Answer :
Let's start by determining the range of the function [tex]\( f(x) = -2\sqrt{x-3} + 8 \)[/tex].
1. Domain of [tex]\( f(x) \)[/tex]:
- The function involves [tex]\( \sqrt{x - 3} \)[/tex].
- The square root function is defined for [tex]\( x - 3 \geq 0 \)[/tex].
- Therefore, [tex]\( x \geq 3 \)[/tex].
2. Analyze [tex]\( f(x) \)[/tex]:
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -2\sqrt{3-3} + 8 = -2 \cdot 0 + 8 = 8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Thus, [tex]\( -2\sqrt{x - 3} \to -\infty \)[/tex]
- Consequently, [tex]\( f(x) = -2\sqrt{x-3} + 8 \to -\infty \)[/tex]
The range of [tex]\( f(x) \)[/tex] is [tex]\( (-\infty, 8] \)[/tex].
Next, we need to check the range for each given [tex]\( g(x) \)[/tex]:
Option 1: [tex]\( g_1(x) = \sqrt{x-3} - 8 \)[/tex]
- Domain: [tex]\( x \geq 3 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ g_1(3) = \sqrt{3-3} - 8 = 0 - 8 = -8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Hence, [tex]\( g_1(x) \to \infty \)[/tex]
- The range of [tex]\( g_1(x) \)[/tex] is [tex]\( [-8, \infty) \)[/tex].
Option 2: [tex]\( g_2(x) = \sqrt{x-3} + 8 \)[/tex]
- Domain: [tex]\( x \geq 3 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ g_2(3) = \sqrt{3-3} + 8 = 0 + 8 = 8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Hence, [tex]\( g_2(x) \to \infty \)[/tex]
- The range of [tex]\( g_2(x) \)[/tex] is [tex]\( [8, \infty) \)[/tex].
Option 3: [tex]\( g_3(x) = -\sqrt{x+3} + 8 \)[/tex]
- Domain: [tex]\( x \geq -3 \)[/tex]
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ g_3(-3) = -\sqrt{-3+3} + 8 = -\sqrt{0} + 8 = 8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x + 3} \to \infty \)[/tex]
- Hence, [tex]\( g_3(x) \to -\infty \)[/tex]
- The range of [tex]\( g_3(x) \)[/tex] is [tex]\( (-\infty, 8] \)[/tex].
Option 4: [tex]\( g_4(x) = -\sqrt{x-3} - 8 \)[/tex]
- Domain: [tex]\( x \geq 3 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ g_4(3) = -\sqrt{3-3} - 8 = -\sqrt{0} - 8 = -8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Hence, [tex]\( g_4(x) \to -\infty \)[/tex]
- The range of [tex]\( g_4(x) \)[/tex] is [tex]\( (-\infty, -8] \)[/tex].
Comparing the ranges of these functions with that of [tex]\( f(x) \)[/tex]:
- [tex]\( f(x) \)[/tex] has range [tex]\( (-\infty, 8] \)[/tex].
- [tex]\( g_1(x) \)[/tex] has range [tex]\( [-8, \infty) \)[/tex].
- [tex]\( g_2(x) \)[/tex] has range [tex]\( [8, \infty) \)[/tex].
- [tex]\( g_3(x) = -\sqrt{x+3} + 8 \)[/tex] has range [tex]\( (-\infty, 8] \)[/tex].
- [tex]\( g_4(x) \)[/tex] has range [tex]\( (-\infty, -8] \)[/tex].
The function with the same range as [tex]\( f(x) \)[/tex] is [tex]\( g_3(x) = -\sqrt{x + 3} + 8 \)[/tex].
So, the correct function is [tex]\( g_3(x) = -\sqrt{x + 3} + 8 \)[/tex].
1. Domain of [tex]\( f(x) \)[/tex]:
- The function involves [tex]\( \sqrt{x - 3} \)[/tex].
- The square root function is defined for [tex]\( x - 3 \geq 0 \)[/tex].
- Therefore, [tex]\( x \geq 3 \)[/tex].
2. Analyze [tex]\( f(x) \)[/tex]:
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -2\sqrt{3-3} + 8 = -2 \cdot 0 + 8 = 8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Thus, [tex]\( -2\sqrt{x - 3} \to -\infty \)[/tex]
- Consequently, [tex]\( f(x) = -2\sqrt{x-3} + 8 \to -\infty \)[/tex]
The range of [tex]\( f(x) \)[/tex] is [tex]\( (-\infty, 8] \)[/tex].
Next, we need to check the range for each given [tex]\( g(x) \)[/tex]:
Option 1: [tex]\( g_1(x) = \sqrt{x-3} - 8 \)[/tex]
- Domain: [tex]\( x \geq 3 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ g_1(3) = \sqrt{3-3} - 8 = 0 - 8 = -8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Hence, [tex]\( g_1(x) \to \infty \)[/tex]
- The range of [tex]\( g_1(x) \)[/tex] is [tex]\( [-8, \infty) \)[/tex].
Option 2: [tex]\( g_2(x) = \sqrt{x-3} + 8 \)[/tex]
- Domain: [tex]\( x \geq 3 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ g_2(3) = \sqrt{3-3} + 8 = 0 + 8 = 8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Hence, [tex]\( g_2(x) \to \infty \)[/tex]
- The range of [tex]\( g_2(x) \)[/tex] is [tex]\( [8, \infty) \)[/tex].
Option 3: [tex]\( g_3(x) = -\sqrt{x+3} + 8 \)[/tex]
- Domain: [tex]\( x \geq -3 \)[/tex]
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ g_3(-3) = -\sqrt{-3+3} + 8 = -\sqrt{0} + 8 = 8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x + 3} \to \infty \)[/tex]
- Hence, [tex]\( g_3(x) \to -\infty \)[/tex]
- The range of [tex]\( g_3(x) \)[/tex] is [tex]\( (-\infty, 8] \)[/tex].
Option 4: [tex]\( g_4(x) = -\sqrt{x-3} - 8 \)[/tex]
- Domain: [tex]\( x \geq 3 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ g_4(3) = -\sqrt{3-3} - 8 = -\sqrt{0} - 8 = -8 \][/tex]
- As [tex]\( x \to \infty \)[/tex]:
- [tex]\( \sqrt{x - 3} \to \infty \)[/tex]
- Hence, [tex]\( g_4(x) \to -\infty \)[/tex]
- The range of [tex]\( g_4(x) \)[/tex] is [tex]\( (-\infty, -8] \)[/tex].
Comparing the ranges of these functions with that of [tex]\( f(x) \)[/tex]:
- [tex]\( f(x) \)[/tex] has range [tex]\( (-\infty, 8] \)[/tex].
- [tex]\( g_1(x) \)[/tex] has range [tex]\( [-8, \infty) \)[/tex].
- [tex]\( g_2(x) \)[/tex] has range [tex]\( [8, \infty) \)[/tex].
- [tex]\( g_3(x) = -\sqrt{x+3} + 8 \)[/tex] has range [tex]\( (-\infty, 8] \)[/tex].
- [tex]\( g_4(x) \)[/tex] has range [tex]\( (-\infty, -8] \)[/tex].
The function with the same range as [tex]\( f(x) \)[/tex] is [tex]\( g_3(x) = -\sqrt{x + 3} + 8 \)[/tex].
So, the correct function is [tex]\( g_3(x) = -\sqrt{x + 3} + 8 \)[/tex].
