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# Domains and Functions

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Let $$f_1(x)=\sqrt{1-x}$$, and for integers $$n\geq2$$, let $$f_n(x)=f_{n-1}(\sqrt{n^2-x}).$$

Let $$N$$ be the largest value of $$n$$ for which the domain of $$f_n$$ is nonempty. For this value of $$N$$, the domain of $$f_N$$ consists of a single point $$\left \{ c \right \}.$$

Compute $$c.$$

Mar 21, 2020

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So I gave up on the question and it showed the answer afterward :( but I'll explain it here:

(Basically, they proceeded to keep bashing out the function and solved for the domains of each function).

The function $$f_{1}(x)=\sqrt{1-x}$$ is defined when $$x\leq1$$

Next, we have $$f_2(x)=f_1(\sqrt{4-x})=\sqrt{1-\sqrt{4-x}}$$ . For this to be defined, we must have $$4-x\ge0$$ or $$x \le 4,$$ and the number $$\sqrt{4-x}$$ must lie in the domain of $$f_1$$ so $$\sqrt{4-x} \le 1,$$ or $$x \ge 3.$$ Thus, the domain of $$f_2(x)$$ is $$[3, 4]$$.

Similarly, for $$f_3(x) = f_2\left(\sqrt{9-x}\right)$$ to be defined, we must have $$x \le 9,$$ and the number $$\sqrt{9-x}$$ must lie in the interval $$[3, 4].$$ Therefore, $$3 \le \sqrt{9-x} \le 4.$$ Squaring all parts of this inequality chain gives $$9 \le 9-x \le 16,$$ and so $$-7 \le x \le 0.$$ Thus, the domain of $$f_3$$ is $$[-7, 0].$$

Similarly, for $$f_4(x) = f_3\left(\sqrt{16-x}\right)$$ to be defined, we must have $$x \le 16,$$ and $$\sqrt{16-x}$$ must lie in the interval $$[-7, 0].$$ But $$\sqrt{16-x}$$ is always nonnegative, so we must have $$\sqrt{16-x} = 0,$$ or $$x=16.$$ Thus, the domain of $$f_4$$ consists of a single point $$\left \{ 16 \right \}.$$

We see, then, that $$f_5(x) = f_4\left(\sqrt{25-x}\right)$$ is defined if and only if $$\sqrt{25-x} = 16,$$ or $$x = 25 - 16^2 = -231.$$ Therefore, the domain of $$f_5$$ is $$\left \{ -231 \right \}.$$

The domain of $$f_6(x)$$ is empty, because $$\sqrt{36-x}$$ can never equal a negative number like $$-231$$ Thus, $$N=5$$ and $$c = \boxed{-231}.$$

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Mar 21, 2020