Domain Analysis: Finding Valid Inputs for (x+4)² - 10¼

Question

Find the positive and negative domains of the function below:

y=(x+4)21014 y=\left(x+4\right)^2-10\frac{1}{4}

Step-by-Step Solution

To determine the positive and negative domains of the function, follow these steps:

  • Step 1: Solve (x+4)2=414(x+4)^2 = \frac{41}{4}.
  • Step 2: This implies x+4=±412x+4 = \pm \frac{\sqrt{41}}{2}.
  • Step 3: Solving these gives the roots: x=8+412x = \frac{-8+\sqrt{41}}{2} and x=8412x = \frac{-8-\sqrt{41}}{2}.
  • Step 4: Divide the real number line using these roots into intervals:
    • Interval 1: x<8412x < \frac{-8-\sqrt{41}}{2}
    • Interval 2: 8412<x<8+412\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}
    • Interval 3: x>8+412x > \frac{-8+\sqrt{41}}{2}
  • Step 5: Test each interval to see where the function is greater or less than zero, using the sign of (x+4)2414(x+4)^2 - \frac{41}{4}.

Testing reveals that:

  • For Interval 1, the function is positive.
  • For Interval 2, the function is negative.
  • For Interval 3, the function is positive.

Thus, the negative domain is 8412<x<8+412 \frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2} and the positive domains are x>8+412 x > \frac{-8+\sqrt{41}}{2} or x<8412 x < \frac{-8-\sqrt{41}}{2} .

Therefore, the correct answer is:

x<0:8412<x<8+412 x < 0 :\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}

x>8+412 x > \frac{-8+\sqrt{41}}{2} or x>0:x<8412 x > 0 : x < \frac{-8-\sqrt{41}}{2}

Answer

x < 0 :\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}

x > \frac{-8+\sqrt{41}}{2} or x > 0 : x < \frac{-8-\sqrt{41}}{2}