+400 The function $f$ satisfies
f(\sqrt{2x} + x) = \frac{1}{x}
for all x \ge \frac{1}{2}. Find f(4).
+1632 The function \(f\) satisfies \(f(\sqrt{2x} + x) = \frac{1}{x}\) for all \(x \ge \frac{1}{2}\), find \(f(4)\).
\(f(4) = f(\sqrt{2x}+x)={1\over{x}}\)
\(\sqrt{2x}+x=4\); \(\sqrt{2x} = -x + 4\); \(2x = x^2 - 8x + 16\); \(x^2-10x+16=0\); \((x-8)(x-2)=0\)
Note that x = 8 i s an extraneous solution, because \(\sqrt{2*8}\) does not equal to \(-8 + 4 = -4\), for a square root can't end up negative.
Thus, our only solution, x = 2. \(f(\sqrt{2(2)}+2)={1\over2}=f(4)\), so our final answer is 1/2.
+1632 The function \(f\) satisfies \(f(\sqrt{2x} + x) = \frac{1}{x}\) for all \(x \ge \frac{1}{2}\), find \(f(4)\).
\(f(4) = f(\sqrt{2x}+x)={1\over{x}}\)
\(\sqrt{2x}+x=4\); \(\sqrt{2x} = -x + 4\); \(2x = x^2 - 8x + 16\); \(x^2-10x+16=0\); \((x-8)(x-2)=0\)
Note that x = 8 i s an extraneous solution, because \(\sqrt{2*8}\) does not equal to \(-8 + 4 = -4\), for a square root can't end up negative.
Thus, our only solution, x = 2. \(f(\sqrt{2(2)}+2)={1\over2}=f(4)\), so our final answer is 1/2.
