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Invertible functions Question 2

renyikouniao

Member
Jun 1, 2013
41
Show that if f(x) is an invertible function then g(x)=f(x)+c is an invertible function.If f^(-1) is known what is g^(-1)

Thank you in advance
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
a) Does adding a constant to a function affect its monotonicity?

b) What effect does the constant have on the range of $g(x)$, and hence on the domain of $g^{-1}(x)$?
 

renyikouniao

Member
Jun 1, 2013
41
a) Does adding a constant to a function affect its monotonicity?

No,it doesn't

b) What effect does the constant have on the range of $g(x)$, and hence on the domain of $g^{-1}(x)$?
I have no idea on this part
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Suppose we have the monotonic function:

\(\displaystyle f(x)=e^x\)

We know its range is \(\displaystyle (0,\infty)\).

So, what is the range then of:

\(\displaystyle g(x)=f(x)+c\)

What relationship exists between the range of a function and the domain of its inverse?
 

renyikouniao

Member
Jun 1, 2013
41
Suppose we have the monotonic function:

\(\displaystyle f(x)=e^x\)

We know its range is \(\displaystyle (0,\infty)\).

So, what is the range then of:

\(\displaystyle g(x)=f(x)+c\)

What relationship exists between the range of a function and the domain of its inverse?

The range of g(x) also is (0,infinity)

The range of f(x) is the domain of g(x),the domain of g(x) is the range of f(x)?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Adding a constant to a function has the effect of shifting it vertically, thereby affecting its range. The range of $g(x)=e^x+c$ is therefore $(c,\infty)$. So then what would the domain of $g^{-1}(x)$ be, and how do we shift the domain of a function?
 

renyikouniao

Member
Jun 1, 2013
41
Adding a constant to a function has the effect of shifting it vertically, thereby affecting its range. The range of $g(x)=e^x+c$ is therefore $(c,\infty)$. So then what would the domain of $g^{-1}(x)$ be, and how do we shift the domain of a function?

the domain of $g^{-1}(x)$ would be $(c,\infty)$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Correct, so how would we shift the domain? How do we shift a function horizontally?
 

renyikouniao

Member
Jun 1, 2013
41

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, how would we shift a function $c$ units to the right?
 

renyikouniao

Member
Jun 1, 2013
41

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Suppose $f(x)$ has a root at $x=c$, or $f(c)=0$. Then $f(x-h)$ will have a root at $x-h=c$, or $x=c+h$. Hence, $f(x-h)$ is the graph of $f(x)$ shifted $h$ units to the right. So how can we apply this to the original problem?
 

renyikouniao

Member
Jun 1, 2013
41
Suppose $f(x)$ has a root at $x=c$, or $f(c)=0$. Then $f(x-h)$ will have a root at $x-h=c$, or $x=c+h$. Hence, $f(x-h)$ is the graph of $f(x)$ shifted $h$ units to the right. So how can we apply this to the original problem?

So..the original function has domain:c,infinity. range:0,h
the inverse function has domain:0,h. range:c,infinity
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's go back to the original question...

a) We are given that $f(x)$ is an invertible function, and then we are asked to then show that $g(x)=f(x)+c$ is also invertible.

While we have observed that adding a constant to a function does not affect its monotonicity, how can we explain/demonstrate this mathematically?

b) We are asked to give $g^{-1}(x)$ in terms of the known $f^{-1}(x)$.

We have observed that the graph of $g(x)$ is shifted vertically from that of $f(x)$, hence the graph of $g^{-1}(x)$ must be horizontal shifted (by the same amount) from that of $f^{-1}(x)$. So how can we state this mathematically?
 

chisigma

Well-known member
Feb 13, 2012
1,704
a) Does adding a constant to a function affect its monotonicity?
It should be noted that f(x) is not requested to be continous, so that monotonicity is not a necessary condition for f(x) to be invertible. For example the function...

$$f(x)=\begin{cases} -2 - x \ \text{if}\ -1< x < 0\\ 0\ \text{if}\ x=0\\ 1 + x\ \text{if}\ 0< x< 1 \end{cases}\ (1)$$

... is invertible even if not monotic...


Kind regards


$\chi$ $\sigma$
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Another example: $$f(x)=\begin{cases} x &\mbox{if}& x\not\in\{0,1\}\\ 0& \text{if}& x=1\\ 1 & \text{if}& x=0 \end{cases}\Rightarrow \exists f^{-1}\;\wedge\;f^{-1}=f$$
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
First, some metamathematical remarks. I admire Mark's patience in explaining the effect of adding a constant on the range of a function, but I find the whole first page hard to follow.

Adding a constant to a function has the effect of shifting it vertically, thereby affecting its range. The range of $g(x)=e^x+c$ is therefore $(c,\infty)$. So then what would the domain of $g^{-1}(x)$ be, and how do we shift the domain of a function?
I am not sure it is necessary to talk about the relationship between the domains of $f^{-1}$ and $g^{-1}$ because if the domain of some function $u$ is $D$ and the domain of some $v$ is $\{x+c\mid x\in D\}$, it obviously does not follow that $v(x)=u(x-c)$.

We have observed that the graph of $g(x)$ is shifted vertically from that of $f(x)$, hence the graph of $g^{-1}(x)$ must be horizontal shifted (by the same amount) from that of $f^{-1}(x)$. So how can we state this mathematically?
I am not sure it is helpful to talk about graph shifting, either. I am pretty sure that if the OP did not know how adding a constant changes the range, s/he would be lost by the fact quoted above. Horizontal graph shifting is counterintuitive because adding a positive constant to the argument leads to the graph's shift to the left rather than to the right.

And it has been pointed out that instead of monotonicity of $g(x)$ we must show its injectivity.

I think it is much easier to focus on algebra involving concrete numbers $x$ and $y$ instead of talking about domains or graphs as a whole.

So, we have $g(x)=f(x)+c$ and we need to find $g^{-1}$. This means we need to solve the equation $f(x)+c=y$ for $x$. It is solved in the usual way: we look at operations applied to $x$ in the left-hand side and reverse them. There are two operations applied to $x$: first $f$ and then adding $c$. Therefore, we first have to subtract $c$ from both sides to neutralize addition, and then ... (use the fact that $f^{-1}$ is known).
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
All I was trying to impart is that if the graph of $f(x)$ is shifted up $c$ units to get $g(x)$, then the graph of $f^{-1}(x)$ will be shifted to the right by $c$ units to get $g^{-1}(x)$. This immediately leads to the result we find from the algebraic approach.