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Sagot :
To solve the problem, we need to find the composite functions [tex]\( (f \circ g)(x) \)[/tex] and [tex]\( (g \circ f)(x) \)[/tex], and determine their respective domains.
### 1. Finding [tex]\( (f \circ g)(x) \)[/tex]
The notation [tex]\( (f \circ g)(x) \)[/tex] represents the function [tex]\( f(g(x)) \)[/tex].
Step-by-Step Solution:
1. Start with the inner function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \sqrt{6 - x} \][/tex]
2. Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f(\sqrt{6 - x}) \][/tex]
3. Evaluate [tex]\( f \)[/tex] at [tex]\( \sqrt{6 - x} \)[/tex]:
[tex]\[ f(\sqrt{6 - x}) = (\sqrt{6 - x})^2 + 2 \][/tex]
4. Simplify:
[tex]\[ (\sqrt{6 - x})^2 = 6 - x \][/tex]
[tex]\[ f(g(x)) = 6 - x + 2 \][/tex]
[tex]\[ f(g(x)) = 8 - x \][/tex]
Thus, [tex]\( (f \circ g)(x) = 8 - x \)[/tex].
Domain of [tex]\( (f \circ g)(x) \)[/tex]:
We need to ensure that the expression inside [tex]\( g(x) \)[/tex], [tex]\( 6 - x \)[/tex], is non-negative because we cannot take the square root of a negative number.
[tex]\[ 6 - x \geq 0 \][/tex]
[tex]\[ x \leq 6 \][/tex]
So, the domain of [tex]\( (f \circ g)(x) \)[/tex] is:
[tex]\[ (-\infty, 6] \][/tex]
### 2. Finding [tex]\( (g \circ f)(x) \)[/tex]
The notation [tex]\( (g \circ f)(x) \)[/tex] represents the function [tex]\( g(f(x)) \)[/tex].
Step-by-Step Solution:
1. Start with the inner function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x^2 + 2 \][/tex]
2. Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(x^2 + 2) \][/tex]
3. Evaluate [tex]\( g \)[/tex] at [tex]\( x^2 + 2 \)[/tex]:
[tex]\[ g(x^2 + 2) = \sqrt{6 - (x^2 + 2)} \][/tex]
4. Simplify:
[tex]\[ 6 - (x^2 + 2) = 6 - x^2 - 2 \][/tex]
[tex]\[ g(x^2 + 2) = \sqrt{4 - x^2} \][/tex]
Thus, [tex]\( (g \circ f)(x) = \sqrt{4 - x^2} \)[/tex].
Domain of [tex]\( (g \circ f)(x) \)[/tex]:
We need to ensure that the expression inside the square root, [tex]\( 4 - x^2 \)[/tex], is non-negative because we cannot take the square root of a negative number.
[tex]\[ 4 - x^2 \geq 0 \][/tex]
[tex]\[ x^2 \leq 4 \][/tex]
[tex]\[ -2 \leq x \leq 2 \][/tex]
So, the domain of [tex]\( (g \circ f)(x) \)[/tex] is:
[tex]\[ [-2, 2] \][/tex]
### Summary
1. [tex]\( (f \circ g)(x) = 8 - x \)[/tex] with domain [tex]\( (-\infty, 6] \)[/tex].
2. [tex]\( (g \circ f)(x) = \sqrt{4 - x^2} \)[/tex] with domain [tex]\( [-2, 2] \)[/tex].
### 1. Finding [tex]\( (f \circ g)(x) \)[/tex]
The notation [tex]\( (f \circ g)(x) \)[/tex] represents the function [tex]\( f(g(x)) \)[/tex].
Step-by-Step Solution:
1. Start with the inner function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \sqrt{6 - x} \][/tex]
2. Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f(\sqrt{6 - x}) \][/tex]
3. Evaluate [tex]\( f \)[/tex] at [tex]\( \sqrt{6 - x} \)[/tex]:
[tex]\[ f(\sqrt{6 - x}) = (\sqrt{6 - x})^2 + 2 \][/tex]
4. Simplify:
[tex]\[ (\sqrt{6 - x})^2 = 6 - x \][/tex]
[tex]\[ f(g(x)) = 6 - x + 2 \][/tex]
[tex]\[ f(g(x)) = 8 - x \][/tex]
Thus, [tex]\( (f \circ g)(x) = 8 - x \)[/tex].
Domain of [tex]\( (f \circ g)(x) \)[/tex]:
We need to ensure that the expression inside [tex]\( g(x) \)[/tex], [tex]\( 6 - x \)[/tex], is non-negative because we cannot take the square root of a negative number.
[tex]\[ 6 - x \geq 0 \][/tex]
[tex]\[ x \leq 6 \][/tex]
So, the domain of [tex]\( (f \circ g)(x) \)[/tex] is:
[tex]\[ (-\infty, 6] \][/tex]
### 2. Finding [tex]\( (g \circ f)(x) \)[/tex]
The notation [tex]\( (g \circ f)(x) \)[/tex] represents the function [tex]\( g(f(x)) \)[/tex].
Step-by-Step Solution:
1. Start with the inner function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x^2 + 2 \][/tex]
2. Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(x^2 + 2) \][/tex]
3. Evaluate [tex]\( g \)[/tex] at [tex]\( x^2 + 2 \)[/tex]:
[tex]\[ g(x^2 + 2) = \sqrt{6 - (x^2 + 2)} \][/tex]
4. Simplify:
[tex]\[ 6 - (x^2 + 2) = 6 - x^2 - 2 \][/tex]
[tex]\[ g(x^2 + 2) = \sqrt{4 - x^2} \][/tex]
Thus, [tex]\( (g \circ f)(x) = \sqrt{4 - x^2} \)[/tex].
Domain of [tex]\( (g \circ f)(x) \)[/tex]:
We need to ensure that the expression inside the square root, [tex]\( 4 - x^2 \)[/tex], is non-negative because we cannot take the square root of a negative number.
[tex]\[ 4 - x^2 \geq 0 \][/tex]
[tex]\[ x^2 \leq 4 \][/tex]
[tex]\[ -2 \leq x \leq 2 \][/tex]
So, the domain of [tex]\( (g \circ f)(x) \)[/tex] is:
[tex]\[ [-2, 2] \][/tex]
### Summary
1. [tex]\( (f \circ g)(x) = 8 - x \)[/tex] with domain [tex]\( (-\infty, 6] \)[/tex].
2. [tex]\( (g \circ f)(x) = \sqrt{4 - x^2} \)[/tex] with domain [tex]\( [-2, 2] \)[/tex].
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