# G composed with F

#### dwsmith

##### Well-known member
$F(x) = x + 5\qquad\qquad G(x) = \frac{|x|}{x}, \ \text{if} \ x\neq 0, \ G(0) = 1$

$G(F(x)) = G(x + 5) = \frac{|x + 5|}{x + 5} = \begin{cases} 1 & \text{if} \ x \geq 0\\ -1 & \text{if} \ x\in (0,-5)\cup (-5,\infty) \end{cases}$

Is the correct for the composition?

#### Ackbach

##### Indicium Physicus
Staff member
I'm not sure I'd quite agree yet. I would do
$$G(F(x))=G(x+5)=\begin{cases}\frac{|x+5|}{x+5}, \quad & x+5\not=0\\ 1, \quad & x+5=0\end{cases},$$
and go from there.

#### dwsmith

##### Well-known member
I'm not sure I'd quite agree yet. I would do
$$G(F(x))=G(x+5)=\begin{cases}\frac{|x+5|}{x+5}, \quad & x+5\not=0\\ 1, \quad & x+5=0\end{cases},$$
and go from there.
View attachment 325
I just graphed it and it looks right.

#### Fantini

MHB Math Helper
I agree with Ackbach. Perhaps it would be better if you just worked with the definition of $G$ as given instead of breaking it into exact expressions.

#### Ackbach

##### Indicium Physicus
Staff member
$F(x) = x + 5\qquad\qquad G(x) = \frac{|x|}{x}, \ \text{if} \ x\neq 0, \ G(0) = 1$

$G(F(x)) = G(x + 5) = \frac{|x + 5|}{x + 5} = \begin{cases} 1 & \text{if} \ x \geq 0\\ -1 & \text{if} \ x\in (0,-5)\cup (-5,\infty) \end{cases}$

Is the correct for the composition?
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.

Staff member