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Sagot :
Let's determine the domain and range of the function [tex]\( f(x) = \frac{x^2 + 5x + 6}{x + 2} \)[/tex].
### Finding the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function [tex]\( f(x) = \frac{x^2 + 5x + 6}{x + 2} \)[/tex], the function is undefined when the denominator is zero because division by zero is undefined.
First, identify the values of [tex]\(x \)[/tex] that make the denominator zero:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = -2 \)[/tex]. Therefore, the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ D: \{ x \in \mathbb{R} \mid x \neq -2 \} \][/tex]
### Finding the Range
To find the range of the function, we simplify it wherever the function is defined. The numerator can be factored as follows:
[tex]\[ x^2 + 5x + 6 = (x + 2)(x + 3) \][/tex]
So the function becomes:
[tex]\[ f(x) = \frac{(x + 2)(x + 3)}{x + 2} \][/tex]
As long as [tex]\( x \neq -2 \)[/tex], the [tex]\( (x + 2) \)[/tex] terms cancel each other out:
[tex]\[ f(x) = x + 3 \quad \text{for} \quad x \neq -2 \][/tex]
Now, this simplified function [tex]\( f(x) = x + 3 \)[/tex] is a linear function. For a linear function in the form [tex]\( y = mx + b \)[/tex], the range is all real numbers except for any specific exclusions that arise from the original function definition.
Without the cancellation, the value at [tex]\( x = -2 \)[/tex] would need to be evaluated in the original form:
[tex]\[ f(-2) = \frac{(-2)^2 + 5(-2) + 6}{-2 + 2} \][/tex]
This clearly leads to an undefined value as it results in division by zero. Upon exclusion, for every other value of [tex]\( x \neq -2 \)[/tex], [tex]\( y \)[/tex] can take any real number value as [tex]\( f(x) = x + 3 \)[/tex] covers all real numbers linearly.
Thus, the range of the function is all real numbers except the value that [tex]\( y \)[/tex] takes when [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2 + 3 = 1 \][/tex]
Therefore, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ R: \{ y \in \mathbb{R} \mid y \neq 1 \} \][/tex]
### Conclusion
By considering the domain and range, the correct answer is:
[tex]\[ D: \{ x \in \mathbb{R} \mid x \neq -2 \} \][/tex]
[tex]\[ R: \{ y \in \mathbb{R} \mid y \neq 1 \} \][/tex]
Hence, the correct option is:
[tex]\[ D:\{x \in R \mid x \neq-2\} ; R:\{y \in R \mid y \neq 1\} \][/tex]
### Finding the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function [tex]\( f(x) = \frac{x^2 + 5x + 6}{x + 2} \)[/tex], the function is undefined when the denominator is zero because division by zero is undefined.
First, identify the values of [tex]\(x \)[/tex] that make the denominator zero:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = -2 \)[/tex]. Therefore, the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ D: \{ x \in \mathbb{R} \mid x \neq -2 \} \][/tex]
### Finding the Range
To find the range of the function, we simplify it wherever the function is defined. The numerator can be factored as follows:
[tex]\[ x^2 + 5x + 6 = (x + 2)(x + 3) \][/tex]
So the function becomes:
[tex]\[ f(x) = \frac{(x + 2)(x + 3)}{x + 2} \][/tex]
As long as [tex]\( x \neq -2 \)[/tex], the [tex]\( (x + 2) \)[/tex] terms cancel each other out:
[tex]\[ f(x) = x + 3 \quad \text{for} \quad x \neq -2 \][/tex]
Now, this simplified function [tex]\( f(x) = x + 3 \)[/tex] is a linear function. For a linear function in the form [tex]\( y = mx + b \)[/tex], the range is all real numbers except for any specific exclusions that arise from the original function definition.
Without the cancellation, the value at [tex]\( x = -2 \)[/tex] would need to be evaluated in the original form:
[tex]\[ f(-2) = \frac{(-2)^2 + 5(-2) + 6}{-2 + 2} \][/tex]
This clearly leads to an undefined value as it results in division by zero. Upon exclusion, for every other value of [tex]\( x \neq -2 \)[/tex], [tex]\( y \)[/tex] can take any real number value as [tex]\( f(x) = x + 3 \)[/tex] covers all real numbers linearly.
Thus, the range of the function is all real numbers except the value that [tex]\( y \)[/tex] takes when [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2 + 3 = 1 \][/tex]
Therefore, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ R: \{ y \in \mathbb{R} \mid y \neq 1 \} \][/tex]
### Conclusion
By considering the domain and range, the correct answer is:
[tex]\[ D: \{ x \in \mathbb{R} \mid x \neq -2 \} \][/tex]
[tex]\[ R: \{ y \in \mathbb{R} \mid y \neq 1 \} \][/tex]
Hence, the correct option is:
[tex]\[ D:\{x \in R \mid x \neq-2\} ; R:\{y \in R \mid y \neq 1\} \][/tex]
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