## Step 1: Find the Derivative and Critical Points Given: \[ r(x) = \frac{x^2}{x^2 - 4} \] First, find the **first derivative** using the quotient rule: \[ r'(x) = \frac{(x^2-4)(2x) - x^2(2x)}{(x^2-4)^2} \] \[ = \frac{2x(x^2 - 4) - 2x x^2}{(x^2-4)^2} \] \[ = \frac{2x(x^2 - 4 - x^2)}{(x^2-4)^2} \] \[ = \frac{2x(-4)}{(x^2-4)^2} \] \[ = \frac{-8x}{(x^2-4)^2} \] Set the numerator to **zero** to find critical points: \[ -8x = \implies x = \] Also, the denominator is zero at \( x = 2 \) and \( x = -2 \), but these are **not in the domain** of \( r(x) \). **Explanation:** We found the critical point at \( x = \). The function is not defined at \( x = 2 \) and \( x = -2 \). --- ## Step 2: Test Intervals for Increasing/Decreasing Pick test points in the intervals: \( (-\infty, -2) \), \( (-2, ) \), \( (, 2) \), \( (2, \infty) \). \[ r'(x) = \frac{-8x}{(x^2-4)^2} \] - **For \( x < -2 \):** Pick \( x = -3 \): \[ r'(-3) = \frac{-8(-3)}{((-3)^2-4)^2} = \frac{24}{(9-4)^2} = \frac{24}{25} > \] So, **increasing**. - **For \( -2 < x < \):** Pick \( x = -1 \): \[ r'(-1) = \frac{-8(-1)}{((-1)^2-4)^2} = \frac{8}{(1-4)^2} = \frac{8}{9} > \] So, **increasing**. - **For \( < x < 2 \):** Pick \( x = 1 \): \[ r'(1) = \frac{-8(1)}{(1-4)^2} = \frac{-8}{9} < \] So, **decreasing**. - **For \( x > 2 \):** Pick \( x = 3 \): \[ r'(3) = \frac{-8(3)}{(9-4)^2} = \frac{-24}{25} < \] So, **decreasing**. **Explanation:** Function increases on \( (-\infty, -2) \) and \( (-2, ) \), decreases on \( (, 2) \) and \( (2, \infty) \). --- ## **Final Answer** - **Increasing on:** \( (-\infty, -2) \) and \( (-2, ) \) - **Decreasing on:** \( (, 2) \) and \( (2, \infty) \) - **Relative maximum at:** \( x = \), since the function changes from increasing to decreasing at this point. There are no relative minima in the domain.