Question 2 ind the natural domain of (a) \( f(x)=\sqrt{\frac{3 x(1-2 x)}{(x-2)(x+3)^{2}}} \)

1. Rewrite the function and note that the square root requires the radicand to be non‐negative and the denominator to be nonzero. We have \[ f(x)=\sqrt{\frac{3x(1-2x)}{(x-2)(x+3)^2}}. \] Since \(3>0\), the sign of the expression under the square root is determined by \[ \frac{x(1-2x)}{(x-2)(x+3)^2}. \] Thus, we require \[ \frac{x(1-2x)}{(x-2)(x+3)^2} \ge 0. \] Also, the denominator must be nonzero: \[ (x-2)(x+3)^2 \neq 0. \] 2. Find the zeros of the numerator and the denominator. - Numerator: \(x(1-2x)=0\) when \[ x=0 \quad \text{or} \quad 1-2x=0\Longrightarrow x=\frac{1}{2}. \] - Denominator: \[ x-2=0\Longrightarrow x=2, \qquad (x+3)^2=0\Longrightarrow x=-3. \] Note that \(x=-3\) and \(x=2\) must be excluded. 3. List all critical points: \[ x=-3,\; 0,\; \frac{1}{2},\; 2. \] These divide the real line into the intervals: \[ (-\infty,-3),\quad (-3,0),\quad (0,\tfrac{1}{2}),\quad (\tfrac{1}{2},2),\quad (2,\infty). \] 4. Determine the sign of \[ \frac{x(1-2x)}{(x-2)(x+3)^2} \] on each interval. Note that \((x+3)^2\) is always positive (except at \(x=-3\), where it is zero). - **Interval \((-\infty,-3)\):** Choose \(x=-4\). Then: \[ x=-4\ (<0),\quad 1-2(-4)=1+8=9\ (>0),\quad x-2=-4-2=-6\ (<0). \] The numerator \(x(1-2x)\) is \((-\cdot +)= -\) and the factor \(x-2\) in the denominator is negative, so \[ \frac{(-)}{(-)} = +. \] - **Interval \((-3,0)\):** Choose \(x=-1\). Then: \[ x=-1\ (<0),\quad 1-2(-1)=1+2=3\ (>0),\quad x-2=-1-2=-3\ (<0). \] The numerator is negative and the factor \(x-2\) is negative, yielding \[ \frac{(-)}{(-)} = +. \] - **Interval \((0,\tfrac{1}{2})\):** Choose \(x=0.25\). Then: \[ x=0.25\ (>0),\quad 1-2(0.25)=1-0.5=0.5\ (>0),\quad x-2=0.25-2=-1.75\ (<0). \] The numerator is positive while the factor \(x-2\) is negative, so \[ \frac{(+)}{(-)} = -. \] - **Interval \((\tfrac{1}{2},2)\):** Choose \(x=1\). Then: \[ x=1\ (>0),\quad 1-2(1)=1-2=-1\ (<0),\quad x-2=1-2=-1\ (<0). \] The numerator is \( (+)\cdot(-)= -\) and the denominator is negative, so \[ \frac{(-)}{(-)} = +. \] - **Interval \((2,\infty)\):** Choose \(x=3\). Then: \[ x=3\ (>0),\quad 1-2(3)=1-6=-5\ (<0),\quad x-2=3-2=1\ (>0). \] The numerator is negative and the denominator is positive, giving \[ \frac{(-)}{(+)} = -. \] 5. Include the points where the numerator is zero (making the fraction zero) provided the denominator is nonzero: - At \(x=0\): The numerator is \(0\) and the denominator equals \((0-2)(0+3)^2 \neq 0\); include \(x=0\). - At \(x=\frac{1}{2}\): The numerator is \(0\) and the denominator \((\frac{1}{2}-2)(\frac{1}{2}+3)^2 \neq 0\); include \(x=\frac{1}{2}\). 6. Combine the intervals where the expression under the square root is nonnegative: - From the sign analysis, the expression is nonnegative on \[ (-\infty,-3) \quad \text{and} \quad (-3,0] \quad \text{and} \quad \left[\frac{1}{2},2\right). \] - Recall \(x=-3\) and \(x=2\) are excluded due to the denominator being zero. 7. Therefore, the natural domain of \( f(x) \) is \[ (-\infty,-3)\cup (-3,0]\cup \left[\frac{1}{2},2\right). \]