# Prove problem help

1. May 25, 2008

### razored

This not homework(self-study book). I do not know where to begin to prove this. This is from "Essentials Calculus" page 45, problem # 11

$$f(x) = \left\{ \begin{array}{rcl}{-1} & \mbox{if}& -\infty < x < -1, \\ x & \mbox{if} & -1\leq x\leq1, \\1 & \mbox{if} & 1 < x <\infty ,\end{array}\right$$
$$g(x) = \frac {1} {2} |x+1 | - \frac {1}{2}|x-1|$$
Prove that $$f(x)\equiv g(x)$$

Latex is awesome First time using it here!

2. May 25, 2008

### rock.freak667

where's your attempt at the proof?

But consider what g(x) is for the conditions for f(x)

3. May 26, 2008

### Dick

Glad you like latex. But, yes, as rock.freak667 points out, split into the three cases that define f.

4. May 26, 2008

### razored

I don't understand ... "split into the three cases that define f." the cases are listed.

I have no idea where to begin to prove this; a profound explanation would be helpful.

5. May 26, 2008

### nicksauce

He means split g(x) into the three cases that define f(x). You must show that when x < -1, g(x) = -1, when -1 < x < 1, g(x) = x, and when x > 1, g(x) = 1.

6. May 26, 2008

### D H

Staff Emeritus
Consider this simpler problem:

$$f(x) = \begin{cases}-x & \text{if}\ x < 0 \\ \phantom{-}x & \text{if}\ x>=0 \end{cases}$$

$$g(x) = |x|$$

For $x>=0$, $g(x)=|x|=x[/tex]. For [itex]x<0$, $g(x)=|x|=-x[/tex]. In both cases, [itex]g(x)=f(x)$. The functions are identical for all x.

You can do the same thing with your $f(x)$ and $g(x)$. In particular, what does $g(x)$ evaluate to in each of the three regions?

7. May 31, 2008

### mistermath

If you want to see how to write it formally out you would do something like:
Case 1: Suppose x < -1 then g(x) = ... = f(x)
Case 2: Suppose x > 1 then g(x) = ... = f(x)
Case 3: Suppose -1<=x<=1 then g(x) = ... = f(x)

Two functions f,g are equal in an interval I if f(x)=g(x) for every x in I. (In your case your interval is all real #'s.)

8. May 31, 2008

### razored

Ok. Figured it out, thanks!