Math archive (10)

Calculus Level 4

If f : R + → R f:\mathbb R^{+}\rightarrow\mathbb R and f ( x ) f ( y ) = f ( x y ) + 2 ( x + y x y + 1 ) f(x)f(y)=f(xy)+2\left(\frac{x+y}{xy}+1\right) for all positive x , y x,y . Find 3 3 times the sum of all possible values of f ( 1 2 ) f\left(\frac{1}{2}\right) .

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The answer is 4.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Anton Amirkhanov
Jul 15, 2020

Having f ( x y ) = f ( x ) f ( y ) − 2 ( x + y x y + 1 ) f(xy)=f(x)f(y)-2(\frac{x+y}{xy}+1) We let x,y=1 and for convenience call f(1) θ θ = θ 2 − 6 \theta=\theta^2-6 solving which θ = 1 ± 5 2 \theta=\frac{1\pm 5}{2} Thus f ( 1 ) = 3 , f ( 1 ) = − 2 f(1)=3, f(1)=-2 Are the 2 possibilities. Now letting y=1 we get f ( x ) = f ( x ) f ( 1 ) − 2 ( 2 + 1 x ) f(x)=f(x)f(1)-2(2+\frac{1}{x}) Solving for f(x) we get f ( x ) = 2 ( 2 + 1 x ) f ( 1 ) − 1 f(x)=\frac{2(2+\frac{1}{x})}{f(1)-1} Plugging in the 2 possible values of f(1) we get the 2 functions f 1 ( x ) = 2 + 1 x , f 2 ( x ) = − 2 3 ( 2 + 1 x ) f_1(x)=2+\frac{1}{x}, f_2(x)=-\frac{2}{3}(2+\frac{1}{x}) Plugging them into the original equation we trivially see that they satisfy it, now plugging in 0.5 we get that f 1 ( 1 2 ) = 4 , f 2 ( 1 2 ) = − 8 3 f_1(\frac{1}{2})=4,f_2(\frac{1}{2})=-\frac{8}{3} Thus 3 ( f 1 ( 1 2 ) + f 2 ( 1 2 ) ) = 3 ( 4 − 8 3 ) = 12 − 8 = 3(f_1(\frac{1}{2})+f_2(\frac{1}{2}))=3(4-\frac{8}{3})=12-8= □ 4 \Box{4}

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