Pregunta
The function \( f \) is defined below. \[ f(x)=\frac{x-1}{x^{2}-2 x+1} \] Find all values of \( x \) that are NOT in the domain of \( f \). If there is more than one value, separate them with commas.
Ask by Barrett Dunn. in the United States
Mar 10,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
The value of \( x \) not in the domain of \( f \) is \( 1 \).
Solución
1. The given function is
\[
f(x)=\frac{x-1}{x^{2}-2x+1}.
\]
2. To find values of \( x \) not in the domain, we need to determine when the denominator is equal to 0, since division by zero is not allowed.
3. Factor the denominator:
\[
x^{2}-2x+1 = (x-1)^2.
\]
4. Set the denominator equal to 0 and solve for \( x \):
\[
(x-1)^2=0.
\]
Taking the square root of both sides gives:
\[
x-1=0.
\]
Therefore:
\[
x=1.
\]
5. The value of \( x \) that is not in the domain of \( f \) is \( 1 \).
Respondido por UpStudy AI y revisado por un tutor profesional
Como
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Mind Expander
To find the values of \( x \) that are not in the domain of the function \( f(x) = \frac{x-1}{x^2 - 2x + 1} \), we need to determine when the denominator is equal to zero, since division by zero is undefined. First, we simplify the denominator: \[ x^2 - 2x + 1 = (x - 1)^2. \] Now, set the denominator equal to zero: \[ (x - 1)^2 = 0. \] Solving this, we get: \[ x - 1 = 0 \implies x = 1. \] Thus, the only value of \( x \) that is not in the domain of \( f(x) \) is: \[ \boxed{1}. \]

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