# Domain and Range problem

The domain of $f$ is the set of all non-negative real number; range is the set $\{-1,0,2,7\}$. What are the domain and range of $A$,$B$,$C$ if:

$A(x)=f(x+2)$

$B(x)=f(x)+2$

$C(x)=2*f(-x)$

and give a convincing argument that the inverse of is not a function.

I have no idea how to start to finding the domain and range. Can someone help me out.

Loosely speaking the domain of $A(x)$ is the set of numbers for which it is possible to compute the result.

In the case of $A(x)$ you should ask: "What numbers can I input to $A$ so that I can compute a result? well since $A(x)=f(x+2)$ and $f(x)$ can only accept non-negative numbers then you reason that $x+2$ should be non negative, i.e. $x+2\ge0$ which is the same as $x$ must be greater or equal than $-2$. So the domain of $A(x)$ are all the real numbers that are greater or equal than $-2$.

With similar reasoning you find that the domain of $B(x)$ is the same as $f(x)$, i.e. the non-negative real numbers. And the domain of $C$ is the non-positive real numbers.

the range of $A(x)$ is the same as $f(x)$. the range of $B(x)$ is the set $\{1,2,4,9\}$ because you add the number $2$ to the result of $f(x)$ and the range of $C(x)$ is the set $\{-2,0,4,14\}$ since you multiply by $2$ the result of $f(x)$.

$f(x)$ cannot have an inverse because your domain is a much bigger set than its range. Since the range of $f(x)$ has only $4$ numbers and the domain of $f(x)$ has much much more (infinitely many more).

• Good answer, but can you explain why $A(x)$ has the same domain as $f(x)$? It is a correct statement, but it's not entirely obvious why it is. Aug 19 '13 at 16:01
• @Ataraxia: I think that you meant that $A$ has the same range as $f$. I agree that a little more explanation there would be good, something like this: for each $y\in\{-1,0,2,7\}$ there is an $x\ge 0$ such that $f(x)=y$, and in that case $x-2\ge-2$, so $x-2$ is in the range of $A$, and $A(x-2)=f\big((x-2)+2\big)=f(x)=y$, so $y$ is in the range of $A$. Aug 19 '13 at 17:10
• Yes, that's what I meant to say, sorry for the confusion. Aug 19 '13 at 17:12