Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of the function given below. $$ f(x)=\frac{9 x}{1-x^{2}} $$ A. There is/are (a) local extremum/extrema at $$ \mathrm{f}(\mathrm{x}) $$ is decreasing on (Use a comma to separate answers as needed.) B. There are no local extrema. Find the intervals where increasing. $$ \mathrm{f}(\mathrm{x}) $$ is concave upward or downward. Select the correct choice below and fill in the answer box $$ \mathrm{f}(\mathrm{x}) $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Type your answer in interval notation. Use a comma to separate answers as needed.) A. $$ \mathrm{f}(\mathrm{x}) $$ is concave downward on $$ \square $$. It is never concave upward. B. $$ \mathrm{f}(\mathrm{x}) $$ is concave upward on $$ \square $$ and concave downward on $$ \square $$. C. $$ \mathrm{f}(\mathrm{x}) $$ is concave upward on $$ \square $$. It is never concave downward.

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Function by following steps:
- step0: Find the local extrema:
  $$f\left(x\right)=\frac{9x}{1-x^{2}}$$
- step1: Find the domain:
  $$f\left(x\right)=\frac{9x}{1-x^{2}},x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)$$
- step2: Find the derivative:
  $$f^{\prime}\left(x\right)=\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}}$$
- step3: Find the domain:
  $$f^{\prime}\left(x\right)=\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}},x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)$$
- step4: Substitute $$f^{\prime}\left(x\right)=0:$$
  $$0=\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}}$$
- step5: Swap the sides:
  $$\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}}=0$$
- step6: The statement is false for any value of $$x:$$
  $$x \notin \mathbb{R}$$
- step7: The function has no local extrema:
  $$\textrm{No local extrema}$$
Analyze the intervals of the function $$ f(x)=\frac{9 x}{1-x^{2}} $$
Function by following steps:
- step0: Find the increasing or decreasing interval:
  $$f\left(x\right)=\frac{9x}{1-x^{2}}$$
- step1: Find the domain:
  $$f\left(x\right)=\frac{9x}{1-x^{2}},x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)$$
- step2: Find the derivative:
  $$f^{\prime}\left(x\right)=\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}}$$
- step3: Find the domain:
  $$f^{\prime}\left(x\right)=\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}},x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)$$
- step4: Substitute $$f^{\prime}\left(x\right)=0:$$
  $$0=\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}}$$
- step5: Swap the sides:
  $$\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}}=0$$
- step6: The statement is false for any value of $$x:$$
  $$x \notin \mathbb{R}$$
- step7: Determine the intervals:
  $$\begin{align}&x<-1\&-11\end{align}$$
- step8: Choose the points:
  $$\begin{align}&x_{1}=-2\&x_{2}=0\&x_{3}=2\end{align}$$
- step9: Find the values of the derivatives:
  $$\begin{align}&f^{\prime}\left(-2\right)=5\&f^{\prime}\left(0\right)=9\&f^{\prime}\left(2\right)=5\end{align}$$
- step10: Calculate:
  $$\begin{align}&x<-1\textrm{ is increasing interval}\&-11\textrm{ is increasing interval}\end{align}$$
- step11: Evaluate:
  $$\begin{align}&\textrm{The increasing interval is}\textrm{ }x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)\&\textrm{No decreasing interval}\end{align}$$
Analyze the inflection points of the function $$ f(x)=\frac{9 x}{1-x^{2}} $$
Function by following steps:
- step0: Find the inflection points:
  $$f\left(x\right)=\frac{9x}{1-x^{2}}$$
- step1: Find the domain:
  $$f\left(x\right)=\frac{9x}{1-x^{2}},x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)$$
- step2: Find the derivative:
  $$f^{\prime}\left(x\right)=\frac{9+9x^{2}}{\left(1-x^{2}\right)^{2}}$$
- step3: Find the second derivative:
  $$f^{\prime\prime}\left(x\right)=\frac{54x+18x^{3}}{\left(1-x^{2}\right)^{3}}$$
- step4: Find the domain:
  $$f^{\prime\prime}\left(x\right)=\frac{54x+18x^{3}}{\left(1-x^{2}\right)^{3}},x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)$$
- step5: Substitute $$f^{\prime\prime}\left(x\right)=0:$$
  $$0=\frac{54x+18x^{3}}{\left(1-x^{2}\right)^{3}}$$
- step6: Swap the sides:
  $$\frac{54x+18x^{3}}{\left(1-x^{2}\right)^{3}}=0$$
- step7: Cross multiply:
  $$54x+18x^{3}=\left(1-x^{2}\right)^{3}\times 0$$
- step8: Simplify the equation:
  $$54x+18x^{3}=0$$
- step9: Factor the expression:
  $$18x\left(3+x^{2}\right)=0$$
- step10: Divide both sides:
  $$x\left(3+x^{2}\right)=0$$
- step11: Separate into possible cases:
  $$\begin{align}&x=0\&3+x^{2}=0\end{align}$$
- step12: Solve the equation:
  $$\begin{align}&x=0\&x \notin \mathbb{R}\end{align}$$
- step13: Find the union:
  $$x=0$$
- step14: Check if the solution is in the defined range:
  $$x=0,x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)$$
- step15: Find the intersection:
  $$x=0$$
- step16: Determine the intervals:
  $$\begin{align}&\left(-1,0\right),\left(0,1\right)\end{align}$$
- step17: Choose the points:
  $$x_{1}=-\frac{1}{2},x_{2}=\frac{1}{2}$$
- step18: Find the value of the second derivative:
  $$f^{\prime\prime}\left(-\frac{1}{2} \right)=-\frac{208}{3},f^{\prime\prime}\left(\frac{1}{2} \right)=\frac{208}{3}$$
- step19: The inflection point is at $$x=0:$$
  $$\begin{align}&f\left(x\right)=\frac{9x}{1-x^{2}},x=0\end{align}$$
- step20: Calculate the function value:
  $$f\left(0\right)=0$$
- step21: The inflection point is $$\left(0,0\right):$$
  $$\begin{align}&\left(0,0\right)\end{align}$$
Let's summarize the pertinent information obtained from analyzing the function $$ f(x) = \frac{9x}{1 - x^2} $$.

### A. Local Extrema
- There are **no local extrema** for the function.

### B. Intervals of Increase
- The function $$ f(x) $$ is increasing on the intervals:
  $$
  x \in (-\infty, -1) \cup (-1, 1) \cup (1, +\infty)
  $$
- There are no intervals where the function is decreasing.

### C. Concavity
- The function has an inflection point at $$ x = 0 $$.
- To determine concavity:
  - The function is **concave upward** on the interval $$ (0, +\infty) $$.
  - The function is **concave downward** on the interval $$ (-\infty, 0) $$.

### Summary of Concavity
- **Concave upward** on $$ (0, +\infty) $$.
- **Concave downward** on $$ (-\infty, 0) $$.

### Final Choices
- **A.** There are no local extrema.
- **B.** $$ f(x) $$ is concave upward on $$ (0, +\infty) $$ and concave downward on $$ (-\infty, 0) $$.

### Graph Sketch
To sketch the graph of $$ f(x) = \frac{9x}{1 - x^2} $$:
1. Identify vertical asymptotes at $$ x = -1 $$ and $$ x = 1 $$ (where the denominator is zero).
2. Mark the inflection point at $$ (0, 0) $$.
3. Plot the increasing intervals and concavity regions.

The graph will show behavior approaching the vertical asymptotes and will change concavity at the inflection point.
- **No local extrema.** - **Increasing on** $$ (-\infty, -1) $$, $$ (-1, 1) $$, and $$ (1, +\infty) $$. - **Concave upward on** $$ (0, +\infty) $$. - **Concave downward on** $$ (-\infty, 0) $$.

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