Question

1. which of the following is the graph of the function $f(x)=\frac{2x}{x^{2}+1}$? you may use

$f(x)=\frac{2(1 - x^{2})}{(x^{2}+1)^{2}}$ and $f(x)=\frac{4x(x^{2}-3)}{(x^{2}+1)^{3}}$.

Explanation:

Step1: Analyze the function at \(x = 0\)

Substitute \(x = 0\) into \(f(x)=\frac{2x}{x^{2}+1}\). Then \(f(0)=\frac{2\times0}{0^{2}+1}=0\).

Step2: Analyze the limit as \(x

ightarrow\pm\infty\)
We know that \(\lim_{x
ightarrow\pm\infty}f(x)=\lim_{x
ightarrow\pm\infty}\frac{2x}{x^{2}+1}\). Divide numerator and denominator by \(x^{2}\) (for \(x
eq0\)), we get \(\lim_{x
ightarrow\pm\infty}\frac{\frac{2}{x}}{1 + \frac{1}{x^{2}}}=0\).

Step3: Analyze the first - derivative \(f^{\prime}(x)=\frac{2(1 - x^{2})}{(x^{2}+1)^{2}}\)

Set \(f^{\prime}(x)=0\), then \(2(1 - x^{2})=0\), which gives \(x=\pm1\). When \(x\in(-\infty,- 1)\), \(f^{\prime}(x)<0\) (function is decreasing); when \(x\in(-1,1)\), \(f^{\prime}(x)>0\) (function is increasing); when \(x\in(1,\infty)\), \(f^{\prime}(x)<0\) (function is decreasing).

Step4: Analyze the second - derivative \(f^{\prime\prime}(x)=\frac{4x(x^{2}-3)}{(x^{2}+1)^{3}}\)

Set \(f^{\prime\prime}(x) = 0\), then \(4x(x^{2}-3)=0\), which gives \(x = 0,x=\pm\sqrt{3}\). The function changes concavity at these points.

Since \(f(0) = 0\), we can eliminate options (b), (d), (e) (because they do not pass through the origin \((0,0)\)). Option (a) has the wrong increasing - decreasing behavior.

Answer:

(c)