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anonymous
 3 years ago
find the values for c such that the function is continuous at all x. FUNCTION BELOW
anonymous
 3 years ago
find the values for c such that the function is continuous at all x. FUNCTION BELOW

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=(e ^{x}2c) if x \ge0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0those go together. I just couldn't figure out how to get them all in at once

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0In 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yep, that i get. So do i set both functions equal to 0 and solve for c or something?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0We 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.

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep cuz it says less than 0

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Ok cool :) so we'll set that equal to the other piece.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do we make the other side a limit as well?

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Yes :) the limits need to agree in order for this function to be continuous.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alrighty, well the left side is equal to just c, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because if i multiply by C , i can use the rule that says sinx/x=1

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0you remembered your identity i take it hehe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep! the right side is giving me fits though lol

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Nah it's still 1, my bad.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well would the right just be e^x2? Or can i not do that cuz i'm thinking of derivative rules?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well the derivative is the limit as x approaches 0, so shouldn't e^x, stay e^x?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0no 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh! well then we're left with \[c=2c\] ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh well i just plugged in e^0 in my calculator and it gave me 1

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0hah XD that's a way to do it i guess! :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol so its actuallly c=12c?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Do 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It looks logical enough to me. Thanks, AGAIN lol

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0This 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:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i might be hunting you down again! lol
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