Find the critical numbers, the intervals on which \( f(x) \) is increasing, the intervals on which \( f(x) \) is decreasing, and the local extrema. Do not graph. \( f(x)=2 x+\frac{8}{x} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice, A. The critical number(s) is(are) (Type integers or simplified fractions. Use a comma to separate answers as needed.) Belect the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is increasing on \( \square \). (Type your answer in interval notation. Type integers or simplified fractions. Use a comma to separate answers as needed.) Be function is never increasing.

To find the critical numbers, intervals of increase and decrease, and local extrema for the function \( f(x) = 2x + \frac{8}{x} \), we will follow these steps: 1. **Find the derivative \( f'(x) \)**. 2. **Set the derivative equal to zero to find critical numbers**. 3. **Determine the sign of the derivative to find intervals of increase and decrease**. 4. **Identify local extrema based on the critical numbers and the sign of the derivative**. ### Step 1: Find the derivative The function is given by: \[ f(x) = 2x + \frac{8}{x} \] To find the derivative \( f'(x) \), we will differentiate each term: \[ f'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}\left(\frac{8}{x}\right) \] Calculating the derivative of \( \frac{8}{x} \) using the power rule: \[ \frac{d}{dx}\left(\frac{8}{x}\right) = \frac{d}{dx}(8x^{-1}) = -8x^{-2} = -\frac{8}{x^2} \] Thus, the derivative is: \[ f'(x) = 2 - \frac{8}{x^2} \] ### Step 2: Set the derivative equal to zero Now, we set the derivative equal to zero to find critical numbers: \[ 2 - \frac{8}{x^2} = 0 \] Solving for \( x \): \[ \frac{8}{x^2} = 2 \] Multiplying both sides by \( x^2 \): \[ 8 = 2x^2 \] Dividing by 2: \[ 4 = x^2 \] Taking the square root: \[ x = 2 \quad \text{or} \quad x = -2 \] ### Step 3: Determine intervals of increase and decrease Next, we will analyze the sign of \( f'(x) \) in the intervals determined by the critical numbers \( x = -2 \) and \( x = 2 \). The intervals to test are \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \). 1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \) \[ f'(-3) = 2 - \frac{8}{(-3)^2} = 2 - \frac{8}{9} = \frac{18}{9} - \frac{8}{9} = \frac{10}{9} > 0 \] 2. **Interval \( (-2, 2) \)**: Choose \( x = 0 \) (note that \( f(x) \) is undefined at \( x = 0 \), so we can choose \( x = -1 \)) \[ f'(-1) = 2 - \frac{8}{(-1)^2} = 2 - 8 = -6 < 0 \] 3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \) \[ f'(3) = 2 - \frac{8}{3^2} = 2 - \frac{8}{9} = \frac{18}{9} - \frac{8}{9} = \frac{10}{9} > 0 \] ### Summary of intervals - \( f'(x) > 0 \) on \( (-\infty, -2) \) and \( (2, \infty) \) (increasing) - \( f'(x) < 0 \) on \( (-2, 2) \) (decreasing) ### Step 4: Identify local extrema - At \( x = -2 \), the function changes from increasing to decreasing, indicating a local maximum. - At \( x = 2 \), the function changes from decreasing to increasing, indicating a local minimum. ### Final Answers 1. **Critical numbers**: - The critical number(s) is(are): \( -2, 2 \) 2. **Intervals of increase**: - The function is increasing on \( (-\infty, -2) \) and \( (2, \infty) \). 3. **Intervals of decrease**: - The function is decreasing on \( (-2, 2) \). 4. **Local extrema**: - Local maximum at \( x = -2 \) - Local minimum at \( x = 2 \) Thus, the answers are: A. The critical number(s) is(are): \( -2, 2 \) B. The function is increasing on \( (-\infty, -2) \cup (2, \infty) \).