Find the critical numbers, the intervals on which \( l(x) \) is incroasing, the inturvals on which \( f(x) \) is decreasing, and the local extroma. Do not graph \[ T(x)=2 x+\frac{8}{x} \] Select the corred choice below and, il necossary, fill in the arrswer box to completo your choice. A. The function is decreasing ont \( \square \) (Type your answer in inte val nolalion Type inlogers or simplifiod fractons. Use a corma lo separale answers as neerded) B. The function is never decreasing. Solect the correct choice below and, il necessary, fill in the answar boxcs lo complete your choice. (Type inlegers or simplifed fractions.) A. The furction has a local minimum \( 1(\square=\square \), und no local maxinum. B. The function has a local maximumn \( f(\square)=\square \) and a local minimum \( f(\square)=\square \). C. The function has a local maximum \( f(\square)=\square \), and no local mirimum,

**Step 1. Determine the Domain** The function is \[ T(x)=2x+\frac{8}{x}. \] Since there is a division by \( x \), the domain is \[ \{x \in \mathbb{R} \mid x \neq 0\}. \] --- **Step 2. Find the Critical Numbers** 1. Compute the derivative: \[ T'(x)=\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(\frac{8}{x}\right)=2-\frac{8}{x^2}. \] 2. Set the derivative equal to zero: \[ 2-\frac{8}{x^2}=0 \quad \Longrightarrow \quad 2=\frac{8}{x^2} \quad \Longrightarrow \quad 2x^2=8 \quad \Longrightarrow \quad x^2=4. \] 3. Solve for \( x \): \[ x=\pm 2. \] Thus, the critical numbers are \( x=-2 \) and \( x=2 \). --- **Step 3. Determine the Intervals of Increase and Decrease using the First Derivative Test** The derivative is \[ T'(x)=2-\frac{8}{x^2}. \] Recall the domain excludes \( x=0 \). We consider the intervals: 1. **Interval \((-\infty, -2)\):** Let \( x=-3 \). Then, \[ T'(-3)=2-\frac{8}{9} \approx 2-0.89=1.11>0. \] So, \( T(x) \) is increasing on \((-\infty,-2)\). 2. **Interval \((-2, 0)\):** Let \( x=-1 \). Then, \[ T'(-1)=2-\frac{8}{1}=2-8=-6<0. \] So, \( T(x) \) is decreasing on \((-2,0)\). 3. **Interval \((0, 2)\):** Let \( x=1 \). Then, \[ T'(1)=2-\frac{8}{1}=2-8=-6<0. \] So, \( T(x) \) is decreasing on \((0,2)\). 4. **Interval \((2, \infty)\):** Let \( x=3 \). Then, \[ T'(3)=2-\frac{8}{9}\approx2-0.89=1.11>0. \] So, \( T(x) \) is increasing on \((2,\infty)\). - **Decreasing Intervals:** The function is decreasing on the union of the intervals \((-2,0)\) and \((0,2)\). --- **Step 4. Identify the Local Extrema** 1. At \( x=-2 \): - For \( x < -2 \), \( T'(x)>0 \) (increasing). - For \( x > -2 \) (but \( x \) still in \((-2,0)\)), \( T'(x)<0 \) (decreasing). This change from increasing to decreasing indicates a **local maximum** at \( x=-2 \). 2. At \( x=2 \): - For \( x < 2 \) (but \( x \) in \((0,2)\)), \( T'(x)<0 \) (decreasing). - For \( x > 2 \), \( T'(x)>0 \) (increasing). This change from decreasing to increasing indicates a **local minimum** at \( x=2 \). 3. Compute \( T(x) \) at the critical values: - At \( x=-2 \): \[ T(-2)=2(-2)+\frac{8}{-2}=-4-4=-8. \] - At \( x=2 \): \[ T(2)=2(2)+\frac{8}{2}=4+4=8. \] --- **Final Answers** 1. **Critical numbers:** \( x=-2 \) and \( x=2 \). 2. **Intervals on which \( T(x) \) is increasing:** \[ (-\infty,-2) \text{ and } (2,\infty). \] 3. **Intervals on which \( T(x) \) is decreasing:** \[ (-2,0) \text{ and } (0,2). \] 4. **Local extrema:** - **Local maximum:** \( T(-2)=-8 \). - **Local minimum:** \( T(2)=8 \). --- **Selection Responses** - For the question regarding when the function is decreasing, select: **A.** The function is decreasing on \( (-2,0) \cup (0,2) \). - For the question regarding local extrema, select: **B.** The function has a local maximum \( T(-2)=-8 \) and a local minimum \( T(2)=8 \).