Answer :
To determine the domain of the function [tex]\( l(x) \)[/tex], we need to consider each piece of the function separately and then combine their domains. The function is defined piecewise as follows:
1. [tex]\( l(x) = \frac{x^2 + 2}{x^2 - 25} \)[/tex] for [tex]\( x > 0 \)[/tex]
2. [tex]\( l(x) = x + \sqrt{x^2 + x} \)[/tex] for [tex]\( x \leq 0 \)[/tex]
Step 1: Find the domain for the first piece, [tex]\(\frac{x^2 + 2}{x^2 - 25}\)[/tex] where [tex]\(x > 0\)[/tex].
- The expression is a rational function which means it's defined for all [tex]\(x > 0\)[/tex] except where the denominator is zero.
- The denominator is zero when [tex]\( x^2 - 25 = 0 \)[/tex].
- Solving for [tex]\(x\)[/tex], we have [tex]\( x^2 = 25 \)[/tex] giving [tex]\( x = \pm 5 \)[/tex].
- Since we are only considering [tex]\( x > 0 \)[/tex], the value [tex]\( x = 5 \)[/tex] must be excluded from the domain.
- Therefore, the domain for this piece is all positive [tex]\( x \)[/tex] except [tex]\( x = 5 \)[/tex]: [tex]\( \left(0, 5\right) \cup \left(5, \infty\right) \)[/tex].
Step 2: Find the domain for the second piece, [tex]\(x + \sqrt{x^2 + x}\)[/tex] where [tex]\(x \leq 0\)[/tex].
- The expression involves a square root, which means the expression inside the root must be non-negative to ensure it's defined for real numbers.
- The condition we need to satisfy is [tex]\( x^2 + x \geq 0 \)[/tex].
- Factor the inequality: [tex]\( x(x + 1) \geq 0 \)[/tex].
- The critical points are [tex]\( x = 0 \)[/tex] and [tex]\( x = -1 \)[/tex].
- Testing intervals around these critical points, we find:
- For [tex]\( x \leq -1 \)[/tex], say [tex]\( x = -2 \)[/tex], [tex]\(x(x+1) = (-2)(-1) = 2 > 0\)[/tex], so the inequality holds.
- For [tex]\( -1 < x \leq 0 \)[/tex], say [tex]\( x = -0.5 \)[/tex], [tex]\(x(x+1) = (-0.5)(0.5) = -0.25 < 0\)[/tex], so the inequality does not hold.
- Therefore, the solution to [tex]\( x(x + 1) \geq 0 \)[/tex] for [tex]\( x \leq 0 \)[/tex] is just [tex]\( x \leq -1\)[/tex].
- The domain for this piece is [tex]\( (-\infty, 0] \)[/tex].
Step 3: Combine the domains from both pieces.
- The domain for the entire function [tex]\( l(x) \)[/tex] is the union of the two parts:
- From the first piece: [tex]\( \left(0, 5\right) \cup \left(5, \infty\right) \)[/tex]
- From the second piece: [tex]\( (-\infty, 0] \)[/tex]
Therefore, the overall domain for [tex]\( l(x) \)[/tex] is:
[tex]\[ (-\infty, 0] \cup \left(0, 5\right) \cup \left(5, \infty\right) \][/tex]
1. [tex]\( l(x) = \frac{x^2 + 2}{x^2 - 25} \)[/tex] for [tex]\( x > 0 \)[/tex]
2. [tex]\( l(x) = x + \sqrt{x^2 + x} \)[/tex] for [tex]\( x \leq 0 \)[/tex]
Step 1: Find the domain for the first piece, [tex]\(\frac{x^2 + 2}{x^2 - 25}\)[/tex] where [tex]\(x > 0\)[/tex].
- The expression is a rational function which means it's defined for all [tex]\(x > 0\)[/tex] except where the denominator is zero.
- The denominator is zero when [tex]\( x^2 - 25 = 0 \)[/tex].
- Solving for [tex]\(x\)[/tex], we have [tex]\( x^2 = 25 \)[/tex] giving [tex]\( x = \pm 5 \)[/tex].
- Since we are only considering [tex]\( x > 0 \)[/tex], the value [tex]\( x = 5 \)[/tex] must be excluded from the domain.
- Therefore, the domain for this piece is all positive [tex]\( x \)[/tex] except [tex]\( x = 5 \)[/tex]: [tex]\( \left(0, 5\right) \cup \left(5, \infty\right) \)[/tex].
Step 2: Find the domain for the second piece, [tex]\(x + \sqrt{x^2 + x}\)[/tex] where [tex]\(x \leq 0\)[/tex].
- The expression involves a square root, which means the expression inside the root must be non-negative to ensure it's defined for real numbers.
- The condition we need to satisfy is [tex]\( x^2 + x \geq 0 \)[/tex].
- Factor the inequality: [tex]\( x(x + 1) \geq 0 \)[/tex].
- The critical points are [tex]\( x = 0 \)[/tex] and [tex]\( x = -1 \)[/tex].
- Testing intervals around these critical points, we find:
- For [tex]\( x \leq -1 \)[/tex], say [tex]\( x = -2 \)[/tex], [tex]\(x(x+1) = (-2)(-1) = 2 > 0\)[/tex], so the inequality holds.
- For [tex]\( -1 < x \leq 0 \)[/tex], say [tex]\( x = -0.5 \)[/tex], [tex]\(x(x+1) = (-0.5)(0.5) = -0.25 < 0\)[/tex], so the inequality does not hold.
- Therefore, the solution to [tex]\( x(x + 1) \geq 0 \)[/tex] for [tex]\( x \leq 0 \)[/tex] is just [tex]\( x \leq -1\)[/tex].
- The domain for this piece is [tex]\( (-\infty, 0] \)[/tex].
Step 3: Combine the domains from both pieces.
- The domain for the entire function [tex]\( l(x) \)[/tex] is the union of the two parts:
- From the first piece: [tex]\( \left(0, 5\right) \cup \left(5, \infty\right) \)[/tex]
- From the second piece: [tex]\( (-\infty, 0] \)[/tex]
Therefore, the overall domain for [tex]\( l(x) \)[/tex] is:
[tex]\[ (-\infty, 0] \cup \left(0, 5\right) \cup \left(5, \infty\right) \][/tex]
