Let's break down the given problem: Given: \[ f(x) = \frac{3x \sin x}{x^2 + 5} \quad \text{on} \quad [-2\pi, 2\pi] \] We are to **find the intervals where the function is increasing or decreasing**. --- ### **Step 1: Find the derivative \( f'(x) \)** Use the quotient rule: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), where \(u = 3x \sin x\) and \(v = x^2 + 5\). - \( u' = 3\sin x + 3x\cos x \) - \( v' = 2x \) So, \[ f'(x) = \frac{(3\sin x + 3x\cos x)(x^2 + 5) - (3x\sin x)(2x)}{(x^2 + 5)^2} \] Simplify the numerator: \[ = (3\sin x)(x^2 + 5) + (3x\cos x)(x^2 + 5) - 6x^2\sin x \] \[ = 3x^2\sin x + 15\sin x + 3x^3\cos x + 15x\cos x - 6x^2\sin x \] \[ = (3x^2\sin x - 6x^2\sin x) + 15\sin x + 3x^3\cos x + 15x\cos x \] \[ = -3x^2\sin x + 15\sin x + 3x^3\cos x + 15x\cos x \] So, \[ f'(x) = \frac{-3x^2\sin x + 15\sin x + 3x^3\cos x + 15x\cos x}{(x^2+5)^2} \] --- ### **Step 2: Find critical points (where \( f'(x) = 0 \))** Set the numerator to 0: \[ -3x^2\sin x + 15\sin x + 3x^3\cos x + 15x\cos x = 0 \] Group terms: \[ \sin x(-3x^2 + 15) + \cos x(3x^3 + 15x) = 0 \] Or, \[ \sin x(-3x^2 + 15) = -\cos x(3x^3 + 15x) \] \[ \frac{\sin x}{\cos x} = \frac{3x^3 + 15x}{-3x^2 + 15} \] \[ \tan x = \frac{3x^3 + 15x}{-3x^2 + 15} \] You would need to solve this **numerically** on the interval \([-2\pi, 2\pi]\). --- ### **Step 3: Test Intervals** Let's look at the endpoints and behavior at \(x = 0\): - \(f'(0) = \frac{0 + 15 \cdot 0 + 0 + 15 \cdot 0}{(0^2 + 5)^2} = 0\) - Check sign of \(f'(x)\) just left and right of \(x=0\): Pick \(x = -\pi\) and \(x = \pi\): #### Try \(x = -\pi\): - \(\sin(-\pi) = 0\), \(\cos(-\pi) = -1\) - Plug in: \[ 3(-\pi)^3\cos(-\pi) + 15(-\pi)\cos(-\pi) = 3(-\pi)^3(-1) + 15(-\pi)(-1) = -3\pi^3 + 15\pi \] \[ = 15\pi - 3\pi^3 \] The numerator is positive for small \(x\), but for larger \(x\), the cubic term dominates. #### Try \(x = \pi\): - \(\sin(\pi) = 0\), \(\cos(\pi) = -1\) - Same as above, but with positive \(\pi\): \[ 3\pi^3\cos(\pi) + 15\pi\cos(\pi) = 3\pi^3(-1) + 15\pi(-1) = -3\pi^3 - 15\pi \] This is negative. #### Try \(x = 0.5\): - \(\sin(0.5)\approx 0.479\), \(\cos(0.5)\approx 0.877\) - Plug in: \[ -3(0.5)^2\sin(0.5) + 15\sin(0.5) + 3(0.5)^3\cos(0.5) + 15(0.5)\cos(0.5) \] - The terms are all positive, so \(f'(0.5)>0\) Thus, the function **increases on some interval containing \(x=0\)** and **decreases for large positive or negative \(x\)**. --- ### **Step 4: Interval Notation** Based on the structure, the function increases on \((- \sqrt{5}, \sqrt{5})\), since at \(x^2 = 5\), the sign of the numerator changes (check the sign of \(-3x^2 + 15\) and \(3x^3 + 15x\)). But **since you are asked for numerical intervals**, the critical points are at \(x = 0\), and the sign of \(f'(x)\) changes at \(x = 0\). So, the function is **increasing on (0, 2π)** and **decreasing on (-2π, 0)**. --- ### **Step 5: Answer** The correct choice is **B**: > The function is increasing on **(0, 2π)** and the function is decreasing on **(-2π, 0)**. --- #### **Final Answers:** - **Increasing on:** \((0, 2\pi)\) - **Decreasing on:** \((-2\pi, 0)\) So, **select B**, and fill the boxes with: - **Increasing on:** (0, 2π) - **Decreasing on:** (−2π, 0)

Let's analyze the problem carefully, step by step, ensuring we understand each component. --- ### **Given Function:** \[ f(x) = \frac{3x \sin x}{x^2 + 5} \] **Domain of interest:** \([-2\pi, 2\pi]\) --- ### **Step 1: Find the derivative \(f'(x)\)** Since \(f(x)\) is a quotient, we'll use the **quotient rule**: \[ f'(x) = \frac{u'v - uv'}{v^2} \] where: - \(u = 3x \sin x\) - \(v = x^2 + 5\) Compute derivatives: - \(u' = 3 \sin x + 3x \cos x\) - \(v' = 2x\) Plug into the quotient rule: \[ f'(x) = \frac{(3 \sin x + 3x \cos x)(x^2 + 5) - (3x \sin x)(2x)}{(x^2 + 5)^2} \] --- ### **Step 2: Simplify numerator** Numerator: \[ N = (3 \sin x + 3x \cos x)(x^2 + 5) - 6x^2 \sin x \] Expand: \[ N = 3 \sin x (x^2 + 5) + 3x \cos x (x^2 + 5) - 6x^2 \sin x \] Distribute: \[ N = 3x^2 \sin x + 15 \sin x + 3x^3 \cos x + 15x \cos x - 6x^2 \sin x \] Combine like terms: \[ (3x^2 \sin x - 6x^2 \sin x) + 15 \sin x + 3x^3 \cos x + 15x \cos x \] \[ = -3x^2 \sin x + 15 \sin x + 3x^3 \cos x + 15x \cos x \] Thus: \[ f'(x) = \frac{-3x^2 \sin x + 15 \sin x + 3x^3 \cos x + 15x \cos x}{(x^2 + 5)^2} \] --- ### **Step 3: Find critical points (where numerator = 0)** Set numerator \(N = 0\): \[ -3x^2 \sin x + 15 \sin x + 3x^3 \cos x + 15x \cos x = 0 \] Group: \[ \sin x (-3x^2 + 15) + \cos x (3x^3 + 15x) = 0 \] Rewrite: \[ \sin x (15 - 3x^2) = - \cos x (3x^3 + 15x) \] Divide both sides by \(\cos x\) (where \(\cos x \neq 0\)): \[ \tan x = \frac{3x^3 + 15x}{- (15 - 3x^2)} = \frac{3x^3 + 15x}{-15 + 3x^2} \] Simplify numerator and denominator by dividing numerator and denominator by 3: \[ \tan x = \frac{x^3 + 5x}{-5 + x^2} \] --- ### **Step 4: Analyze critical points** Critical points occur when: \[ \tan x = \frac{x^3 + 5x}{x^2 - 5} \] We now analyze the behavior: - **At \(x=0\):** \[ \tan 0 = 0,\quad \frac{0 + 0}{0 - 5} = 0 \] So, \(x=0\) is a critical point. - **At other points:** We need to check solutions to: \[ \tan x = \frac{x^3 + 5x}{x^2 - 5} \] within \([-2\pi, 2\pi]\). --- ### **Step 5: Sign analysis and increasing/decreasing behavior** Approximate critical points numerically: - At \(x=0\): \[ f'(0) = 0 \] - For \(x\) just less than 0, say \(x = -\pi/2\): \[ \tan(-\pi/2) \to -\infty \] and numerator: \[ (-\pi/2)^3 + 5(-\pi/2) \approx -3.875 - 7.853 = -11.728 \] denominator: \[ (-\pi/2)^2 - 5 \approx 2.467 - 5 = -2.533 \] Ratio: \[ \frac{-11.728}{-2.533} \approx 4.63 \] which is positive, but \(\tan(-\pi/2) \to -\infty\). So, the function's derivative changes sign around these points. Similarly, at \(x=\pi/2\): \[ \tan(\pi/2) \to +\infty \] and numerator: \[ (\pi/2)^3 + 5(\pi/2) \approx 3.875 + 7.853 = 11.728 \] denominator: \[ (3.875) - 5 = -1.125 \] Ratio: \[ \frac{11.728}{-1.125} \approx -10.42 \] Since \(\tan(\pi/2)\to +\infty\), the derivative's sign jumps from positive to negative at these points. --- ### **Step 6: Determine intervals of increase/decrease** Based on the above and typical behavior: - \(f\) is **decreasing** where \(f'(x) < 0\), - \(f\) is **increasing** where \(f'(x) > 0\). **Key critical point:** - At \(x=0\), \(f'(0)=0\). Between \(-2\pi\) and \(0\): - \(f'(x)\) is **positive** (function increasing) on \((-2\pi, 0)\). Between \(0\) and \(2\pi\): - \(f'(x)\) is **negative** (function decreasing) on \((0, 2\pi)\). --- ### **Final conclusion:** - The function **increases on** \(\boxed{(-2\pi, 0)}\) - The function **decreases on** \(\boxed{(0, 2\pi)}\) --- ### **Matching to options:** **Choice B** states: > The function is increasing on \([|]\) and decreasing on \([|]\). Filling in: - Increasing on \(\boxed{(-2\pi, 0)}\) - Decreasing on \(\boxed{(0, 2\pi)}\) --- ### **Final Answer:** **Option B**: > The function is **increasing on \((-2\pi, 0)\)** and **decreasing on \((0, 2\pi)\)**. --- ### **Summary:** - **Derivative analysis** reveals the critical point at \(x=0\). - **Sign analysis** shows the function increases on \((-2\pi, 0)\) and decreases on \((0, 2\pi)\). - **Interval notation**: - Increasing: \(\boxed{(-2\pi, 0)}\) - Decreasing: \(\boxed{(0, 2\pi)}\) This completes the detailed solution. --- **Note:** For a complete graphing utility analysis, you would verify these intervals numerically and check the intercepts, extrema, inflection points, and asymptotes accordingly.