+0  
 
0
11
1
avatar+400 

The function $f$ satisfies
f(\sqrt{2x} + x) = \frac{1}{x}
for all x \ge \frac{1}{2}.  Find f(4).

 Feb 16, 2024

Best Answer 

 #1
avatar+1622 
+2

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.

 Feb 16, 2024
 #1
avatar+1622 
+2
Best Answer

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.

proyaop Feb 16, 2024

0 Online Users