- Thread starter
- #1

$G(F(x)) = G(x + 5) = \frac{|x + 5|}{x + 5} = \begin{cases}

1 & \text{if} \ x \geq 0\\

-1 & \text{if} \ x\in (0,-5)\cup (-5,\infty)

\end{cases}$

Is the correct for the composition?

- Thread starter dwsmith
- Start date

- Thread starter
- #1

$G(F(x)) = G(x + 5) = \frac{|x + 5|}{x + 5} = \begin{cases}

1 & \text{if} \ x \geq 0\\

-1 & \text{if} \ x\in (0,-5)\cup (-5,\infty)

\end{cases}$

Is the correct for the composition?

- Admin
- #2

- Jan 26, 2012

- 4,202

$$G(F(x))=G(x+5)=\begin{cases}\frac{|x+5|}{x+5}, \quad & x+5\not=0\\

1, \quad & x+5=0\end{cases},$$

and go from there.

- Thread starter
- #3

View attachment 325

$$G(F(x))=G(x+5)=\begin{cases}\frac{|x+5|}{x+5}, \quad & x+5\not=0\\

1, \quad & x+5=0\end{cases},$$

and go from there.

I just graphed it and it looks right.

- Feb 29, 2012

- 342

- Admin
- #5

- Jan 26, 2012

- 4,202

Your definition has a conflict in it. Suppose $x=1$. Then it satisfies $x\geq 0$ as well as being in the interval $(-5,\infty)$. So $G \circ F$ would evaluate both to $+1$ and $-1$. Therefore, the composition you have defined there is not a function, but a relation. Either that, or it's an ill-defined function.

$G(F(x)) = G(x + 5) = \frac{|x + 5|}{x + 5} = \begin{cases}

1 & \text{if} \ x \geq 0\\

-1 & \text{if} \ x\in (0,-5)\cup (-5,\infty)

\end{cases}$

Is the correct for the composition?

- Admin
- #6

- Jan 26, 2012

- 4,202

What exact commands did you execute to produce this graph?View attachment 325

I just graphed it and it looks right.