Here's the question you clicked on:
gabie1121
find the values for c such that the function is continuous at all x. FUNCTION BELOW
\[f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}\]
\[f(x)=(e ^{x}-2c) if x \ge0\]
those go together. I just couldn't figure out how to get them all in at once
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.
Yep, that i get. So do i set both functions equal to 0 and solve for c or something?
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.
\[\large \lim_{x \rightarrow 0^-}f(x) \quad = \quad \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x}\]
Understand why our function is sin(cx)/x when we're approaching from the left?
The tiny negative (that looks like an exponent) is letting us know we're approaching from the left.
yep cuz it says less than 0
Ok cool :) so we'll set that equal to the other piece.
do we make the other side a limit as well?
\[\large \lim_{x \rightarrow 0^-}f(x) \qquad \qquad = \qquad \qquad \lim_{x \rightarrow 0^+}f(x)\]\[\large \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x} \qquad = \qquad \lim_{x \rightarrow 0^+}e^x-2c\]
Yes :) the limits need to agree in order for this function to be continuous.
alrighty, well the left side is equal to just c, right?
because if i multiply by C , i can use the rule that says sinx/x=1
you remembered your identity i take it hehe
yep! the right side is giving me fits though lol
Or would it be -1 since we're coming from the left? Hmm I didn't think about that. lemme check real quick.
Nah it's still 1, my bad.
well would the right just be e^x-2? Or can i not do that cuz i'm thinking of derivative rules?
well the derivative is the limit as x approaches 0, so shouldn't e^x, stay e^x?
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!
oh! well then we're left with \[c=2c\] ?
Woops! Recall that if we have a 0 in the exponent, what will that change our base to?
oh well i just plugged in e^0 in my calculator and it gave me 1
hah XD that's a way to do it i guess! :D
lol so its actuallly c=1-2c?
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
c=1/3! , and I don't have one yet. I will monday so its not a huge deal
It looks logical enough to me. Thanks, AGAIN lol
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:
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
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
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
i might be hunting you down again! lol