Question

For what value of c is the function defined below continuous on (-\infty,\infty)?
f(x) = \begin{cases} x^2 - c^2, & x < 4, \\ c x + 20, & x \geq 4. \end{cases}

Answer 1

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.

What does it mean for f to be continuous at 4, what is the definition?

Answer 2

What is the definition of continuity:
A function, f, is continuous at x = 4 if .... what?

Answer 3

No idea?

Answer 4

A function, f, is continuous at x = 4 if
\[\lim_{x \rightarrow 4} \ f(x) = f(4)\]

Answer 5

So now you need to find the c for which this is true. Let me know if you get stuck.

Answer 6

Do I plug in 4 to one of the functions then use substitution? Im confused

Answer 7

What is f(4)?

And when does the limit as x-->4 of f(x) exist?

Answer 8

Is 4 where the function is continuous, and wouldn't the limit exist at 4?

Answer 9

The point is this. x = 4 is the only place where we have a question mark.

So how to resolve the question? We go back to definitions.

Answer 10

So we need to find c so that indeed the limit as x -->4 of f(x) equals f(4). Because BY DEFINITION, that is what we need for continuity at x = 4 and therefore for the entire real line, (-infty, +infty)

Answer 11

So to find out what value of c is required to make

limit as x --> 4 f(x) = f(4)

we had better understand and evaluate both sides of this equation.

So first, what is f(4) ?

Answer 12

Well, when x = 4, the second part of the definition of f(x) comes into play.

Hence f(4) = 4c + 20

Following?

Answer 13

So you took cx+20 and plugged 4 in for c?

Answer 14

No x = 4

Answer 15

oh, yeah meant the other way around

Answer 16

ugg .. the site is so slow

Answer 17

plugged 4 in for x in cx+20...

Answer 18

Ok, so that's f(4). We want the limit as x --> 4 of f(x) equal to that.

Answer 19

Now what is that limit? Well, for that limit to exist the left-hand limit and the right-hand limit have to be the same.

Answer 20

I.e., as we approach from below 4 and from above 4.

Answer 21

In notation we write respectively\[\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)\]

Answer 22

Now the second of these is easy, because for x > 4, f(x) = cx + 20. hence

limit as x --> 4+ (i.e., from above, from the right) of f(x) is just
4x + 20

Answer 23

4c + 20 sorry

Answer 24

following?

Answer 25

Yeah so far I think so

Answer 26

On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
\[\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2\]

Answer 27

Thus these two limits, the one from above and below are equal if and only if

4c + 20 = 16 - c^2

Or in other words, the limit as x --> 4 of f(x) exists if and only if

4c + 20 = 16 - c^2

Answer 28

Now you need to find the values of c for which this is true.

Answer 29

I.e., for what values of c is
c^2 + 4c + 4 = 0

Answer 30

So, for what values of c is that true?

Answer 31

Do i need to solve for c in any way?

Answer 32

However you like, it's a quadratic equation and by now I'm sure you've seen hundreds of them.

Answer 33

So, quick, I have to go so.

Answer 34

-2?

Answer 35

Yes, c = -2.

Now for c = -2, the limit as x --> 4 f(x) exists and is equal to .... what?

Answer 36

quick now, you've got the formula above

Answer 37

is it -2 as well?

Answer 38

No. Look at either of the expressions for the limit:
x --> 4+ the limit of f(x) is 4c + 20
x --> 4- the limit of f(x) is 16 - c^2

Hence for c = -2, the limit as x-->4 of f(x) is ....

Answer 39

for c = -2, the limit as x-->4 of f(x) is 12 because
4c + 20 = -8 + 20 = 12
and 16 - c^2 = 16 - 4 = 12
also

Now what is f(4) if c = -2?

Answer 40

Remember f(4) = 4c + 20
= ....., what?

Answer 41

still there?

Answer 42

Yeah, would it be 12?

Answer 43

Yes.

Answer 44

Therefore if c = -2,

\[\lim_{x \rightarrow 4} f(x) = 12 = f(4)\]

Answer 45

That is to say, if c = -2, f(x) is continuous at x = 4.

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers (-infty, +infty)

Answer 46

****

So now, do yourself a favor. Take a blank piece of paper and work the entire problem again from the beginning.

Then take another piece of blank piece of paper and work the entire problem without looking at the first piece or our discussion here.

When you can do that, then you really understand this problem.

Answer 47

OK, thanks a lot, I'll do that for sure

Answer 48

'good answer' appreciated

Answer 49

Crap yeah was going to do that, thanks again