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Sagot :
Let's analyze the given functions [tex]\( f(x) = -2x - 7 \)[/tex] and [tex]\( g(x) = \sqrt{-x - 2} \)[/tex] to determine the compositions [tex]\( f \circ g(x) \)[/tex] and [tex]\( g \circ f(x) \)[/tex], as well as their respective domains.
### 1. Finding [tex]\( f \circ g(x) \)[/tex]
To find [tex]\( f \circ g(x) \)[/tex], we substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\sqrt{-x - 2}\right) \][/tex]
Now, substitute [tex]\( \sqrt{-x - 2} \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\sqrt{-x - 2}\right) = -2\left(\sqrt{-x - 2}\right) - 7 \][/tex]
So,
[tex]\[ f \circ g(x) = -2\sqrt{-x - 2} - 7 \][/tex]
### 2. Domain of [tex]\( f \circ g(x) \)[/tex]
The domain of [tex]\( f \circ g(x) \)[/tex] is determined by the domain of [tex]\( g(x) \)[/tex] and any restrictions from [tex]\( f(x) \)[/tex].
For [tex]\( g(x) = \sqrt{-x - 2} \)[/tex] to be real, the expression inside the square root must be non-negative:
[tex]\[ -x - 2 \geq 0 \][/tex]
[tex]\[ -x \geq 2 \][/tex]
[tex]\[ x \leq -2 \][/tex]
Thus, the domain of [tex]\( f \circ g(x) \)[/tex] is [tex]\( (-\infty, -2] \)[/tex].
### 3. Finding [tex]\( g \circ f(x) \)[/tex]
To find [tex]\( g \circ f(x) \)[/tex], we substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(-2x - 7) \][/tex]
Now, substitute [tex]\(-2x - 7\)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(-2x - 7) = \sqrt{-(-2x - 7) - 2} \][/tex]
[tex]\[ g(-2x - 7) = \sqrt{2x + 7 - 2} \][/tex]
[tex]\[ g(-2x - 7) = \sqrt{2x + 5} \][/tex]
So,
[tex]\[ g \circ f(x) = \sqrt{2x + 5} \][/tex]
### 4. Domain of [tex]\( g \circ f(x) \)[/tex]
The domain of [tex]\( g \circ f(x) \)[/tex] is determined by the domain of [tex]\( f(x) \)[/tex] and any restrictions from [tex]\( g(x) \)[/tex].
For [tex]\( g(x) = \sqrt{2x + 5} \)[/tex] to be real, the expression inside the square root must be non-negative:
[tex]\[ 2x + 5 \geq 0 \][/tex]
[tex]\[ 2x \geq -5 \][/tex]
[tex]\[ x \geq -\frac{5}{2} \][/tex]
Thus, the domain of [tex]\( g \circ f(x) \)[/tex] is [tex]\(\left[-\frac{5}{2}, \infty \right) \)[/tex].
### Summary
- [tex]\( f \circ g(x) = -2\sqrt{-x - 2} - 7 \)[/tex]
- The domain of [tex]\( f \circ g(x) \)[/tex] is [tex]\( (-\infty, -2] \)[/tex].
- [tex]\( g \circ f(x) = \sqrt{2x + 5} \)[/tex]
- The domain of [tex]\( g \circ f(x) \)[/tex] is [tex]\(\left[-\frac{5}{2}, \infty \right) \)[/tex].
### 1. Finding [tex]\( f \circ g(x) \)[/tex]
To find [tex]\( f \circ g(x) \)[/tex], we substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\sqrt{-x - 2}\right) \][/tex]
Now, substitute [tex]\( \sqrt{-x - 2} \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\sqrt{-x - 2}\right) = -2\left(\sqrt{-x - 2}\right) - 7 \][/tex]
So,
[tex]\[ f \circ g(x) = -2\sqrt{-x - 2} - 7 \][/tex]
### 2. Domain of [tex]\( f \circ g(x) \)[/tex]
The domain of [tex]\( f \circ g(x) \)[/tex] is determined by the domain of [tex]\( g(x) \)[/tex] and any restrictions from [tex]\( f(x) \)[/tex].
For [tex]\( g(x) = \sqrt{-x - 2} \)[/tex] to be real, the expression inside the square root must be non-negative:
[tex]\[ -x - 2 \geq 0 \][/tex]
[tex]\[ -x \geq 2 \][/tex]
[tex]\[ x \leq -2 \][/tex]
Thus, the domain of [tex]\( f \circ g(x) \)[/tex] is [tex]\( (-\infty, -2] \)[/tex].
### 3. Finding [tex]\( g \circ f(x) \)[/tex]
To find [tex]\( g \circ f(x) \)[/tex], we substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(-2x - 7) \][/tex]
Now, substitute [tex]\(-2x - 7\)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(-2x - 7) = \sqrt{-(-2x - 7) - 2} \][/tex]
[tex]\[ g(-2x - 7) = \sqrt{2x + 7 - 2} \][/tex]
[tex]\[ g(-2x - 7) = \sqrt{2x + 5} \][/tex]
So,
[tex]\[ g \circ f(x) = \sqrt{2x + 5} \][/tex]
### 4. Domain of [tex]\( g \circ f(x) \)[/tex]
The domain of [tex]\( g \circ f(x) \)[/tex] is determined by the domain of [tex]\( f(x) \)[/tex] and any restrictions from [tex]\( g(x) \)[/tex].
For [tex]\( g(x) = \sqrt{2x + 5} \)[/tex] to be real, the expression inside the square root must be non-negative:
[tex]\[ 2x + 5 \geq 0 \][/tex]
[tex]\[ 2x \geq -5 \][/tex]
[tex]\[ x \geq -\frac{5}{2} \][/tex]
Thus, the domain of [tex]\( g \circ f(x) \)[/tex] is [tex]\(\left[-\frac{5}{2}, \infty \right) \)[/tex].
### Summary
- [tex]\( f \circ g(x) = -2\sqrt{-x - 2} - 7 \)[/tex]
- The domain of [tex]\( f \circ g(x) \)[/tex] is [tex]\( (-\infty, -2] \)[/tex].
- [tex]\( g \circ f(x) = \sqrt{2x + 5} \)[/tex]
- The domain of [tex]\( g \circ f(x) \)[/tex] is [tex]\(\left[-\frac{5}{2}, \infty \right) \)[/tex].
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