Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}\]
\[f(x)=(e ^{x}-2c) if x \ge0\]
those go together. I just couldn't figure out how to get them all in at once

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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.
And solve for c.
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?
Yes very good ^^
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?
Not 0 silly! :O
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
yah 1 :3
lol so its actuallly c=1-2c?
Yah looks good c:
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question