So I have the function $f(x)=\ln\tan^{-1}x$. The inverse is $g(x)=\tan(e^x)$.
I am given that $a=\ln \frac{5\pi}{4}$ and then I have to verify that $f(\tan(e^a)) \neq a$, which was easy enough. I then have to explain why this fact does not contradict the definition of an inverse function.
Is it sufficient to say because $f(x)$ is one to one? I know that by symmetry the composition of $f(x)$ and $g(x)$ in any order has to equal x but from above that's clearly not the case.
Is the one to one case sufficient or is there something I have to add?
Could it also be the fact that $a$ isn't even in the domain of the original function and hence can't be considered when you're trying to contradict the definition?
I also know that $\frac{5\pi}{4}$ is not in the domain of $\tan^{-1}x$ so I had to subtract $\pi$ from the angle to get an equivalent angle.
Thanks!
I continue from your previous question.
For a function $f:x\mapsto y$ to be invertible with inverse $f^{-1}:y\mapsto x$, it is not enough to give the functions themselves. A domain needs to be specified for both functions such that the inverse relations hold.
In this case, $f$ has the domain $(0,\frac\pi2)$ and $f^{-1}=g$ has the domain $(-\infty,\ln\frac\pi2)$. However, you put in $a=\ln\frac{5\pi}4$, which is outside the domain of $g$, into $g$. Therefore $f(g(a))$ does not necessarily equal $a$, and it doesn't here.