I have the function f(x) equal to 3/x+(-2x + 6)/[x(x-2)] when x different to 0, and 2,
and f(x) equals 9 when x = 0
I have to show that f(x) has a removable discontinuity at x=0
I'm stucked with the transformation of these fractions and it leads me nowhere, can you show me please the first few steps?
I also tried to put it in LaTeX but I got something wrong, (see below)
[tex]f(x)=\left\{\begin{array}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }\x \ne 0, 2\\9, & \mbox{if}x=0\end{array}\right[tex]
Try
[tex]f(x)=\begin{cases}{\frac{3}{x}+\frac{-2x + 6}{x(x-2)}}, &\mbox{ if }x \ne 0, 2\\9, & \mbox{ if }x=0\end{cases}\right[/tex]
It comes out as
I don't recall how to center the 9.
-Dan