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Let f(x) = (x^2 -5x - 14)/(x-7). Identify another function that agrees with f(x) at all but one point.
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Could you define a piecewise function? e.g \[f(x)=(x^2-5x-14)/(x-7) for ( -\infty,2) and (2,\infty)\] so the function does not include x=2, yet agrees everywhere else?
yes, that sounds correct
wait, shouldn't it be 7 instead of 2? Since it would make the function undefined?
The original function itself does not contain 7 in domain..So, your answer would be something like this (-infinity,2)U(2,infinity)-{7}
The domain of the function in the question is R-{7} So, your answer satisfies every value except at x=2
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the function you are looking for is \(g(x)=x-2\)
scratch that, it is \(g(x)=x+2\) sorry
Toatally forgot it @satellite73
*totally
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