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

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To analyze the function $$ f(x) = \frac{9x}{1 - x^2} $$, we will follow these steps: ### Step 1: Identify Vertical Asymptotes Vertical asymptotes occur where the denominator is zero. Set the denominator equal to zero: $$ 1 - x^2 = 0 $$ Solving for $$ x $$: $$ x^2 = 1 \implies x = \pm 1 $$ Thus, the vertical asymptotes are at $$ x = -1 $$ and $$ x = 1 $$. ### Step 2: Determine Intervals of Increase and Decrease To find where $$ f(x) $$ is increasing or decreasing, we need to find the derivative $$ f'(x) $$ and analyze its sign. Using the quotient rule: $$ f'(x) = \frac{(1 - x^2)(9) - (9x)(-2x)}{(1 - x^2)^2} $$ Simplifying the numerator: $$ f'(x) = \frac{9(1 - x^2 + 2x^2)}{(1 - x^2)^2} = \frac{9(1 + x^2)}{(1 - x^2)^2} $$ Since $$ 1 + x^2 > 0 $$ for all $$ x $$, the sign of $$ f'(x) $$ depends on the denominator $$ (1 - x^2)^2 $$, which is always positive except at the vertical asymptotes. Thus, $$ f'(x) > 0 $$ for $$ x < -1 $$ and $$ x > 1 $$, and $$ f'(x) < 0 $$ for $$ -1 < x < 1 $$. ### Step 3: Summary of Increasing and Decreasing Intervals - **Increasing**: $$ (-\infty, -1) $$ and $$ (1, \infty) $$ - **Decreasing**: $$ (-1, 1) $$ ### Step 4: Find Local Extrema Local extrema occur at critical points where $$ f'(x) = 0 $$ or is undefined. Since $$ f'(x) $$ does not equal zero for any $$ x $$ (it is always positive or negative), there are no local extrema. ### Final Answers 1. **Vertical Asymptotes**: $$ x = -1, x = 1 $$ 2. **Intervals of Increase and Decrease**: - $$ f(x) $$ is increasing on $$ (-\infty, -1), (1, \infty) $$. - It is decreasing on $$ (-1, 1) $$. 3. **Local Extrema**: - There are no local extrema. ### Graph Sketch The graph of $$ f(x) $$ will show vertical asymptotes at $$ x = -1 $$ and $$ x = 1 $$, with the function increasing on the intervals $$ (-\infty, -1) $$ and $$ (1, \infty) $$, and decreasing on $$ (-1, 1) $$. The function approaches infinity as it nears the asymptotes. This analysis provides a comprehensive understanding of the function's behavior. - **Vertical Asymptotes**: $$ x = -1, x = 1 $$ - **Intervals of Increase and Decrease**: - Increasing: $$ (-\infty, -1) $$ and $$ (1, \infty) $$ - Decreasing: $$ (-1, 1) $$ - **Local Extrema**: None

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