The function f satisfies
f(\sqrt{2x} + x) = \frac{1}{x}
for all x \ge \frac{1}{2}. Find f(4).
The function f satisfies f(√2x+x)=1x for all x≥12, find f(4).
f(4)=f(√2x+x)=1x
√2x+x=4; √2x=−x+4; 2x=x2−8x+16; x2−10x+16=0; (x−8)(x−2)=0
Note that x = 8 i s an extraneous solution, because √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(√2(2)+2)=12=f(4), so our final answer is 1/2.
The function f satisfies f(√2x+x)=1x for all x≥12, find f(4).
f(4)=f(√2x+x)=1x
√2x+x=4; √2x=−x+4; 2x=x2−8x+16; x2−10x+16=0; (x−8)(x−2)=0
Note that x = 8 i s an extraneous solution, because √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(√2(2)+2)=12=f(4), so our final answer is 1/2.