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#### renyikouniao

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- Jun 1, 2013

- 41

Thank you in advance

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- Jun 1, 2013

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Thank you in advance

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- Jun 1, 2013

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I have no idea on this parta) 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 $g^{-1}(x)$?

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

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?

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- Jun 1, 2013

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

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)?

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- Jun 1, 2013

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

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- Jun 1, 2013

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Correct, so how would we shift the domain? How do we shift a function horizontally?

shift c units horizontally?

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- Jun 1, 2013

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I don't get this...Yes, how would we shift a function $c$ units to the right?

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- Jun 1, 2013

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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

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- #14

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?

- Feb 13, 2012

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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...a) Does adding a constant to a function affect its monotonicity?

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

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

Kind regards

$\chi$ $\sigma$

- Jan 29, 2012

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- Jan 30, 2012

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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 $v(x)=u(x-c)$.

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.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?

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).

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