Question

what is the points of discontinuity? Are they all removable? y=(x-5)/x^2-6x+5

Answer 1

HI!!!factor an cancel
the discontinuity you can cancel is "removeable' the other is not

Answer 2

\[\frac{x-5}{(x-5)(x-1)}\] is the start so you see it is not defined at \(5,1\)

Answer 3

then when you cancel you will "remove" the discontinuity at \(5\)

Answer 4

I still don't get it? can you write it out please

Answer 5

0/0 is undefined if use x=5 in eq. above you will have (5-5)/(5-5)(5-1)
0/(0*4)=0/0

Answer 6

From graph above, we see that at x=5 or (x-5)=0, the function equals "0/0" so is undefined. It is therefore discontinuous at x=5, as mentioned by @AJ01 .

The vertical asymptote at x=1 is evidently another discontinuity, which is not removable, since the limits of \(f(x_+)\ne f(x_-)\).

On the other hand, the discontinuity at x=5 has the property that \(f(x_+)= f(x_-)\)=1/4, so we can "remove" the discontinuity by redefining the function as a piecewise function:
g(x)=\(\dfrac{x-5}{x^2-6x+5}\) for \(x\in (-\infty,5)\cup(5,\infty)\)
and
g(x)=1/4 for x=5

Remember that the original f(x) is still undefined at x=5. It is "removed" only by redefining f(x), thus filling the gap.