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Sagot :
To determine the domain of the given function
[tex]\[ f(x) = \sin \left(\sqrt{1-x^2}\right) + \sqrt{x+2} + \frac{1}{\log_{10}(x+1)}, \][/tex]
we need to ensure that each term within the function is well-defined for values of [tex]\( x \)[/tex].
### 1. Analyze [tex]\(\sin \left(\sqrt{1-x^2}\right)\)[/tex]:
The argument of the sine function is [tex]\(\sqrt{1-x^2}\)[/tex]. For the square root function to be defined, the expression inside the square root must be non-negative:
[tex]\[ 1 - x^2 \geq 0. \][/tex]
Solving this inequality:
[tex]\[ 1 - x^2 \geq 0 \implies x^2 \leq 1 \implies -1 \leq x \leq 1. \][/tex]
This provides the first constraint: [tex]\( x \in [-1, 1] \)[/tex].
### 2. Analyze [tex]\(\sqrt{x+2}\)[/tex]:
For the square root function to be defined, the argument must be non-negative:
[tex]\[ x + 2 \geq 0 \implies x \geq -2. \][/tex]
This provides the second constraint: [tex]\( x \in [-2, \infty) \)[/tex].
### 3. Analyze [tex]\(\frac{1}{\log_{10}(x+1)}\)[/tex]:
For this term to be defined and non-zero, the logarithm function in the denominator must be non-zero and its argument must be greater than zero (since the logarithm of a non-positive number is undefined and negative logarithms are not allowed as denominators):
[tex]\[ x + 1 > 0 \implies x > -1. \][/tex]
Additionally, we need to ensure that [tex]\(\log_{10}(x+1) \neq 0\)[/tex]:
[tex]\[ \log_{10}(x+1) = 0 \implies x+1 = 10^0 \implies x+1 = 1 \implies x = 0. \][/tex]
Hence, [tex]\( x \neq 0 \)[/tex].
Combining these constraints:
- From [tex]\(\sin \left(\sqrt{1-x^2}\right)\)[/tex]: [tex]\( x \in [-1, 1] \)[/tex].
- From [tex]\(\sqrt{x+2}\)[/tex]: [tex]\( x \geq -2 \)[/tex].
- From [tex]\(\frac{1}{\log_{10}(x+1)}\)[/tex]: [tex]\( x > -1\)[/tex] and [tex]\( x \neq 0 \)[/tex].
### 4. Combine all conditions:
To combine these conditions, we need the intersection of:
- [tex]\(x \in [-1, 1]\)[/tex]
- [tex]\(x \geq -1\)[/tex]
- [tex]\(x \neq 0\)[/tex]
The intersection of [tex]\(x \in [-1, 1]\)[/tex] and [tex]\(x \geq -1\)[/tex] is [tex]\(x \in [-1, 1]\)[/tex].
Excluding [tex]\(x = 0\)[/tex] from this interval gives:
[tex]\[ x \in [-1, 1] \setminus \{0\} = \{-1 \leq x < 0\} \cup \{0 < x \leq 1\}.\][/tex]
Therefore, the domain of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{[-1, 0) \cup (0, 1]}. \][/tex]
[tex]\[ f(x) = \sin \left(\sqrt{1-x^2}\right) + \sqrt{x+2} + \frac{1}{\log_{10}(x+1)}, \][/tex]
we need to ensure that each term within the function is well-defined for values of [tex]\( x \)[/tex].
### 1. Analyze [tex]\(\sin \left(\sqrt{1-x^2}\right)\)[/tex]:
The argument of the sine function is [tex]\(\sqrt{1-x^2}\)[/tex]. For the square root function to be defined, the expression inside the square root must be non-negative:
[tex]\[ 1 - x^2 \geq 0. \][/tex]
Solving this inequality:
[tex]\[ 1 - x^2 \geq 0 \implies x^2 \leq 1 \implies -1 \leq x \leq 1. \][/tex]
This provides the first constraint: [tex]\( x \in [-1, 1] \)[/tex].
### 2. Analyze [tex]\(\sqrt{x+2}\)[/tex]:
For the square root function to be defined, the argument must be non-negative:
[tex]\[ x + 2 \geq 0 \implies x \geq -2. \][/tex]
This provides the second constraint: [tex]\( x \in [-2, \infty) \)[/tex].
### 3. Analyze [tex]\(\frac{1}{\log_{10}(x+1)}\)[/tex]:
For this term to be defined and non-zero, the logarithm function in the denominator must be non-zero and its argument must be greater than zero (since the logarithm of a non-positive number is undefined and negative logarithms are not allowed as denominators):
[tex]\[ x + 1 > 0 \implies x > -1. \][/tex]
Additionally, we need to ensure that [tex]\(\log_{10}(x+1) \neq 0\)[/tex]:
[tex]\[ \log_{10}(x+1) = 0 \implies x+1 = 10^0 \implies x+1 = 1 \implies x = 0. \][/tex]
Hence, [tex]\( x \neq 0 \)[/tex].
Combining these constraints:
- From [tex]\(\sin \left(\sqrt{1-x^2}\right)\)[/tex]: [tex]\( x \in [-1, 1] \)[/tex].
- From [tex]\(\sqrt{x+2}\)[/tex]: [tex]\( x \geq -2 \)[/tex].
- From [tex]\(\frac{1}{\log_{10}(x+1)}\)[/tex]: [tex]\( x > -1\)[/tex] and [tex]\( x \neq 0 \)[/tex].
### 4. Combine all conditions:
To combine these conditions, we need the intersection of:
- [tex]\(x \in [-1, 1]\)[/tex]
- [tex]\(x \geq -1\)[/tex]
- [tex]\(x \neq 0\)[/tex]
The intersection of [tex]\(x \in [-1, 1]\)[/tex] and [tex]\(x \geq -1\)[/tex] is [tex]\(x \in [-1, 1]\)[/tex].
Excluding [tex]\(x = 0\)[/tex] from this interval gives:
[tex]\[ x \in [-1, 1] \setminus \{0\} = \{-1 \leq x < 0\} \cup \{0 < x \leq 1\}.\][/tex]
Therefore, the domain of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{[-1, 0) \cup (0, 1]}. \][/tex]
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