Question

question 7 of 16 < view policies current attempt in progress find all critical points and then use the first - derivative test to determine local maxima and minima. $f(x)=\frac{x}{x^{2}+1}$ enter the critical points in increasing order. if there is no local maximum or local minimum, enter na. $x = $ $x = $ the local maximum is at $x = $ the local minimum is at $x = $ etextbook and media save for later attempts: 0 of 15 used using multiple attempts will impact your score. 5% score reduction after attempt 5

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the quotient - rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = x\), \(u^\prime=1\), \(v=x^{2}+1\), \(v^\prime = 2x\). Then \(f^\prime(x)=\frac{1\cdot(x^{2}+1)-x\cdot(2x)}{(x^{2}+1)^{2}}=\frac{x^{2}+1 - 2x^{2}}{(x^{2}+1)^{2}}=\frac{1 - x^{2}}{(x^{2}+1)^{2}}\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(\frac{1 - x^{2}}{(x^{2}+1)^{2}} = 0\). Since \((x^{2}+1)^{2}>0\) for all real \(x\), we solve \(1 - x^{2}=0\). Factoring gives \((1 - x)(1 + x)=0\), so \(x=-1\) or \(x = 1\).

Step3: Use the first - derivative test

Choose test points in the intervals \((-\infty,-1)\), \((-1,1)\) and \((1,\infty)\).
For the interval \((-\infty,-1)\), let \(x=-2\). Then \(f^\prime(-2)=\frac{1-(-2)^{2}}{((-2)^{2}+1)^{2}}=\frac{1 - 4}{(4 + 1)^{2}}=-\frac{3}{25}<0\), so \(f(x)\) is decreasing on \((-\infty,-1)\).
For the interval \((-1,1)\), let \(x = 0\). Then \(f^\prime(0)=\frac{1-0}{(0 + 1)^{2}}=1>0\), so \(f(x)\) is increasing on \((-1,1)\).
For the interval \((1,\infty)\), let \(x = 2\). Then \(f^\prime(2)=\frac{1-2^{2}}{(2^{2}+1)^{2}}=\frac{1 - 4}{(4 + 1)^{2}}=-\frac{3}{25}<0\), so \(f(x)\) is decreasing on \((1,\infty)\).
Since \(f(x)\) changes from decreasing to increasing at \(x=-1\), \(f(-1)\) is a local minimum. Since \(f(x)\) changes from increasing to decreasing at \(x = 1\), \(f(1)\) is a local maximum.

Answer:

\(x=-1\)
\(x = 1\)
The local maximum is at \(x = 1\)
The local minimum is at \(x=-1\)