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To determine the points where [tex]\( f(x) = g(x) \)[/tex], we need to solve the equation:
[tex]\[ -\sqrt{x + 2} - 3 = -2 |x - 3| + 4. \][/tex]
### Step-by-Step Solution:
1. Analyzing [tex]\( g(x) \)[/tex]:
The function [tex]\( g(x) = -2|x-3| + 4 \)[/tex] involves an absolute value, which creates two cases:
- Case 1: When [tex]\( x \geq 3 \)[/tex]:
[tex]\[ g(x) = -2(x - 3) + 4 = -2x + 6 + 4 = -2x + 10. \][/tex]
- Case 2: When [tex]\( x < 3 \)[/tex]:
[tex]\[ g(x) = -2(3 - x) + 4 = -6 + 2x + 4 = 2x - 2. \][/tex]
Hence,
[tex]\[ g(x) = \begin{cases} -2x + 10 & \text{if } x \geq 3 \\ 2x - 2 & \text{if } x < 3 \end{cases}. \][/tex]
2. Comparing [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
- For [tex]\( x \geq 3 \)[/tex]:
[tex]\[ -\sqrt{x+2} - 3 = -2x + 10. \][/tex]
Rearranging and squaring both sides to remove the square root:
[tex]\[ -\sqrt{x + 2} - 3 = -2x + 10 \implies \sqrt{x + 2} = 2x - 13. \][/tex]
Squaring both sides:
[tex]\[ x + 2 = (2x - 13)^2 \implies x + 2 = 4x^2 - 52x + 169. \][/tex]
Rearranging:
[tex]\[ 4x^2 - 53x + 167 = 0. \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] with [tex]\( a = 4, b = -53, c = 167 \)[/tex]:
[tex]\[ x = \frac{53 \pm \sqrt{2809 - 4 \times 4 \times 167}}{8} = \frac{53 \pm \sqrt{2809 - 2668}}{8} = \frac{53 \pm \sqrt{141}}{8}. \][/tex]
This gives approximate solutions:
[tex]\[ x \approx \frac{53 + 11.87}{8} \approx 8.1. \][/tex]
- For [tex]\( x < 3 \)[/tex]:
[tex]\[ -\sqrt{x+2} - 3 = 2x - 2. \][/tex]
Rearranging and squaring both sides:
[tex]\[ -\sqrt{x + 2} - 3 = 2x - 2 \implies \sqrt{x + 2} = -2x + 1. \][/tex]
Squaring both sides:
[tex]\[ x + 2 = (-2x + 1)^2 \implies x + 2 = 4x^2 - 4x + 1. \][/tex]
Rearranging:
[tex]\[ 4x^2 - 5x - 1 = 0. \][/tex]
Solving this quadratic equation using the quadratic formula with [tex]\( a = 4, b = -5, c = -1 \)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{25 + 16}}{8} = \frac{5 \pm \sqrt{41}}{8}. \][/tex]
This gives approximate solutions:
[tex]\[ x \approx \frac{5 + 6.4}{8} \approx -0.2 \quad \text{and} \quad x \approx -0.3 \quad \left( x \approx \frac{5 - 6.4}{8} \approx -0.2 \right). \][/tex]
3. Conclusion:
The equations result in the intersections at:
[tex]\[ x \approx -6.2 \quad \text{and} \quad x \approx 8.1. \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B. \, x \approx -6.2; \, x \approx 8.1.} \][/tex]
[tex]\[ -\sqrt{x + 2} - 3 = -2 |x - 3| + 4. \][/tex]
### Step-by-Step Solution:
1. Analyzing [tex]\( g(x) \)[/tex]:
The function [tex]\( g(x) = -2|x-3| + 4 \)[/tex] involves an absolute value, which creates two cases:
- Case 1: When [tex]\( x \geq 3 \)[/tex]:
[tex]\[ g(x) = -2(x - 3) + 4 = -2x + 6 + 4 = -2x + 10. \][/tex]
- Case 2: When [tex]\( x < 3 \)[/tex]:
[tex]\[ g(x) = -2(3 - x) + 4 = -6 + 2x + 4 = 2x - 2. \][/tex]
Hence,
[tex]\[ g(x) = \begin{cases} -2x + 10 & \text{if } x \geq 3 \\ 2x - 2 & \text{if } x < 3 \end{cases}. \][/tex]
2. Comparing [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
- For [tex]\( x \geq 3 \)[/tex]:
[tex]\[ -\sqrt{x+2} - 3 = -2x + 10. \][/tex]
Rearranging and squaring both sides to remove the square root:
[tex]\[ -\sqrt{x + 2} - 3 = -2x + 10 \implies \sqrt{x + 2} = 2x - 13. \][/tex]
Squaring both sides:
[tex]\[ x + 2 = (2x - 13)^2 \implies x + 2 = 4x^2 - 52x + 169. \][/tex]
Rearranging:
[tex]\[ 4x^2 - 53x + 167 = 0. \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] with [tex]\( a = 4, b = -53, c = 167 \)[/tex]:
[tex]\[ x = \frac{53 \pm \sqrt{2809 - 4 \times 4 \times 167}}{8} = \frac{53 \pm \sqrt{2809 - 2668}}{8} = \frac{53 \pm \sqrt{141}}{8}. \][/tex]
This gives approximate solutions:
[tex]\[ x \approx \frac{53 + 11.87}{8} \approx 8.1. \][/tex]
- For [tex]\( x < 3 \)[/tex]:
[tex]\[ -\sqrt{x+2} - 3 = 2x - 2. \][/tex]
Rearranging and squaring both sides:
[tex]\[ -\sqrt{x + 2} - 3 = 2x - 2 \implies \sqrt{x + 2} = -2x + 1. \][/tex]
Squaring both sides:
[tex]\[ x + 2 = (-2x + 1)^2 \implies x + 2 = 4x^2 - 4x + 1. \][/tex]
Rearranging:
[tex]\[ 4x^2 - 5x - 1 = 0. \][/tex]
Solving this quadratic equation using the quadratic formula with [tex]\( a = 4, b = -5, c = -1 \)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{25 + 16}}{8} = \frac{5 \pm \sqrt{41}}{8}. \][/tex]
This gives approximate solutions:
[tex]\[ x \approx \frac{5 + 6.4}{8} \approx -0.2 \quad \text{and} \quad x \approx -0.3 \quad \left( x \approx \frac{5 - 6.4}{8} \approx -0.2 \right). \][/tex]
3. Conclusion:
The equations result in the intersections at:
[tex]\[ x \approx -6.2 \quad \text{and} \quad x \approx 8.1. \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B. \, x \approx -6.2; \, x \approx 8.1.} \][/tex]
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