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G composed with F

dwsmith

Well-known member
Feb 1, 2012
1,673
$F(x) = x + 5\qquad\qquad G(x) = \frac{|x|}{x}, \ \text{if} \ x\neq 0, \ G(0) = 1$




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

Is the correct for the composition?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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
Feb 1, 2012
1,673
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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
Jan 26, 2012
4,191
$F(x) = x + 5\qquad\qquad G(x) = \frac{|x|}{x}, \ \text{if} \ x\neq 0, \ G(0) = 1$




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

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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191