Trying to find limit involving continuity concept

[Torshi]

Homework Statement

Find c such that the function f(x) { x^2-9 while x≤ c and 6x-18 x > c } is continuous everywhere.

Homework Equations

Given above. Basic algebra.

The Attempt at a Solution

I made a number line. Showing that x^2-9 is approaching from the left side and 6x-18 is approaching from the right side given the designation of the inequality.

If there were numbers, I could easily do it, but the "c" is throwing me off.

An attempt to a similar problem with numbers would simply be chug and plug and knowing which side it comes from left or right, and if there was a variable "a" in there as well you can substitute.

I have no idea how to start the problem, would it become c^2-9 and 6c-18? and try to solve for c?

 

[SithsNGiggles]

$f(x)=\begin{cases} x^2-9 & \mbox{if } x \leq c\\ 6x-18 & \mbox{if } x>c\end{cases}$

For $f(x)$ to be continuous at $c$, you must satisfy
$\displaystyle\lim_{x\to c^{+}} f(x) = \displaystyle\lim_{x\to c^{-}} f(x)$

That is,
$\displaystyle\lim_{x\to c^{+}} x^2-9 = \displaystyle\lim_{x\to c^{-}} 6x-18$

That's basically what (it seems) you've done already, but this is the reasoning behind the step you've taken.

 

[SammyS]

Homework Statement

Find c such that the function f(x) { x^2-9 while x≤ c and 6x-18 x > c } is continuous everywhere.

Homework Equations

Given above. Basic algebra.

The Attempt at a Solution

I made a number line. Showing that x^2-9 is approaching from the left side and 6x-18 is approaching from the right side given the designation of the inequality.

If there were numbers, I could easily do it, but the "c" is throwing me off.

An attempt to a similar problem with numbers would simply be chug and plug and knowing which side it comes from left or right, and if there was a variable "a" in there as well you can substitute.

I have no idea how to start the problem, would it become c^2-9 and 6c-18? and try to solve for c?

IMO: It's more important to know why a method works than to know a method without understanding it.

So, here are some questions:

What is necessary in order that f(x) be continuous at x = c ?

What is $\displaystyle \lim_{x\to c^-}\,x^2-9\ ?$

What is $\displaystyle \lim_{x\to c^+}\,6x-18\ ?$

 

[Torshi]

$f(x)=\begin{cases} x^2-9 & \mbox{if } x \leq c\\ 6x-18 & \mbox{if } x>c\end{cases}$

For $f(x)$ to be continuous at $c$, you must satisfy
$\displaystyle\lim_{x\to c^{+}} f(x) = \displaystyle\lim_{x\to c^{-}} f(x)$

That is,
$\displaystyle\lim_{x\to c^{+}} x^2-9 = \displaystyle\lim_{x\to c^{-}} 6x-18$

That's basically what (it seems) you've done already, but this is the reasoning behind the step you've taken.

But I don't know how to move on from there.. I end up getting something like x/x = √-3

 

[Torshi]

IMO: It's more important to know why a method works than to know a method without understanding it.

So, here are some questions:

What is necessary in order that f(x) be continuous at x = c ?

What is $\displaystyle \lim_{x\to c^-}\,x^2-9\ ?$

What is $\displaystyle \lim_{x\to c^+}\,6x-18\ ?$

do you have to set the two limits equal?

 

[SammyS]

But I don't know how to move on from there.. I end up getting something like x/x = √-3

If you solve c²-9 =6c-18 for c,

how do you get x/x = √-3 ?

There is no x in c²-9 =6c-18 , and x/x = 1 .

 

[SammyS]

do you have to set the two limits equal?

Yes, which should be clear if you answer the question about the function being continuous at x = c .

 

[Torshi]

If you solve c²-9 =6c-18 for c,

how do you get x/x = √-3 ?

There is no x in c²-9 =6c-18 , and x/x = 1 .

I put x because the other poster. But, essentially wouldn't it be:

c^2-9 = 6c-18
c^2=6c-9
c^2/c = -3
..idk what to do ... if c was a number in the inequality then i could easily do this problem

alright hold on..

 

[Torshi]

bump

nvm got it.

Since f is continuous everywhere, it is continuous at x = c.
So the left-hand limit must equal the right-hand limit. Thus:
c^2 - 9 = 6c - 18
c^2 - 6c + 9 = 0
(c - 3)^2 = 0
c = 3

thank you!

 

[SithsNGiggles]

do you have to set the two limits equal?

Yes, that's what I was suggesting.