Question

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

Answer 1

\[f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}\]

Answer 2

\[f(x)=(e ^{x}-2c) if x \ge0\]

Answer 3

those go together. I just couldn't figure out how to get them all in at once

Answer 4

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.

Answer 5

Yep, that i get. So do i set both functions equal to 0 and solve for c or something?

Answer 6

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.

Answer 7

\[\large \lim_{x \rightarrow 0^-}f(x) \quad = \quad \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x}\]

Answer 8

Understand why our function is sin(cx)/x when we're approaching from the left?

Answer 9

The tiny negative (that looks like an exponent) is letting us know we're approaching from the left.

Answer 10

yep cuz it says less than 0

Answer 11

Ok cool :) so we'll set that equal to the other piece.

Answer 12

And solve for c.

Answer 13

do we make the other side a limit as well?

Answer 14

\[\large \lim_{x \rightarrow 0^-}f(x) \qquad \qquad = \qquad \qquad \lim_{x \rightarrow 0^+}f(x)\]\[\large \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x} \qquad = \qquad \lim_{x \rightarrow 0^+}e^x-2c\]

Answer 15

Yes :) the limits need to agree in order for this function to be continuous.

Answer 16

alrighty, well the left side is equal to just c, right?

Answer 17

Yes very good ^^

Answer 18

because if i multiply by C , i can use the rule that says sinx/x=1

Answer 19

you remembered your identity i take it hehe

Answer 20

yep! the right side is giving me fits though lol

Answer 21

Or would it be -1 since we're coming from the left? Hmm I didn't think about that. lemme check real quick.

Answer 22

Nah it's still 1, my bad.

Answer 23

well would the right just be e^x-2? Or can i not do that cuz i'm thinking of derivative rules?

Answer 24

well the derivative is the limit as x approaches 0, so shouldn't e^x, stay e^x?

Answer 25

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!

Answer 26

oh! well then we're left with \[c=2c\] ?

Answer 27

Woops! Recall that if we have a 0 in the exponent, what will that change our base to?

Answer 28

Not 0 silly! :O

Answer 29

oh well i just plugged in e^0 in my calculator and it gave me 1

Answer 30

hah XD that's a way to do it i guess! :D

Answer 31

yah 1 :3

Answer 32

lol so its actuallly c=1-2c?

Answer 33

Yah looks good c:

Answer 34

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

Answer 35

c=1/3! , and I don't have one yet. I will monday so its not a huge deal

Answer 36

It looks logical enough to me. Thanks, AGAIN lol

Answer 37

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:

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

i might be hunting you down again! lol