Answer:
[tex]g(x)=\begin{cases}f(x) \ \ \text{ if } \ x \ne 1\\ -1 \ \ \ \ \text{ if }\ x=1\end{cases}[/tex]
Step-by-step explanation:
We are given the function:
[tex]f(x) = \dfrac{x^2-3x+2}{x-1}[/tex]
and know that it has a removable discontinuity (undefined point) at x = 1.
We are trying to replicate this function in [tex]g(x)[/tex] using a piecewise definition while removing its discontinuity at x = 1.
[tex]g(x)=\begin{cases}f(x) \ \ \text{ if } \ x \ne 1\\ ? \ \ \ \ \ \ \,\text{ if }\ x=1\end{cases}[/tex]
To solve for out the y-value that would make f(x) continuous at x = 1, we can take the limit of the function as it approaches that x-value:
[tex]? = \displaystyle \lim_{x\,\to \,1}f(x)[/tex]
[tex]\displaystyle = \lim_{x\,\to \,1}\!\left(\dfrac{x^2-3x+2}{x-1}\right)[/tex]
↓ factoring the numerator
[tex]\displaystyle = \lim_{x\,\to \,1}\!\left(\dfrac{(x-2)(x-1)}{x-1}\right)[/tex]
↓ canceling the [tex](x-1)[/tex] that appears in both the numerator and denominator of the fraction
[tex]\displaystyle = \lim_{x\,\to \,1}(x-2)[/tex]
↓ evaluating the limit
[tex]= 1 - 2[/tex]
↓ executing the subtraction
[tex]= - 1[/tex]
So, the piecewise definition for [tex]g(x)[/tex] is:
[tex]\boxed{g(x)=\begin{cases}f(x) \ \ \text{ if } \ x \ne 1\\ -1 \ \ \ \ \text{ if }\ x=1\end{cases}}[/tex]
