how to find the domain of (fofof)(x)
where f(x)=sqrt(x-1)
Book answer is [2,infinity).
The domain of f(g(x)) is the intersection of the domain of f with the domain of g. Extending the concept to three functions, the domain of f(g(h(x)) is the intersection of the domain of f with the domain of g with the domain of h.
$\displaystyle f(x) = \sqrt{x - 1}$
The domain of $\displaystyle f(x)$ is all $\displaystyle x$ such that $\displaystyle x - 1 \geq 0$. This implies that $\displaystyle x \geq 1$.
$\displaystyle f(f(x)) = \sqrt{\sqrt{x - 1} - 1}$
The domain of $\displaystyle f(f(x))$ is all $\displaystyle x$ such that $\displaystyle \sqrt{x - 1} - 1 \geq 0$ and $\displaystyle x \geq 1$ (from the domain of $\displaystyle f(x)$). Let's solve the inequality in terms of $\displaystyle x$.
$\displaystyle \sqrt{x - 1} - 1 \geq 0$
$\displaystyle \implies \sqrt{x - 1} \geq 1$
$\displaystyle \implies x -1 \geq 1$
$\displaystyle \implies x \geq 2$
The intersection of $\displaystyle x \geq 1$ and $\displaystyle x \geq 2$ is $\displaystyle x \geq 2$, so the domain of $\displaystyle f(f(x))$ is $\displaystyle x \geq 2$.
$\displaystyle f(f(f(x))) = \sqrt{\sqrt{\sqrt{x - 1} - 1} - 1}$.
The domain of $\displaystyle f(f(f(x)))$ is all $\displaystyle x$ such that $\displaystyle \sqrt{\sqrt{x - 1} - 1} - 1 \geq 0$ and $\displaystyle x \geq 2$ (from the domain of $\displaystyle f(f(x))$). Let's solve the inequality in terms of $\displaystyle x$.
$\displaystyle \sqrt{\sqrt{x - 1} - 1} - 1 \geq 0$
$\displaystyle \implies \sqrt{\sqrt{x - 1} - 1} \geq 1$
$\displaystyle \implies \sqrt{x - 1} - 1 \geq 1$
$\displaystyle \implies \sqrt{x - 1} \geq 2$
$\displaystyle \implies x - 1 \geq 4$
$\displaystyle \implies x \geq 5$
The intersection of $\displaystyle x \geq 2$ and $\displaystyle x \geq 5$ is $\displaystyle x \geq 5$, so the domain of $\displaystyle f(f(f(x)))$ is $\displaystyle x \geq 5$.
I believe the answer in your book is incorrect assuming I did not err.