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Sagot :
To determine which functions have the same domain as [tex]\( y = 2\sqrt{x} \)[/tex], we need to first understand the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
1. Domain of [tex]\( y = 2\sqrt{x} \)[/tex]:
- The expression inside the square root, [tex]\( x \)[/tex], must be non-negative.
- Therefore, [tex]\( x \geq 0 \)[/tex].
- The domain of [tex]\( y = 2\sqrt{x} \)[/tex] is [tex]\( [0, \infty) \)[/tex].
Next, let's analyze the domain of each given function:
1. [tex]\( y = \frac{1}{x} \)[/tex]:
- The function is undefined at [tex]\( x = 0 \)[/tex].
- The domain is [tex]\( x \neq 0 \)[/tex], which means it is defined for [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex].
- This is not the same as the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
2. [tex]\( y = x^2 \)[/tex]:
- [tex]\( x \)[/tex] can take any real number.
- The function is defined for all [tex]\( x \in \mathbb{R} \)[/tex].
- Though this function includes the domain [tex]\( [0, \infty) \)[/tex], it is also defined for negative values, which goes beyond the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
3. [tex]\( y = \log(x) \)[/tex]:
- The natural logarithm function is defined for [tex]\( x > 0 \)[/tex].
- The domain is [tex]\( (0, \infty) \)[/tex].
- This is not the same as the domain of [tex]\( y = 2\sqrt{x} \)[/tex], as it does not include [tex]\( x = 0 \)[/tex].
4. [tex]\( y = x \)[/tex]:
- [tex]\( x \)[/tex] can take any real number.
- The function is defined for all [tex]\( x \in \mathbb{R} \)[/tex].
- Similar to [tex]\( y = x^2 \)[/tex], it is also defined for negative values, which is beyond the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
5. [tex]\( y = 1 \)[/tex]:
- This is a constant function.
- It is defined for all [tex]\( x \in \mathbb{R} \)[/tex].
- This function, being defined everywhere, does include the domain [tex]\( [0, \infty) \)[/tex], but is broader than the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
Given the domains discussed:
- Functions with domains that fully match [tex]\( [0, \infty) \)[/tex] are:
- [tex]\( y = 2\sqrt{x} \)[/tex] itself.
- [tex]\( y = 1 \)[/tex] (since it is defined everywhere).
So, the functions that have the same domain as [tex]\( y = 2\sqrt{x} \)[/tex] are:
- [tex]\( y = 2\sqrt{x} \)[/tex]
- [tex]\( y = 1 \)[/tex]
Thus, functions [tex]\( y = 2\sqrt{x} \)[/tex] and [tex]\( y = 1 \)[/tex] have domains that match exactly or include the domain of [tex]\( y = 2\sqrt{x} \)[/tex], meeting the criteria.
1. Domain of [tex]\( y = 2\sqrt{x} \)[/tex]:
- The expression inside the square root, [tex]\( x \)[/tex], must be non-negative.
- Therefore, [tex]\( x \geq 0 \)[/tex].
- The domain of [tex]\( y = 2\sqrt{x} \)[/tex] is [tex]\( [0, \infty) \)[/tex].
Next, let's analyze the domain of each given function:
1. [tex]\( y = \frac{1}{x} \)[/tex]:
- The function is undefined at [tex]\( x = 0 \)[/tex].
- The domain is [tex]\( x \neq 0 \)[/tex], which means it is defined for [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex].
- This is not the same as the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
2. [tex]\( y = x^2 \)[/tex]:
- [tex]\( x \)[/tex] can take any real number.
- The function is defined for all [tex]\( x \in \mathbb{R} \)[/tex].
- Though this function includes the domain [tex]\( [0, \infty) \)[/tex], it is also defined for negative values, which goes beyond the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
3. [tex]\( y = \log(x) \)[/tex]:
- The natural logarithm function is defined for [tex]\( x > 0 \)[/tex].
- The domain is [tex]\( (0, \infty) \)[/tex].
- This is not the same as the domain of [tex]\( y = 2\sqrt{x} \)[/tex], as it does not include [tex]\( x = 0 \)[/tex].
4. [tex]\( y = x \)[/tex]:
- [tex]\( x \)[/tex] can take any real number.
- The function is defined for all [tex]\( x \in \mathbb{R} \)[/tex].
- Similar to [tex]\( y = x^2 \)[/tex], it is also defined for negative values, which is beyond the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
5. [tex]\( y = 1 \)[/tex]:
- This is a constant function.
- It is defined for all [tex]\( x \in \mathbb{R} \)[/tex].
- This function, being defined everywhere, does include the domain [tex]\( [0, \infty) \)[/tex], but is broader than the domain of [tex]\( y = 2\sqrt{x} \)[/tex].
Given the domains discussed:
- Functions with domains that fully match [tex]\( [0, \infty) \)[/tex] are:
- [tex]\( y = 2\sqrt{x} \)[/tex] itself.
- [tex]\( y = 1 \)[/tex] (since it is defined everywhere).
So, the functions that have the same domain as [tex]\( y = 2\sqrt{x} \)[/tex] are:
- [tex]\( y = 2\sqrt{x} \)[/tex]
- [tex]\( y = 1 \)[/tex]
Thus, functions [tex]\( y = 2\sqrt{x} \)[/tex] and [tex]\( y = 1 \)[/tex] have domains that match exactly or include the domain of [tex]\( y = 2\sqrt{x} \)[/tex], meeting the criteria.
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