Question

Let c be a constant, and let f(x) be the following function:
(x^2+6x-16
(3x^2-6x+2 (fraction) ,x=/=2
f(x)=(
(7x-5 , x=2
Find all values of c (if any exist) which make f(x) continuous at x=2.

Answer 1

OK.... that is just hard to read. Lets see what we can do with that first.

Answer 2

Oh sorry I forgot to mention I already solved the fraction part. I came out with 2.

Answer 3

I'm just having some trouble with the 7c-5 part. How can I take the limit of 7c-5 when I'm only given x...?

Answer 4

You are looking to see where it makes the function continuous.

Answer 5

Yea...

Answer 6

is the \(3x^2-6x+2\) over \(3x^2-6x+2\) or? Just trying to get a grasp on the problem first.

Answer 7

Yea sorry.

Answer 8

And all that is at \(x\ne 2\)

Answer 9

Wait wait... it's... (x^2+6x-16)/(3x^2-7x+2) at x=/=2

Answer 10

Is that clearer?

Answer 11

\(f(x)\begin{cases}
\dfrac{x^2+6x-16}{3x^2-7x+2} & \text{ if } x\ne 2 \\
7x-5 & \text{ if } x= 2
\end{cases}\)

Answer 12

Yea. That's it. I wasn't sure how to do all the symbols and stuff.

Answer 13

No. The bottom is 7c-5

Answer 14

Right click that and look at show math as tex.

Answer 15

OK. Not much of a difference.

Answer 16

Have you taken the limit of:
\(\dfrac{x^2+6x-16}{3x^2-7x+2}\)

Answer 17

Yea and I got 2...

Answer 18

But if x=/=2 did I do something wrong?

Answer 19

well, it is:
\(\lim\limits_{x\rightarrow2 }\dfrac{x^2+6x-16}{3x^2-7x+2}\)
that you need to look at.

That is because your goal is "Find all values of c (if any exist) which make f(x) continuous at x=2." For it to be continuous, there must NOT be a gap ar x=2.

Answer 20

If you just plug in you get 0/0. So I factored to get ((x+8)(x-2))/((3x-1)(x-2)), then cancelled the two x-2's and got (x+8)/(3x-1), then when I plugged 2 in I got 10/5, or 2. Is that right?

Answer 21

\(\lim\limits_{x\rightarrow2 }\dfrac{x^2+6x-16}{3x^2-7x+2}\)

\(\lim\limits_{x\rightarrow2 }\dfrac{(x+8)\cancel{(x-2)}}{\cancel{(x-2)}(3x-1)}\)

\(\lim\limits_{x\rightarrow2 }\dfrac{(x+8)}{(3x-1)}\)

Answer 22

And then you plug in 2 for all the x's... right?

Answer 23

Yep. And that gets you, as you said, 2. So, (2,2) is a hole, or undefined. And you want to fill it. Now, what does that mean the second equation must equal at x=2?

Answer 24

c=2....? The c throws me off...

Answer 25

Or... oh wait. 7c-5=2? So then solve for c?

Answer 26

7c=7, c=1?

Answer 27

Yah.

Answer 28

So then what would the answer be?

Answer 29

Is it (2,1)....? Or...

Answer 30

c=1, as you said.

Answer 31

That's the final answer?

Answer 32

Oh I get it. Find all values of c. So c=1 is the answer...

Answer 33

Yep. 1. See, you are looking for the value of c that makes f(x)=2 a valid equation. That is it.

Answer 34

Thanks so much. I've been stuck on this for a while. I need to start taking better notes. :P

Answer 35

np. Limits are an odd concept. They are not where you are, but where you are going. Wait until you learn that \(\frac{1}{3}=\infty\) in one respect.

Answer 36

Lol I can't wait. :P Thanks again.

Answer 37

It has to do with limits. If you take .3 + .03 + .003 + ... and so on, you get .3 that repeats until infinity. And that is 1/3rd. You take a limit of a sum function to prove this.

Answer 38

Oh. o.o That's weird. I don't like math lol.

Answer 39

I'm gonna close this question and open a new one, lol. Got a few more problems I can't finish.

Answer 40

Kk.