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Sagot :
Let's solve the given problem step-by-step.
### Step 1: Define the Functions
We have:
[tex]\[ f(x) = 5x - 6 \][/tex]
[tex]\[ g(x) = x - 9 \][/tex]
### Step 2: Determine [tex]\( (f + g)(x) \)[/tex]
To find [tex]\( (f + g)(x) \)[/tex]:
[tex]\[ (f + g)(x) = f(x) + g(x) \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f + g)(x) = (5x - 6) + (x - 9) \][/tex]
Combine like terms:
[tex]\[ (f + g)(x) = 5x + x - 6 - 9 = 6x - 15 \][/tex]
### Step 3: Determine the Domain of [tex]\(f + g\)[/tex]
Both [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] are linear functions, which are defined for all real numbers. Therefore, their sum is also defined for all real numbers.
So, the domain of [tex]\(f + g\)[/tex] is:
[tex]\[ \boxed{(-\infty, \infty)} \][/tex]
### Step 4: Determine [tex]\( (f - g)(x) \)[/tex]
To find [tex]\( (f - g)(x) \)[/tex]:
[tex]\[ (f - g)(x) = f(x) - g(x) \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f - g)(x) = (5x - 6) - (x - 9) \][/tex]
Distribute the negative sign and combine like terms:
[tex]\[ (f - g)(x) = 5x - 6 - x + 9 = 4x + 3 \][/tex]
### Step 5: Determine [tex]\( (f \cdot g)(x) \)[/tex]
To find [tex]\( (f \cdot g)(x) \)[/tex]:
[tex]\[ (f \cdot g)(x) = f(x) \cdot g(x) \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f \cdot g)(x) = (5x - 6) \cdot (x - 9) \][/tex]
Expand the product:
[tex]\[ (f \cdot g)(x) = 5x^2 - 45x - 6x + 54 = 5x^2 - 51x + 54 \][/tex]
### Step 6: Determine [tex]\( \frac{f}{g}(x) \)[/tex]
To find [tex]\( \frac{f}{g}(x) \)[/tex]:
[tex]\[ \frac{f}{g}(x) = \frac{f(x)}{g(x)} \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ \frac{f}{g}(x) = \frac{5x - 6}{x - 9} \][/tex]
The domain of this function is all real numbers except where [tex]\(g(x) = 0\)[/tex]. Solving for [tex]\(x\)[/tex]:
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]
So, the domain of [tex]\(\frac{f}{g}\)[/tex] is:
[tex]\[ (-\infty, 9) \cup (9, \infty) \][/tex]
### Summary
Based on the given problem and the calculations:
1. [tex]\( (f + g)(x) = 6x - 15 \)[/tex]
- Domain: [tex]\( (-\infty, \infty) \)[/tex]
2. [tex]\( (f - g)(x) = 4x + 3 \)[/tex]
3. [tex]\( (f \cdot g)(x) = 5x^2 - 51x + 54 \)[/tex]
4. [tex]\( \frac{f}{g}(x) = \frac{5x - 6}{x - 9} \)[/tex]
- Domain: [tex]\( (-\infty, 9) \cup (9, \infty) \)[/tex]
Given the specific part of the question:
> [tex]$( f + g )( x )=$[/tex] [tex]\( \boxed{6x - 15} \)[/tex]
> The domain of [tex]$f + g$[/tex] is [tex]\( \boxed{(-\infty, \infty)} \)[/tex]
> [tex]$( f - g )( x )=$[/tex] [tex]\( \boxed{4x + 3} \)[/tex]
### Step 1: Define the Functions
We have:
[tex]\[ f(x) = 5x - 6 \][/tex]
[tex]\[ g(x) = x - 9 \][/tex]
### Step 2: Determine [tex]\( (f + g)(x) \)[/tex]
To find [tex]\( (f + g)(x) \)[/tex]:
[tex]\[ (f + g)(x) = f(x) + g(x) \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f + g)(x) = (5x - 6) + (x - 9) \][/tex]
Combine like terms:
[tex]\[ (f + g)(x) = 5x + x - 6 - 9 = 6x - 15 \][/tex]
### Step 3: Determine the Domain of [tex]\(f + g\)[/tex]
Both [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] are linear functions, which are defined for all real numbers. Therefore, their sum is also defined for all real numbers.
So, the domain of [tex]\(f + g\)[/tex] is:
[tex]\[ \boxed{(-\infty, \infty)} \][/tex]
### Step 4: Determine [tex]\( (f - g)(x) \)[/tex]
To find [tex]\( (f - g)(x) \)[/tex]:
[tex]\[ (f - g)(x) = f(x) - g(x) \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f - g)(x) = (5x - 6) - (x - 9) \][/tex]
Distribute the negative sign and combine like terms:
[tex]\[ (f - g)(x) = 5x - 6 - x + 9 = 4x + 3 \][/tex]
### Step 5: Determine [tex]\( (f \cdot g)(x) \)[/tex]
To find [tex]\( (f \cdot g)(x) \)[/tex]:
[tex]\[ (f \cdot g)(x) = f(x) \cdot g(x) \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f \cdot g)(x) = (5x - 6) \cdot (x - 9) \][/tex]
Expand the product:
[tex]\[ (f \cdot g)(x) = 5x^2 - 45x - 6x + 54 = 5x^2 - 51x + 54 \][/tex]
### Step 6: Determine [tex]\( \frac{f}{g}(x) \)[/tex]
To find [tex]\( \frac{f}{g}(x) \)[/tex]:
[tex]\[ \frac{f}{g}(x) = \frac{f(x)}{g(x)} \][/tex]
Substitute the definitions of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ \frac{f}{g}(x) = \frac{5x - 6}{x - 9} \][/tex]
The domain of this function is all real numbers except where [tex]\(g(x) = 0\)[/tex]. Solving for [tex]\(x\)[/tex]:
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]
So, the domain of [tex]\(\frac{f}{g}\)[/tex] is:
[tex]\[ (-\infty, 9) \cup (9, \infty) \][/tex]
### Summary
Based on the given problem and the calculations:
1. [tex]\( (f + g)(x) = 6x - 15 \)[/tex]
- Domain: [tex]\( (-\infty, \infty) \)[/tex]
2. [tex]\( (f - g)(x) = 4x + 3 \)[/tex]
3. [tex]\( (f \cdot g)(x) = 5x^2 - 51x + 54 \)[/tex]
4. [tex]\( \frac{f}{g}(x) = \frac{5x - 6}{x - 9} \)[/tex]
- Domain: [tex]\( (-\infty, 9) \cup (9, \infty) \)[/tex]
Given the specific part of the question:
> [tex]$( f + g )( x )=$[/tex] [tex]\( \boxed{6x - 15} \)[/tex]
> The domain of [tex]$f + g$[/tex] is [tex]\( \boxed{(-\infty, \infty)} \)[/tex]
> [tex]$( f - g )( x )=$[/tex] [tex]\( \boxed{4x + 3} \)[/tex]
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