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
Best Answer
By definition, the domain of a sum of functions is the intersection of their domains.
So, for example, if $f(x) = \frac{1}{x}$ and $g(x) = -\frac{1}{x}$, then the domain of $h(x) = f(x) + g(x)$ is "all $x\neq 0$", because in order to be able to compute $f(x)$ and to be able to compute $g(x)$, you need to $x\neq 0$.
However, if you attempt to compute a formula for $h(x)$ you get $h(x)=0$, which is defined everywhere. That's the problem: when you simplify, you may allow new values that were not allowed before the simplification. (Same thing happens with composition: if you compose $f$ with itself and "simplify", you get $h(x) = x$, which is defined everywhere, even though $f(f(x))$ cannot be computed at $0$, because $f$ is not defined at $0$).
In other words: to compute domains of sums, differences, products, quotients, and compositions, you need to look at the domains of the original functions, not at the simplified formula you might get after substitution and doing algebra.
This is what happened with your computation. When you went from $\ln(2x-1)-\ln(x-1)$ to $\ln\left(\frac{2x-1}{x-1}\right)$, you "simplified" and added possibilities: in order to be able to compute both $\ln(2x-1)$ and $\ln(x-1)$, you need both $2x-1$ and $x-1$ to be positive. But to compute $\ln\left(\frac{2x-1}{x-1}\right)$, you only need both to have the same sign. That's where the extraneous $(-\infty,\frac{1}{2})$ come from.
To compute the domain correctly, you need to intersect the domain of $f(x)$ and the domain of $f(\frac{x}{2})$.
The domain of $f(x)$ is $(\frac{1}{2}\infty)$. The domain of $f(\frac{x}{2})$ is $(1,\infty)$. So the domain of $h(x)$ is the intersection of these two domains.