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# Removing discontinuities (factoring)

AP.CALC:
LIM‑2 (EU)
,
LIM‑2.C (LO)
,
LIM‑2.C.1 (EK)
,
LIM‑2.C.2 (EK)

## Video transcript

the function f of X is equal to 6x squared plus 18x plus 12 over x squared minus 4 is not defined at X is equal to positive or negative 2 and we see that Y that is if X is equal to positive or negative 2 then x squared is going to be equal to positive 4 and 4 minus 4 is 0 and then we're going to have a 0 in the denominator and that's not defined we don't know what happens when you divide or we've never defined what happens when you divide by 0 so they say what value should be assigned to F of negative 2 to make f of X continuous at that point so to think about that let's try to actually simplify f of X so f of X I'll just rewrite it is equal to actually let me just start simplifying right from the get-go so in the numerator I can factor out a 6 out of every one of those terms so it's 6 times x squared plus 3x plus 2 over and the denominator this is a difference of squares this is X plus 2 times X minus 2 and then we can factor this expression up here so this is going to be equal to 6 times let me do it in a different color so we think of two numbers that if I take their product I get two if I take their sum I get three the most obvious one is two and one so this is 6 times X plus 2 times X plus one when you take the product there you'll get x squared plus 3x plus 2 and then all of that all of that over X plus 2 times X minus 2 now if we know that X does not equal negative 2 then we can divide both the numerator and the denominator by X plus 2 the reason why I'm making that constraint is that if X were to be equal to negative 2 that X plus 2 is going to be equal to 0 and you won't be able to do that you can't we don't know what it means to divide something by 0 so we could say that this is going to be equal to so if we we can divide the numerator and the denominator by X plus 2 but we have to assume that X is not equal to negative 2 so this is equal to 6 6 times so we're going to have we're going to divide by X plus 2 in the numerator X plus 2 in the denominator so it's going to be 6 times X plus 1 X plus 1 over X minus 2 over X minus 2 and we have to put the constraint here because now we've changed it now this expression over here is actually defined at x equals negative 2 but in order to be equivalent to the original function we have to we have to constrain it so we will say 4 for X not equal not equal to negative 2 and it's also obvious that X can't be equal to 2 here this one also isn't defined at positive 2 because you're dividing by 0 so you could say for X does not equal to positive or negative 2 if you want to make it very explicit but they asked us what can we assign F of negative 2 to make the function continuous at the point well the function is completely equivalent to this to this expression except that the function is not defined at x equals negative 2 so that's why we have to put that constraint here if we wanted this if we wanted this to be the same thing as our original function but if we want if we wanted to re-engineer the function so it is continuous at that point well then we just have to set f of X equal to whatever this expression would have been when X is equal to negative 2 so let's think about that let's think about that so 6 6 times negative 2 plus 1 over negative 2 minus minus 2 is equal to this is 6 times negative 1 so it's negative 6 over negative 4 which is equal to 3 halves so if we redefine f of X if we say if we say f of X is equal to f of X is equal to 6x squared plus 18x plus 12 over x squared minus 4 4 for X not equal positive or negative 2 and it's equal to 3 halves 4 x equals negative two now this function is going to be the exact same thing as this right over here this f of X this new one this new definition this extended definition of our original one is now equivalent to this expression is equal to six times X plus one over X minus 2 but just to answer the question what value should be assigned to F of negative two to make f of X continuous at that point well f of X should be or F of negative two should be three halves