Questions: Find the absolute extrema if they exist, as well as all values of x where they occur, for the function f(x) = 4x / (x^2 + 2). Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute maximum is , which occurs at x = - (Round the absolute maximum to two decimal places as needed. Type an exact answer for the value of x where the maximum occurs. Use a comma to separat B. There is no absolute maximum.

Solution Steps

To find the absolute extrema of the function \( f(x) = \frac{4x}{x^2 + 2} \), we need to follow these steps:

1. Find the critical points: Compute the derivative of the function and set it to zero to find critical points.
2. Evaluate the function at critical points and endpoints: Since the function is defined for all real numbers, consider the behavior as \( x \to \pm \infty \).
3. Determine the absolute extrema: Compare the function values at the critical points and endpoints to find the absolute maximum and minimum.

To find the absolute extrema, we first need to find the critical points of the function \( f(x) = \frac{4x}{x^2 + 2} \). We do this by finding the derivative and setting it equal to zero.

The derivative of \( f(x) \) using the quotient rule is: \[ f'(x) = \frac{(x^2 + 2)(4) - (4x)(2x)}{(x^2 + 2)^2} \]

Simplifying the numerator: \[ = \frac{4x^2 + 8 - 8x^2}{(x^2 + 2)^2} = \frac{-4x^2 + 8}{(x^2 + 2)^2} \]

Setting the derivative equal to zero to find critical points: \[ -4x^2 + 8 = 0 \]

Solving for \( x \): \[ -4x^2 = -8 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2} \]

Since the function is defined for all real numbers, we consider the critical points \( x = \sqrt{2} \) and \( x = -\sqrt{2} \).

Evaluate \( f(x) \) at these points: \[ f(\sqrt{2}) = \frac{4\sqrt{2}}{(\sqrt{2})^2 + 2} = \frac{4\sqrt{2}}{2 + 2} = \frac{4\sqrt{2}}{4} = \sqrt{2} \]

\[ f(-\sqrt{2}) = \frac{4(-\sqrt{2})}{(-\sqrt{2})^2 + 2} = \frac{-4\sqrt{2}}{2 + 2} = \frac{-4\sqrt{2}}{4} = -\sqrt{2} \]

The function \( f(x) \) approaches zero as \( x \to \pm \infty \). Therefore, the absolute maximum and minimum must occur at the critical points.

• The absolute maximum value is \( \sqrt{2} \) at \( x = \sqrt{2} \).
• The absolute minimum value is \(-\sqrt{2}\) at \( x = -\sqrt{2} \).

Final Answer