Find the cntical numbers, the intervals on which \( f(x) \) is increasing, the inlervals on which \( f(x) \) is decreasing, and the local extrema. Do nol graph. \[ f(x)=2 x+\frac{8}{x} \] Select the correct choice below and, if necessary, fill in the answer box to complete your choce. A. The function is decreasing on \( \square \) . (Type your answer in interval notation. Type integers or simplilied fractions. Use a comma lo separale answers as needed.) B. The function is never decreasing. Select the correct choice below and, if necessary, fill in the airswer boxes to complele your choice. (Type infegers or simplified fractions.) A. The function has a local minimum \( \mathrm{f}(\square)=\square \), and no local maximum. B. The function hes a local maximum f( \( \square \) \( = \) \( \square \) and a local minimum f( \( \square \) ) \( = \) \( \square \) C. The function has a local maximum f( \( \square \) \( 7= \) \( \square \) , and no local minimum. D. The function has no local exirema.

To find the critical numbers of the function \( f(x) = 2x + \frac{8}{x} \), we first need to compute the derivative \( f'(x) \) and set it equal to zero. Calculating the derivative: \[ f'(x) = 2 - \frac{8}{x^2} \] Setting the derivative equal to zero for critical points: \[ 2 - \frac{8}{x^2} = 0 \implies \frac{8}{x^2} = 2 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2 \] Next, we determine the intervals for increasing and decreasing by checking test points around the critical points: 1. For \( x < -2 \): Choose \( x = -3 \): \[ f'(-3) = 2 - \frac{8}{9} > 0 \quad (\text{Increasing}) \] 2. For \( -2 < x < 2 \): Choose \( x = 0 \) (not valid here since \( f(x) \) is undefined at zero): Instead, take \( x = -1 \): \[ f'(-1) = 2 - 8 < 0 \quad (\text{Decreasing}) \] 3. For \( x > 2 \): Choose \( x = 3 \): \[ f'(3) = 2 - \frac{8}{9} > 0 \quad (\text{Increasing}) \] Thus, the function is: - Increasing on \( (-\infty, -2) \) and \( (2, \infty) \) - Decreasing on \( (-2, 2) \) Now, let's summarize the findings for local extrema: - The function has a local minimum at \( x = 2 \): \[ f(2) = 2(2) + \frac{8}{2} = 4 + 4 = 8 \] - The function has no local maximum since it only changes from increasing to decreasing at \( x = -2 \), but does not achieve a maximum point. So the answers are: A. The function is decreasing on \( (-2, 2) \). A. The function has a local minimum \( f(2)=8 \), and no local maximum.