Part 1 of 5
Points: 0 of 1 increasing or decreasing; c) give the coordinates of any points of inflection; d) identify intervals where the function is concave up or concave down, and e) sketch the graph.
\[
q(x)=\frac{x}{x^{2}+9}
\]
a) What are the coordinates of the relative extrema? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
A. The relative minimum point(s) is/are $\square$ and there are no relative maximum point(s).
(Simplify your answer. Use integers or fractions for any numbers in the expression. Type an ordered pair. Use a comma to separate answers as needed.)
B. The relative minimum point(s) is/are $\square$ and the relative maximum point(s) is/are
(Simplify your answers. Use integers or fractions for any numbers in the expression. Type an ordered pair. Use a comma to separate answers as needed.)
C. The relative maximum point(s) is/are $\square$ and there are no relative minimum point(s).
(Simplify your answer. Use integers or fractions for any numbers in the expression. Type an ordered pair. Use a comma to separate answers as needed.)
D. There are no relative minimum points and there are no relative maximum points.

Step 1 ：To find the relative extrema, we first need to find the derivative of the function \(q(x)\).

Step 2 ：Using the quotient rule, we get \(q'(x) = \frac{(x^{2}+9)\cdot1 - x\cdot2x}{(x^{2}+9)^{2}} = \frac{9-x^{2}}{(x^{2}+9)^{2}}\).

Step 3 ：Setting \(q'(x)\) equal to zero gives us \(9-x^{2} = 0\), which simplifies to \(x^{2} = 9\).

Step 4 ：Solving for \(x\) gives us \(x = 3\) and \(x = -3\).

Step 5 ：We then substitute these values into the original function to find the corresponding \(y\) values. For \(x = 3\), we get \(q(3) = \frac{3}{3^{2}+9} = \frac{3}{18} = \frac{1}{6}\). For \(x = -3\), we get \(q(-3) = \frac{-3}{(-3)^{2}+9} = \frac{-3}{18} = -\frac{1}{6}\).

Step 6 ：So, the relative extrema are at the points \((3, \frac{1}{6})\) and \((-3, -\frac{1}{6})\).

Step 7 ：To determine whether these points are relative minima or maxima, we need to examine the sign of the second derivative at these points.

Step 8 ：The second derivative of \(q(x)\) is \(q''(x) = \frac{2x(9-x^{2}) + 4x^{2}(x^{2}-9)}{(x^{2}+9)^{3}} = \frac{18x^{2} - 2x^{4}}{(x^{2}+9)^{3}}\).

Step 9 ：Substituting \(x = 3\) into \(q''(x)\), we get \(q''(3) = \frac{18(3)^{2} - 2(3)^{4}}{(3^{2}+9)^{3}} = \frac{162 - 162}{216} = 0\).

Step 10 ：Substituting \(x = -3\) into \(q''(x)\), we get \(q''(-3) = \frac{18(-3)^{2} - 2(-3)^{4}}{((-3)^{2}+9)^{3}} = \frac{162 - 162}{216} = 0\).

Step 11 ：Since the second derivative is zero at both points, we cannot determine whether these points are relative minima or maxima using the second derivative test.

Step 12 ：However, by examining the graph of the function, we can see that the function is decreasing for \(x < -3\), increasing for \(-3 < x < 3\), and decreasing for \(x > 3\).

Step 13 ：Therefore, the point \((-3, -\frac{1}{6})\) is a relative maximum and the point \((3, \frac{1}{6})\) is a relative minimum.

Step 14 ：So, the answer is B. The relative minimum point is \((3, \frac{1}{6})\) and the relative maximum point is \((-3, -\frac{1}{6})\).

Step 15 ：\(\boxed{\text{B. The relative minimum point is }(3, \frac{1}{6})\text{ and the relative maximum point is }(-3, -\frac{1}{6})\)