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!
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 $k$.