# Range and domain of $x\mapsto 1-f(x+1)$, knowing those of $f$

## Problem:

Let $$f$$ be a function which has domain $$D_f=[-1,2]$$ and range $$=[0,1]$$. What are the domain and range of the function $$g$$ defined by $$g(x) = 1-f(x+1)$$?

## My thinking:

If the domain of $$f$$ is $$[-1,2]$$, then the domain of $$x\mapsto f(x+1)$$ is $$[0,3]$$ (adding 1 to both the extreme limits of domain of $$f$$).

And because the range of $$f$$ is $$[0,1]$$, the range of $$1- f(x+1)$$ is $$[-1,0]$$ (subtracting 1 from both the extreme limits of range of $$f$$).

But this is not the range and domain.

I have been taught that when a change occurs inside the function, only the domain changes. And when the change is outside the function, the range changes. Is this incorrect?

## Question:

What would be the range and domain of $$g$$, and how is my thinking incorrect?

Your thinking is partly correct. Remember that the domain is the set of possible values you can put in for $$x$$. If the domain of $$f$$ is $$[-1,2]$$ then the domain of $$f_1(x) := f(x+1)$$ is $$[-2, 1]$$, shifted one to the left. Take a look at the borders: If you put in $$-2$$ you get $$f_1(-2)=f(-2+1)=f(-1)$$ and this is exactly the left side of the domain of $$f$$.

So you don't replace $$x$$ with the domain but your final output should be the domain $$[-1,2]$$. What you have done here is: $$f([-1,2] + 1) = f([0, 3])$$ but you should do the reverse: $$f([-1,2]) = f(x+1) \Rightarrow x+1 \in [-1,2] \Rightarrow x \in [-2,1]$$ informally speaking.

For the range, if the range of $$f$$ is $$[0,1]$$ then the range of $$-f$$ is $$[-1,0]$$ and thus the range of $$1-f$$ is $$[0,1]$$ again. In this case it doesn't matter if you put in $$x$$ or $$x+1$$ as you are only interested in the domain.

So your thinking is somehow correct but you need to think from the other direction. Similarly that $$(x-d)^2$$ moves the quadratic curve by $$d$$ values to the right, although intuitively subtraction is the left direction.

• Generally speaking, the range would also have been changed for the new function right? Its only in this particular case the range hasn't changed. Mar 23 at 15:21
• Exactly, this is just a coincidence. Also it can not only be shifted, it could also stretched. E.g. take $g(x)=f(2x)$, then the domain would be $[-0.5,1]$ much shorter than $[-1,2]$. Mar 23 at 15:26
• Okay, got it. Thanks a lot! Mar 23 at 15:26

$$f$$ is only defined for $$x\in[-1,2]$$. Thus, $$g(3)=1-f(4)$$ does not make sense.

The correct domain for $$g$$ would be $$[-2,1]$$. Now, if $$x\in[-2,1]$$, $$x+1\in[-1,2]$$ and $$g(x)=1-f(x+1)$$ is fine.

Regarding the range, if you shift $$f$$ to the left one unit its range remains $$[0,1]$$, and as the image of $$[0,1]$$ by $$1-x$$ is again $$[0,1]$$, we reach the range of $$g$$ is $$[0,1]$$.

$$x\in D_g\iff x+1\in D_f\iff x\in D_f-1=[0,3]-1=[-1,2].$$

$$\operatorname{range}(g)=1-\operatorname{range}(x\mapsto f(x+1))=1-\operatorname{range}(f)=1-[0,1]=1+[-1,0]=[0,1].$$