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The functions $f$ and $g$ are defined as follows:

$$f(x)=(2e^x-1)^2+2, x\in\mathbb R$$ $$g(x)=(x-1)^2-1, x\ge k$$

Find the range of values of $k$ such that both functions $gf$ and $g^{-1}$ exists.

I tried to do this question using this method:

Range of $f=[2,\infty)$

Range of $g=[1,\infty)$

In order for the function $gf$ to exist, The Range of $f\subseteq$ Domain of $g$

Therefore $k\le2$ for $gf$ to exist.

For $g^{-1}$ to exist, $k\le1$ or $k\ge1$ as the minium point of $g$ is $(1,-1)$ (The function $g$ needs to be a one-one function for its inverse to exist.)

Therefore, for both functions to exist, The domain of $g$ has to be $(k\le2) \land (k\le1) \iff (k\le1)$ OR $1\le k\le2$

Therefore the domain of $g$ can be $(-\infty, 1]$ or $[1,2]$

Both seem to be the correct answer but both of them at the same time contradicts the question as what they have written is $x\ge k$

Will appreciate it if someone can check if my answer is correct. If the answer is not correct, please do point out my mistakes in solving this question, thank you!

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  • $\begingroup$ I made some edits as I have found out some of my mistakes, seems like both domains can be accepted as solutions. $\endgroup$ – Derp Jun 14 '16 at 11:48
  • $\begingroup$ Range of $g$ is $[-1,\infty]$ $\endgroup$ – Piquito Jun 14 '16 at 12:16
  • $\begingroup$ Edited, thanks. $\endgroup$ – Derp Jun 14 '16 at 12:56
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The domain of $g$ is $[k,\infty)$. In order of $g^{-1}$ to exist, $k$ can not be less than $1$. Because if $k<1$, the domain of $g$ would contain some open interval around the minimum at $x=1$, and hence it would not have inverse.

The existence of $gf$ implies $k\le 2$, as you have found properly.

Then, the conditions for $k$ are $k\ge 1$ and $k\le 2$. That is, if $1\le k\le 2$, both conditions are satisfied.

Don't mix it up with the domain of $g$. The domain of $g$ is not the set of possible values for $k$.

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  • $\begingroup$ If $k\ge 2$, would that mean that $gf$ will not exist? Because the Range of $f$ is $[2, \infty)$ $\endgroup$ – Derp Jun 14 '16 at 11:53
  • $\begingroup$ Sorry, I confused the sign. Fixed. $\endgroup$ – ajotatxe Jun 14 '16 at 11:54
  • $\begingroup$ I think the question is pretty confusing because they stated to find the range of values of $k$, but essentially am I safe to say that this question just wants me to find the Domain of $g$? This is because $g^{-1}$ can exist if its domain is either $(-\infty, 1]$ or $[1, \infty)$ $\endgroup$ – Derp Jun 14 '16 at 11:58
  • $\begingroup$ Oh right, looks like I got it, thanks for the help! $\endgroup$ – Derp Jun 14 '16 at 12:05

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