Prove f(x)=g(x) - Essentials of Calculus p.45, Problem 11

[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!

 

[rock.freak667]

where's your attempt at the proof?

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

 

[Dick]

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

 

[razored]

where's your attempt at the proof?

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

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

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.

 

[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.

 

[D H]

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 $x>=0$, [itex]g(x)=|x|=x[/tex]. For $x<0$, [itex]g(x)=|x|=-x[/tex]. In both cases, $g(x)=f(x)$. 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?

 

[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.)

 

[razored]

Ok. Figured it out, thanks!