# Homework Help: Domain of a Radial Function

1. Jan 16, 2010

### crono_

1. The problem statement, all variables and given/known data

Find the domain of the function.

h(x) = $$\sqrt{4 - x}$$ + $$\sqrt{x^2 - 1}$$

2. Relevant equations
a2 - b2 = (a - b) (a + b)

I think that's all...

3. The attempt at a solution

For starters, I know we can't have a negative number in the radical.

I looked at the 2nd radical first and broke it up...

$$\sqrt{x^2 - 1}$$ = $$\sqrt{(x + 1) (x - 1)}$$

Then I looked at the first radical...

$$\sqrt{4 - x}$$

$$\sqrt{-x + 4}$$

$$\sqrt{- (x - 4)}$$

So I'm left with...

= $$\sqrt{- (x - 4)}$$ + $$\sqrt{(x + 1) (x - 1)}$$

Then found the zero's, but are those necessary with a radical question? I know we can't have zero as a denominator but a zero in a radical is fine.

x = 1
x = -1
x = 4

Now I feel stuck though...I think it's the addition of the two radicals that's throwing me off. If there was just one then it would probably be simpler.

2. Jan 16, 2010

### vela

Staff Emeritus
You're making things hard on yourself. ;)

You said, "For starters, I know we can't have a negative number in the radical." That's all you need. You want both quantities in the radical to be non-negative, so you have:

$$4-x \ge 0$$

and

$$x^2-1 \ge 0$$

Find the conditions on $x$ to satisfy each inequality. The domain of the function are those values of $x$ which satisfy both inequalities simultaneously.

3. Jan 16, 2010

### crono_

It's a habit. :P

Okay...

4 - x $$\geq$$ 0

4 $$\geq$$ x

x2 - 1 $$\geq$$ 0

x2 $$\geq$$ 1

x $$\geq$$ $$\sqrt{1}$$ or just 1

So, now...

1 $$\leq$$ x $$\leq$$ 4

D: [1,4] ???

4. Jan 16, 2010

### vela

Staff Emeritus
So far so good.

Not quite. Remember that $\sqrt{x^2} = |x|$. In other words, you missed a bunch of negative solutions, like $x=-2$ since $(-2)^2=4\geq 1$.

That's one piece. You need to fix your mistake above to find the rest of the domain.

5. Jan 16, 2010

### crono_

I don't know...

The answer is supposed to be (- infinity, -1] U [1,4], but I'm not making sense of that. Anything squared, positive or negative, will end up being positive. Meaning that -22 (from your example) will be positive and therefore part of the domain. But I'm not sure I even understand what I'm saying. If I were presented with a similar question, then I'd be back to square one again.

Would it be possible for you to elaborate a little bit? Let's assume that I'm drawing a complete blank here...

6. Jan 16, 2010

### vela

Staff Emeritus
You have to find all the solutions to $x^2\geq 1$. The positive answers $x\geq 1$ are the ones you normally think of, but there are negative ones too, namely $x\leq -1$.

It's similar to when you solve $x^2=1$. There are two solutions, +1 and -1. In your case, instead of the equals sign, you have an inequality instead. You just have to be careful you get the directions right.

7. Jan 16, 2010

### Mentallic

It's easiest to understand by looking at a graph.
Take the function $f(x)=x^2-1$. Now, to find where $x^2-1\geq0$ just look at where the function f(x) is above the x-axis. This occurs at $x\geq 1$ AND $x\leq -1$.

Also, you can test this by just using random values of x:

test $x^2-1\geq 0$

x=2, $2^2-1=4\geq 0$ therefore true.

x=-3 $(-3)^2-1=9-1=8\geq 0$ therefore also true. This is why you missed the entire interval of (-$\infty$,-1]

8. Jan 17, 2010

### crono_

Okay, it's clicking. Making the graph helped a fair bit. Would you always recommend looking at a graph if you're stuck? Or do you find that to be too time consuming?

Once I have f(x) = x2 - 1 graphed, would it make sense to put f(x) = 4 - x on the same graph? If they intersect with each other, is that...okay? Is rewriting f(x) = 4 - x as f(x) = -x + 4 acceptable?

"In your case, instead of the equals sign, you have an inequality instead. You just have to be careful you get the directions right."

I believe I did get the directions incorrect as I had both:

x $$\geq$$ 1

and

x $$\geq$$ -1

That led some confusion.

Sorry for all the questions. Math generally goes in one ear and then out the other. I don't use it enough for it to stick.

9. Jan 17, 2010

### vela

Staff Emeritus
Graph away! Graphing can be incredibly helpful, even when you're not stuck, to understand what's going on and to give you another way of looking at a problem that you might have overlooked.

In this case, I don't think putting them on the same graph would buy you anything.