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Sagot :
To determine the domain of each function, we need to establish the set of all possible values of [tex]\( x \)[/tex] for which the function is defined.
### i. [tex]\( y = x^2 + 2 \)[/tex]
For this function:
[tex]\[ y = x^2 + 2 \][/tex]
- The expression [tex]\( x^2 + 2 \)[/tex] involves a polynomial function [tex]\( x^2 \)[/tex].
- Polynomial functions are defined for all real numbers since squaring any real number always yields a real result.
- Therefore, there are no restrictions on the values of [tex]\( x \)[/tex].
Domain: All real numbers.
### ii. [tex]\( y = \sqrt{x^2 + 9} \)[/tex]
For this function:
[tex]\[ y = \sqrt{x^2 + 9} \][/tex]
- The expression inside the square root is [tex]\( x^2 + 9 \)[/tex].
- For a square root function to be defined in the set of real numbers, the expression inside the square root must be non-negative. However, [tex]\( x^2 \)[/tex] is always non-negative (since squaring any real number results in a non-negative value), and adding 9 (which is positive) ensures that [tex]\( x^2 + 9 \)[/tex] is always positive.
- Therefore, [tex]\( x^2 + 9 \)[/tex] is always greater than or equal to 9, meaning the square root function can take any real number input without restriction.
Domain: All real numbers.
### iii. [tex]\( y = \frac{4x}{(x-1)(x+3)} \)[/tex]
For this function:
[tex]\[ y = \frac{4x}{(x-1)(x+3)} \][/tex]
- This function is a rational function, which means it’s defined for all values of [tex]\( x \)[/tex] except where the denominator is zero, as division by zero is undefined.
- To find the values of [tex]\( x \)[/tex] that make the denominator zero, set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (x-1)(x+3) = 0 \][/tex]
This gives us two equations:
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
- These solutions [tex]\( x = 1 \)[/tex] and [tex]\( x = -3 \)[/tex] are the points where the denominator becomes zero.
Domain: All real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -3 \)[/tex].
In summary, the domains for each function are as follows:
1. [tex]\( y = x^2 + 2 \)[/tex]: All real numbers.
2. [tex]\( y = \sqrt{x^2 + 9} \)[/tex]: All real numbers.
3. [tex]\( y = \frac{4x}{(x-1)(x+3)} \)[/tex]: All real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -3 \)[/tex].
### i. [tex]\( y = x^2 + 2 \)[/tex]
For this function:
[tex]\[ y = x^2 + 2 \][/tex]
- The expression [tex]\( x^2 + 2 \)[/tex] involves a polynomial function [tex]\( x^2 \)[/tex].
- Polynomial functions are defined for all real numbers since squaring any real number always yields a real result.
- Therefore, there are no restrictions on the values of [tex]\( x \)[/tex].
Domain: All real numbers.
### ii. [tex]\( y = \sqrt{x^2 + 9} \)[/tex]
For this function:
[tex]\[ y = \sqrt{x^2 + 9} \][/tex]
- The expression inside the square root is [tex]\( x^2 + 9 \)[/tex].
- For a square root function to be defined in the set of real numbers, the expression inside the square root must be non-negative. However, [tex]\( x^2 \)[/tex] is always non-negative (since squaring any real number results in a non-negative value), and adding 9 (which is positive) ensures that [tex]\( x^2 + 9 \)[/tex] is always positive.
- Therefore, [tex]\( x^2 + 9 \)[/tex] is always greater than or equal to 9, meaning the square root function can take any real number input without restriction.
Domain: All real numbers.
### iii. [tex]\( y = \frac{4x}{(x-1)(x+3)} \)[/tex]
For this function:
[tex]\[ y = \frac{4x}{(x-1)(x+3)} \][/tex]
- This function is a rational function, which means it’s defined for all values of [tex]\( x \)[/tex] except where the denominator is zero, as division by zero is undefined.
- To find the values of [tex]\( x \)[/tex] that make the denominator zero, set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (x-1)(x+3) = 0 \][/tex]
This gives us two equations:
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
- These solutions [tex]\( x = 1 \)[/tex] and [tex]\( x = -3 \)[/tex] are the points where the denominator becomes zero.
Domain: All real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -3 \)[/tex].
In summary, the domains for each function are as follows:
1. [tex]\( y = x^2 + 2 \)[/tex]: All real numbers.
2. [tex]\( y = \sqrt{x^2 + 9} \)[/tex]: All real numbers.
3. [tex]\( y = \frac{4x}{(x-1)(x+3)} \)[/tex]: All real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -3 \)[/tex].
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