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Mathematics 15 Online
OpenStudy (anonymous):

Find the constant "a", or the constants "a" and "b", such that the function is continuous on the entire real line. g(x) = {(4sin(x)/x, x<0), (a-2x, x >=0)}

OpenStudy (anonymous):

This is killing me... I can't figure out how to do this

OpenStudy (jdoe0001):

well...

OpenStudy (jdoe0001):

\(\large g(x) \begin{cases} \cfrac{4sin(x)}{x}&x<0\\ a-2x&x\ge 0 \end{cases}\) right?

OpenStudy (anonymous):

Yes

OpenStudy (jdoe0001):

I assume this isn't calculus?

OpenStudy (anonymous):

Calc 1

OpenStudy (jdoe0001):

ahhh ok

OpenStudy (jdoe0001):

well.. the thing is if we know where the function without the variables arrives, when x = 0 then we simply need to make the 2nd one arrive at that same point if both of them arrive at the same point when x = 0, then we have removed the discontinuity

OpenStudy (anonymous):

It's for my AP Calculus BC course... I've looked through examples of similar problems but it's not clicking for me.

OpenStudy (jdoe0001):

so we'd like to know where \(\bf \cfrac{4sin(x)}{x}\) ends up at when x = 0

OpenStudy (jdoe0001):

and ..... I'd say \(\bf g(x) \begin{cases} \cfrac{4sin(x)}{x}&x<0\\ a-2x&x\ge 0 \end{cases} \\ \quad \\ lim_{x\to 0}\quad \cfrac{4sin(x)}{x}\implies lim_{x\to 0}\ 4\cdot lim_{x\to 0}\ \cfrac{sin(x)}{x}\) any ideas?

OpenStudy (jdoe0001):

say that gives you the value of "something" then we know it arrives at "something" when x = 0 so we then set a - 2x = "something" and then both of them arrive at that same "something" value and thus the discontinuity would be removed

OpenStudy (anonymous):

I'm not sure how to write limits on here, but the limit of 4sin(x)/x as x->0 is equal to 4, correct?

OpenStudy (anonymous):

So then a=2x+4?

OpenStudy (jdoe0001):

yeap because \(\bf g(x) \begin{cases} \cfrac{4sin(x)}{x}&x<0\\ a-2x&x\ge 0 \end{cases} \\ \quad \\ lim_{x\to 0}\quad \cfrac{4sin(x)}{x}\implies lim_{x\to 0}\ 4\cdot {\color{brown}{ lim_{x\to 0}\ \cfrac{sin(x)}{x}}}=1\to 4\cdot 1\to 4\)

OpenStudy (anonymous):

That's L'Hospital's rule?

OpenStudy (jdoe0001):

so now we know it arrives to 4 when x = 0 so a - 2x = 4 setting x = 0 then a - 2(0) = 4

OpenStudy (jdoe0001):

\(\bf {\color{brown}{ lim_{x\to 0}\ \cfrac{sin(x)}{x}}}=1\) <---? nope, is just from the "squeeze theorem"

OpenStudy (anonymous):

hmm

OpenStudy (jdoe0001):

https://www.youtube.com/watch?v=VIHupMeoDxA <--- using the squeeze theorem

OpenStudy (anonymous):

Oh, yeah I just remembered that proof. So, a=4 would be correct then?

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