervals \(-\infty < x < -\frac{1}{\sqrt{3}}\), \(-\frac{1}{\sqrt{3}} < x < 0\), \(0 < x < \frac{1}{\sqrt{3}}\), and \(\frac{1}{\sqrt{3}} < x < \infty\) and substitute into the second derivative to determine the concavity. If the second derivative is positive, the function is concave up on that interval. If it's negative, the function is concave down. This results in the function being concave up on \(-\infty < x < -\frac{1}{\sqrt{3}}\) and \(0 < x < \frac{1}{\sqrt{3}}\), and concave down on \(-\frac{1}{\sqrt{3}} < x < 0\) and \(\frac{1}{\sqrt{3}} < x < \infty\).

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