# Piecewise functions

1. Sep 27, 2009

### fghtffyrdmns

1. The problem statement, all variables and given/known data
We started piecewise functions in calculus and I'm finding it rather difficult. This question has had me stumped for a while. I'm quite well at finding domain and range, however :).

For what value of k is the following a continuous function

2. Relevant equations

$$f(x)= \frac {x-2}{\sqrt {7x+2} -\sqrt{6x+4}}$$ $$if x \geq {\frac{-2}{7}} and x \neq 2$$
f(x) = k, if x=2

I hope this makes sense.
3. The attempt at a solution

The problem I am having is how I can find k - the constanst. I can't $$\frac {x-2}{\sqrt {7x+2} -\sqrt{6x+4}} = k$$ since k cannot equal to 2.

2. Sep 27, 2009

### Bohrok

You want to cancel out or get rid of the denominator and rewrite the expression in a way where you don't get undefined when x=2. Try multiplying by the conjugate of the denominator.

3. Sep 27, 2009

### fghtffyrdmns

I was thinking of just rationalizing it. But what do I do with the k then?

4. Sep 27, 2009

### Bohrok

If you graph f(x) on a graphing calculator, you would see that the graph clearly approaches a value as x gets close to 2. That is because when you plug x=2 into the function, you get 0/0 which usually, but not always, means that the function actually approaches a value as x gets close to the value that it can't be, the one that makes the function undefined. Once you find the value k that f(x) gets close to as x approaches 2, that will be the value to make the function continuous.

Think of x(x-1)/(x-1) which is undefined at x=1 and its graph has a "hole" there. If I do some canceling, I get just x. Then I can let x=1 and see that the graph of x(x-1)/(x-1), which coincides with the graph of x, gets close to 1 as x approaches 1, just like the graph of x when x gets close to 1.

5. Sep 27, 2009

### fghtffyrdmns

So if I rationalize it, to remove the square roots, can I equate the 2? Then use x = -2/7?

6. Sep 27, 2009

### fghtffyrdmns

I rationalized it:

$$\frac {x-2 \sqrt {7x+2} + x-2\sqrt{6x+4}}{x-2}$$

$$\sqrt {7x+2} + x-2\sqrt{6x+4} = k$$

Substituting x=2, I get k=4.

7. Sep 28, 2009

### Bohrok

k isn't 4. After multiplying by the conjugate of the denominator and simplifying, the x-2 should cancel; looks like you didn't distribute correctly.

$$\frac{x - 2}{\sqrt{7x + 2} - \sqrt{6x + 4}} * \frac{\sqrt{7x + 2} + \sqrt{6x + 4}}{\sqrt{7x + 2} + \sqrt{6x + 4}} = \frac{(x - 2)(\sqrt{7x + 2} + \sqrt{6x + 4})}{7x + 2 - (6x + 4)} = \frac{(x - 2)(\sqrt{7x + 2} + \sqrt{6x + 4})}{x - 2}$$

Last edited: Sep 28, 2009
8. Sep 28, 2009

### fghtffyrdmns

Ohhh, you don't multiply the x-2 in to the conjugate? I understand. I got the x-2 on both the top and bottom, but I multiplied through. I see what I did wrong, thank you.