1. Compute the domain of the real-valued function $$f(x)=\sqrt{3-\sqrt{5-\sqrt{x}}}.$$
2. If $f(x)=ax^4-bx^2+x+5$ and $f(-3)=2$, then what is the value of $f(3)$?
+9464 1. \(f(x)=\sqrt{3-\sqrt{5-\sqrt{x}}}\)
\(\sqrt{x}\) is part of the function so the domain has the restriction:
x ≥ 0
\(\sqrt{5-\sqrt{x}}\) is part of the function so the domain has the restriction:
\(5-\sqrt{x}\geq0\\ 5\geq\sqrt{x}\\ 25\geq x\\ \mathbf{x\leq 25}\)
\(\sqrt{3-\sqrt{5-\sqrt x}}\) is part of the function so the domain has the restriction:
\(3-\sqrt{5-\sqrt x}\geq0\\ 3\geq\sqrt{5-\sqrt x}\\ 9\geq5-\sqrt x\\ 9+\sqrt x\geq5\)
\(\sqrt x\geq-4\) This is true for all real numbers.
The domain restrictions are: x ≥ 0 and x ≤ 25
So the domain is just
0 ≤ x ≤ 25
