1. domain of (fofof)(x)

how to find the domain of (fofof)(x)
where f(x)=sqrt(x-1)

2. 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.

$f(x) = \sqrt{x - 1}$

The domain of $f(x)$ is all $x$ such that $x - 1 \geq 0$. This implies that $x \geq 1$.

$f(f(x)) = \sqrt{\sqrt{x - 1} - 1}$

The domain of $f(f(x))$ is all $x$ such that $\sqrt{x - 1} - 1 \geq 0$ and $x \geq 1$ (from the domain of $f(x)$). Let's solve the inequality in terms of $x$.

$\sqrt{x - 1} - 1 \geq 0$

$\implies \sqrt{x - 1} \geq 1$

$\implies x -1 \geq 1$

$\implies x \geq 2$

The intersection of $x \geq 1$ and $x \geq 2$ is $x \geq 2$, so the domain of $f(f(x))$ is $x \geq 2$.

$f(f(f(x))) = \sqrt{\sqrt{\sqrt{x - 1} - 1} - 1}$.

The domain of $f(f(f(x)))$ is all $x$ such that $\sqrt{\sqrt{x - 1} - 1} - 1 \geq 0$ and $x \geq 2$ (from the domain of $f(f(x))$). Let's solve the inequality in terms of $x$.

$\sqrt{\sqrt{x - 1} - 1} - 1 \geq 0$

$\implies \sqrt{\sqrt{x - 1} - 1} \geq 1$

$\implies \sqrt{x - 1} - 1 \geq 1$

$\implies \sqrt{x - 1} \geq 2$

$\implies x - 1 \geq 4$

$\implies x \geq 5$

The intersection of $x \geq 2$ and $x \geq 5$ is $x \geq 5$, so the domain of $f(f(f(x)))$ is $x \geq 5$.

I believe the answer in your book is incorrect assuming I did not err.

3. Misinterpreted the question.

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f(x)=x/sqrt1 x^2.find fofof(x)

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