Answer :
To determine the domain of the function \( f(x) = \frac{x-1}{x^2+1} \), we need to identify all the possible values of \( x \) that can be input into the function without causing any mathematical issues, particularly focusing on points where the function might be undefined.
1. Identify the potential source of undefined behavior:
- The most common issue that can make a function undefined is division by zero. Therefore, we need to check if the denominator of the function can ever be zero.
2. Analyze the denominator:
- The denominator of our function is \( x^2 + 1 \).
- We need to determine when \( x^2 + 1 = 0 \).
3. Solve for zeros in the denominator:
- Set the denominator equal to zero:
[tex]\[ x^2 + 1 = 0 \][/tex]
- Solve for \( x \):
[tex]\[ x^2 = -1 \][/tex]
Here, we see that \( x^2 = -1 \) has no real solutions because the square of a real number is always non-negative, and adding 1 to a non-negative number is always positive.
4. Conclusion:
- Since \( x^2 + 1 \) is always positive for all real numbers \( x \) (it can never be zero), there are no values of \( x \) that will make the denominator equal to zero.
Thus, the function \( f(x) = \frac{x-1}{x^2+1} \) has no restrictions on \( x \). This means that \( f(x) \) is defined for all real numbers.
The domain of the function [tex]\( f(x) = \frac{x-1}{x^2+1} \)[/tex] is all real numbers.
1. Identify the potential source of undefined behavior:
- The most common issue that can make a function undefined is division by zero. Therefore, we need to check if the denominator of the function can ever be zero.
2. Analyze the denominator:
- The denominator of our function is \( x^2 + 1 \).
- We need to determine when \( x^2 + 1 = 0 \).
3. Solve for zeros in the denominator:
- Set the denominator equal to zero:
[tex]\[ x^2 + 1 = 0 \][/tex]
- Solve for \( x \):
[tex]\[ x^2 = -1 \][/tex]
Here, we see that \( x^2 = -1 \) has no real solutions because the square of a real number is always non-negative, and adding 1 to a non-negative number is always positive.
4. Conclusion:
- Since \( x^2 + 1 \) is always positive for all real numbers \( x \) (it can never be zero), there are no values of \( x \) that will make the denominator equal to zero.
Thus, the function \( f(x) = \frac{x-1}{x^2+1} \) has no restrictions on \( x \). This means that \( f(x) \) is defined for all real numbers.
The domain of the function [tex]\( f(x) = \frac{x-1}{x^2+1} \)[/tex] is all real numbers.
