Mathematics - Functions Question with Solution | TestHub

Given the functional equation $f(x)+2f(1/x)=x+1/x$ (1). Replacing $x$ with $1/x$ in equation (1), we get $f(1/x)+2f(x)=1/x+x$ (2). Multiply equation (2) by 2: $2f(1/x)+4f(x)=2(x+1/x)$ (3). Subtract equation (1) from equation (3): $(2f(1/x)+4f(x))-(f(x)+2f(1/x))=2(x+1/x)-(x+1/x)$. This simplifies to $3f(x)=x+1/x$, so $f(x)=\frac{1}{3}\left(x+\frac{1}{x}\right)$. For $x\in {\mathbb{R}}^{+}$, by the AM-GM inequality, $x+\frac{1}{x}\ge 2\sqrt{x\cdot \frac{1}{x}}=2$. Therefore, $f(x)\ge \frac{1}{3}(2)=\frac{2}{3}$. The range of $f(x)$ is $[2/3,\infty )$.