<p>i have an ap calc bc question...</p>
<p>i am supposed to find the constant a such that the function is continuous on the entire real line:</p>
<p>g(x) = (x^2 - a^2) / (x-a), given that x is not equal to a
8 , given that x = a</p>
<p>could someone please give me the answer to this? i thik i've got it, but i'm not sure.</p>
<p>Well, if a = 0 and x!=a, then x!=0, this means: (x-a) = x - 0 != 0.
This makes function not have any zeros in denominator but then function is absent at the moment when x=a (by definition) -- zeroed x.
You may try manipulating with all negative numbers which make function basically generate zero at the moment when x=(-a) due to that squaring in the numerator. However, there might be another lacks which I'm not yet aware of so don't trust me completely :)
For example, let's say a=-1.
This means, if x=1: (1 -1)/2 = 0, and so the y is the same if x=a=-1, which would be:
(1-1)/-2 = 0. I can't see any possibility of division by zero here.</p>
<p>Am I reading your post wrong, cause I didn't see any factorials. Anyway, your question is kind of confusing. You say x is always equal to a, so g(x) will always be undefined.</p>
<p>If you look at it another way, it looks like a derivative. f'(x)=lim(a-->x) of f(x)-f(a)/x-a . In this case, f(x) would equal x^2, so g(x) would equal 2x, which is continuous for all real values of x anyway. Maybe you could put the exact question up.</p>
<p>
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You say x is always equal to a, so g(x) will always be undefined.
[/quote]
Uh, is 0/0 undefined?</p>
<p>Read the original post. g(x)=8 when x=a.</p>
<p>see my answer under ap test prep</p>
<p>First of all, 0/0 is DEFINITELY undefined, anything over zero is. Second, it doesn't say g(a)=8 it doesn't even look like it says that. Anyway, g(a) would be 0/0 if you took the time to look at the equation.</p>
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First of all, 0/0 is DEFINITELY undefined, anything over zero is.
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Pardon me, it indeed is undefined -- whuuuh :)
[quote]
Read the original post. g(x)=8 when x=a.
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I can't see that, can you restate the question?</p>
<p>thanks stupak. i got your reply in the other thread.</p>