Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( f(x)=\frac{8 x}{x^{2}-1} \) Find the location of any local extrema of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. There is a local maximum at \( x=\square \) and there is a local minimum at \( x=\square \). (Type integers or decimals. Use a comma to separate answers as needed.) B. There is a local maximum at \( x=\square \). There is no local minimum. (Type an integer or a decimal. Use a comma to separate answers as needed.) C. There is a local minimum at \( x=\square \). There is no local maximum. (Type an integer or a decimal. Use a comma to separate answers as needed.) D. There are no local extrema.

To analyze the function \( f(x)=\frac{8x}{x^{2}-1} \), we start by identifying critical points where the derivative is zero or undefined. The first derivative can be found using the quotient rule, yielding \( f'(x) = \frac{8(x^2 - 1) - 8x(2x)}{(x^2 - 1)^2} \). After simplifying, we set the numerator to zero to find potential extrema. The critical points occur when \( 8(x^2 - 1 - 2x^2) = 0 \) leads to \( -8x^2 + 8 = 0 \), or \( x^2 = 1 \). This means \( x = \pm 1 \). Evaluating the second derivative at these points will help determine the nature of the extrema. By analyzing the function's behavior around these points, you find that there's a local maximum at \( x = -1 \) and a local minimum at \( x = 1\). So, the correct choice is: A. There is a local maximum at \( x=-1 \) and there is a local minimum at \( x=1 \).