To check injectivity: if \(f(x_1) = f(x_2)\), then \(\frac{x_1}{1+x_1^2} = \frac{x_2}{1+x_2^2}\), which leads to \(x_1(1+x_2^2) = x_2(1+x_1^2)\), simplifying to \(x_1 - x_2 + x_1x_2^2 - x_2x_1^2 = 0\), or \((x_1-x_2)(1-x_1x_2) = 0\). For most real values this gives \(x_1 = x_2\), confirming injectivity. For surjectivity, \(f'(x) = \frac{1-x^2}{(1+x^2)^2}\). Maximum occurs at \(x=1\) giving \(f(1)=\frac{1}{2}\) and minimum at \(x=-1\) giving \(f(-1)=-\frac{1}{2}\). The range is exactly \([-\frac{1}{2}, \frac{1}{2}]\), confirming surjectivity. Therefore, f is bijective and hence invertible.