(1 point) Determine where the absolute extrema of $f(x)=\frac{4 x}{x^{2}+1}$ on the interval $[-4,0]$ occur. 1. The absolute maximum occurs at $x=$ 2. The absolute minimum occurs at $x=$ Note: You can earn partial credit on this problem.

Step 1 ：\(f'(x) = \frac{d}{dx}(\frac{4x}{x^2+1})\) Using the quotient rule for differentiation, we get: \(f'(x) = \frac{(x^2+1)(4) - 4x(2x)}{(x^2+1)^2} = \frac{4x^2+4 - 8x^2}{(x^2+1)^2} = \frac{-4x^2+4}{(x^2+1)^2}\)

Step 2 ：Find the critical points by setting the derivative equal to zero and solving for x. \(0 = \frac{-4x^2+4}{(x^2+1)^2}\) This equation is only true when the numerator is zero, so we solve \(-4x^2+4 = 0\) for x: \(4x^2 = 4\) \(x^2 = 1\) \(x = \pm 1\) However, only \(x = -1\) is in the interval \([-4,0]\), so \(x = -1\) is the only critical point in the interval.

Step 3 ：Evaluate the function at the critical point and at the endpoints of the interval. \(f(-4) = \frac{4(-4)}{(-4)^2+1} = \frac{-16}{17}\) \(f(-1) = \frac{4(-1)}{(-1)^2+1} = -2\) \(f(0) = \frac{4(0)}{(0)^2+1} = 0\)

Step 4 ：Compare these values to find the absolute maximum and minimum. The absolute maximum value of \(f(x)\) on the interval \([-4,0]\) is \(0\) and it occurs at \(x = 0\). The absolute minimum value of \(f(x)\) on the interval \([-4,0]\) is \(-2\) and it occurs at \(x = -1\). So, the absolute maximum occurs at \(x = 0\) and the absolute minimum occurs at \(x = -1\).

Step 5 ：\(\boxed{\text{The absolute maximum occurs at } x = 0 \text{ and the absolute minimum occurs at } x = -1}\)