Let f(x)=(2x^2+5x-7)/(x-1)

show that F(x) has a removable discontinuity at x=1 and determine what value for F(1) would make f(x) continuous at x=1
I'm not sure how to factor to solve for f(1)..

2 answers

factoring the numerator ... 2x^2+5x-7 = (2x + 7) (x - 1)

F(x) = 2x + 7 ... F(1) = ?

the discontinuity is zero divided by zero
... (x - 1) / (x - 1) for f(1)

f(x) and F(x) are very similar, with an important difference
f(x) = 2x+7
everywhere except at x=1.
so, if we define f(1) = 9 then f(x) will be continuous everywhere.
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