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# Function

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

Feb 16, 2024

#1
+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
+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.

proyaop Feb 16, 2024