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 \(c.\)
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 \(c = \boxed{-231}.\)
