1. ## Continuity problem

Find constant A and B so that the given function will be continuous for all x values.

Piece wise function f(x) = a(x)-4/x-2 if x does not equal 2, and b x = 2

So confused lol

2. Be careful with your parentheses there. $a x-(4/x)-2$ is very different from $a x-4/(x-2).$ Which one did you mean?

Generally, I'd simply use the calculus definition of continuity here: make

$\displaystyle{\lim_{x\to 2}f(x)=f(2)}$

happen.

3. It's the second one, ax-4/(x-2). I can't just substitute 2 in as it is is because it would give me 0 on the bottom so I don't know how to get rid of that on the bottom

4. Originally Posted by Nikhiln25
Find constant A and B so that the given function will be continuous for all x values.

Piece wise function f(x) = a(x)-4/x-2 if x does not equal 2, and b x = 2

So confused lol

What is a(x)? Is it a*x? If yes, it should be ax^2 (I think..)

5. Originally Posted by Also sprach Zarathustra
What is a(x)? Is it a*x? If yes, it should be ax^2 (I think..)
Yes a times x, not ax^2

The problem in the book is given as

ax-4
-----
x-2 if x=/= 2

b if x=2

6. Originally Posted by Nikhiln25
Yes a times x, not ax^2

The problem in the book is given as

ax-4
-----
x-2 if x=/= 2

b if x=2
You mean $\displaystyle f(x) = \left \{ \begin{array}{ll} \frac {ax - 4}{x - 2} & \text{ if } x \ne 2 \\ & \\ b & \text{ if } x = 2 \end{array} \right.$ I take it.

Note that you want $\displaystyle ax - 4$ to have a factor of $\displaystyle x - 2$ in it. So...?

7. Originally Posted by Nikhiln25
It's the second one, ax-4/(x-2).
Based on your subsequent post, this is incorrect. You mean (ax-4)/(x-2).

Using universally accepted order of operations,

ax-4/x-2 means $\displaystyle ax - \frac{4}{x} - 2$

and

ax-4/(x-2) means $\displaystyle ax - \frac{4}{x-2}$

and

(ax-4)/(x-2) means $\displaystyle \frac{ax-4}{x-2}$

It wouldn't be a bad idea to learn how to use LaTeX either. This code

$$f(x)=\begin{cases}\dfrac{ax-4}{x-2}&,\ x\ne2\\\\b&,\ x=2\end{cases}$$

produces

$f(x)=\begin{cases}\dfrac{ax-4}{x-2}&,\ x\ne2\\\\b&,\ x=2\end{cases}$

As to your question.. like Ackbeet said, you need $\displaystyle \lim_{x \to 2} \frac{ax-4}{x-2}=b$. In order to have this you will need to follow Jhevon's advice.

8. Hint:

it's a=2

9. Hey, do like I do- erase the mistake so you can pretend you never made it!