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
Be careful with your parentheses there. $\displaystyle a x-(4/x)-2$ is very different from $\displaystyle a x-4/(x-2).$ Which one did you mean?
Generally, I'd simply use the calculus definition of continuity here: make
$\displaystyle \displaystyle{\lim_{x\to 2}f(x)=f(2)}$
happen.
You mean $\displaystyle \displaystyle f(x) = \left \{ \begin{array}{ll} \frac {ax - 4}{x - 2} & \text{ if } x \ne 2 \\ & \\ b & \text{ if } x = 2 \end{array} \right.$ I take it.
Note that you want $\displaystyle \displaystyle ax - 4$ to have a factor of $\displaystyle \displaystyle x - 2$ in it. So...?
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 $\displaystyle \displaystyle ax - \frac{4}{x} - 2$
and
ax-4/(x-2) means $\displaystyle \displaystyle ax - \frac{4}{x-2}$
and
(ax-4)/(x-2) means $\displaystyle \displaystyle \frac{ax-4}{x-2}$
It wouldn't be a bad idea to learn how to use LaTeX either. This code
[tex]f(x)=\begin{cases}\dfrac{ax-4}{x-2}&,\ x\ne2\\\\b&,\ x=2\end{cases}[/tex]
produces
$\displaystyle f(x)=\begin{cases}\dfrac{ax-4}{x-2}&,\ x\ne2\\\\b&,\ x=2\end{cases}$
As to your question.. like Ackbeet said, you need $\displaystyle \displaystyle \lim_{x \to 2} \frac{ax-4}{x-2}=b$. In order to have this you will need to follow Jhevon's advice.