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gabie1121
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find the values for c such that the function is continuous at all x. FUNCTION BELOW
 one year ago
 one year ago
gabie1121 Group Title
find the values for c such that the function is continuous at all x. FUNCTION BELOW
 one year ago
 one year ago

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gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
\[f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}\]
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
\[f(x)=(e ^{x}2c) if x \ge0\]
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
those go together. I just couldn't figure out how to get them all in at once
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
In order for this function to be continuous, we need to pick a \(c\) value that makes the two pieces connect together nicely at \(x=0\). Imagine railroad tracks, the track needs to be continuous, and can't have any sharp corners or the train will fly off the tracks.
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
Yep, that i get. So do i set both functions equal to 0 and solve for c or something?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
We want to look at them in the limit. We want to see what is happening when we get closer and closer to 0 from the left side. And also what is happening when we get closer and closer to 0 from the right side. For this function to be continuous, they must be approaching the same value. So we'll set these limits equal to one another.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large \lim_{x \rightarrow 0^}f(x) \quad = \quad \lim_{x \rightarrow 0^} \frac{\sin(cx)}{x}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Understand why our function is sin(cx)/x when we're approaching from the left?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
The tiny negative (that looks like an exponent) is letting us know we're approaching from the left.
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
yep cuz it says less than 0
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Ok cool :) so we'll set that equal to the other piece.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
And solve for c.
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
do we make the other side a limit as well?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large \lim_{x \rightarrow 0^}f(x) \qquad \qquad = \qquad \qquad \lim_{x \rightarrow 0^+}f(x)\]\[\large \lim_{x \rightarrow 0^} \frac{\sin(cx)}{x} \qquad = \qquad \lim_{x \rightarrow 0^+}e^x2c\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yes :) the limits need to agree in order for this function to be continuous.
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
alrighty, well the left side is equal to just c, right?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yes very good ^^
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
because if i multiply by C , i can use the rule that says sinx/x=1
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
you remembered your identity i take it hehe
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
yep! the right side is giving me fits though lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Or would it be 1 since we're coming from the left? Hmm I didn't think about that. lemme check real quick.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Nah it's still 1, my bad.
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
well would the right just be e^x2? Or can i not do that cuz i'm thinking of derivative rules?
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
well the derivative is the limit as x approaches 0, so shouldn't e^x, stay e^x?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
no we're not thinking of this as a derivative :) We're looking at the limit and saying to ourselves, "If I plug x=0 directly into this function, does it cause a problem?" If the answer is no, then we can do just that!
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
oh! well then we're left with \[c=2c\] ?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Woops! Recall that if we have a 0 in the exponent, what will that change our base to?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Not 0 silly! :O
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
oh well i just plugged in e^0 in my calculator and it gave me 1
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
hah XD that's a way to do it i guess! :D
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
lol so its actuallly c=12c?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yah looks good c:
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Do you by chance have an answer key that we can check this against? This is one of those annoying problems that it's easy to make a mistake on c: lol
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
c=1/3! , and I don't have one yet. I will monday so its not a huge deal
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
It looks logical enough to me. Thanks, AGAIN lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
This type of problem becomes a little bit harder when they throw `2 unknown constants` at you. Because then you have to also look at the limits of their derivatives. But this was a good problem to get a feel for the concept c:
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
well i just started calc1 this semester, so i haven't learned much yet. Just dipping my toes in the water. I have to take all the way through calculus 3 though, bleh
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Hmm you'll do quite well, I can tell. You seem quite smart. You're very quick on remember how to do little steps. Calc 2 is a doozy!! Power Series made me want to rip my hair out! :O
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, i'm told i'll want to murder myself with Calc 2. Not looking forward to it, but thanks! That makes me feel a little better
 one year ago

gabie1121 Group TitleBest ResponseYou've already chosen the best response.0
i might be hunting you down again! lol
 one year ago
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