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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?

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3 Answers 3

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Haider
    Mar 23 at 15:21
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    $\begingroup$ 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 $[-1,2]$. $\endgroup$
    – LegNaiB
    Mar 23 at 15:26
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    $\begingroup$ Okay, got it. Thanks a lot! $\endgroup$
    – Haider
    Mar 23 at 15:26
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$f$ 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 $[0,1]$.

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

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

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