Let $\space f(x)=-2+\ln (2x-1) \space$ which domain is $(\frac{1}{2},+\infty )$. Now let $h(x)=f(x)-f(\frac{x}{2}) \space $. What is the domain of $h$?
I started to calculate $h$: $\begin{align*} h(x)&=-2+\ln(2x-1)- \left[-2+\ln(\frac{2x}{2}-1) \right]\\ &=-2+ \ln(2x-1)+2-\ln(x-1)\\ &=\ln(2x-1)-\ln(x-1)\\ &=\ln \left (\frac{2x-1}{x-1}\right) \end{align*}$
Then, I set up a table and in a row I put $2x-1$ and in the next row I put $x-1$. I found the zeros of these functions and put them in the header. Then I multiplied the sign of the two functions.
Where the numerator or the denominator are zero, $h$ is not define. Where the sign is negative $h$ is also not define. So the domain of $h$ is $( -\infty,\frac{1}{2})\cup( 1,+\infty)$
But my book says that the domain is only $( 1,+\infty)$. If the book's solution is correct, why this happen?
Thanks