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)}
This is killing me... I can't figure out how to do this
well...
\(\large g(x) \begin{cases} \cfrac{4sin(x)}{x}&x<0\\ a-2x&x\ge 0 \end{cases}\) right?
Yes
I assume this isn't calculus?
Calc 1
ahhh ok
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
It's for my AP Calculus BC course... I've looked through examples of similar problems but it's not clicking for me.
so we'd like to know where \(\bf \cfrac{4sin(x)}{x}\) ends up at when x = 0
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?
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
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?
So then a=2x+4?
yeap because \(\bf g(x) \begin{cases} \cfrac{4sin(x)}{x}&x<0\\ a-2x&x\ge 0 \end{cases} \\ \quad \\ lim_{x\to 0}\quad \cfrac{4sin(x)}{x}\implies lim_{x\to 0}\ 4\cdot {\color{brown}{ lim_{x\to 0}\ \cfrac{sin(x)}{x}}}=1\to 4\cdot 1\to 4\)
That's L'Hospital's rule?
so now we know it arrives to 4 when x = 0 so a - 2x = 4 setting x = 0 then a - 2(0) = 4
\(\bf {\color{brown}{ lim_{x\to 0}\ \cfrac{sin(x)}{x}}}=1\) <---? nope, is just from the "squeeze theorem"
hmm
Oh, yeah I just remembered that proof. So, a=4 would be correct then?
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