Let f:[−2+1,2+1]→[−2+12,2+12]f: [-\sqrt{2}+1, \sqrt{2}+1] \rightarrow \left[-\frac{\sqrt{2}+1}{2}, \frac{\sqrt{2}+1}{2}\right]…

Problem

Let f:[−2+1,2+1]→[−2+12,2+12]f: [-\sqrt{2}+1, \sqrt{2}+1] \rightarrow \left[-\frac{\sqrt{2}+1}{2}, \frac{\sqrt{2}+1}{2}\right] be a function defined by f(x)=1−x1+x2f(x) = \frac{1-x}{1+x^2}. If f−1(x)={−1+λ(4x−4x2+1)2x,x≠0,μ,x=0f^{-1}(x) = \begin{cases} \frac{-1+\lambda(\sqrt{4x-4x^2+1})}{2x}, & x \neq 0, \\ \mu, & x = 0 \end{cases}, then λ+μ\lambda + \mu is:
Intuitive explanation

To find the inverse function, you need to solve y=1−x1+x2y = \frac{1-x}{1+x^2} for xx in terms of yy. This is a rational equation that will rearrange into a quadratic in xx—use the quadratic formula to express xx as a function of yy. The constants λ\lambda and μ\mu emerge from matching the formula's structure to your solution: μ\mu is simply what f−1(0)f^{-1}(0) equals (the xx-value where f(x)=0f(x) = 0), and λ\lambda is the coefficient in front of the square root term that makes the formula work.

Intuitive explanation

To find the inverse function, you need to solve y=1−x1+x2y = \frac{1-x}{1+x^2} for xx in terms of yy. This is a rational equation that will rearrange into a quadratic in xx—use the quadratic formula to express xx as a function of yy. The constants λ\lambda and μ\mu emerge from matching the formula's structure to your solution: μ\mu is simply what f−1(0)f^{-1}(0) equals (the xx-value where f(x)=0f(x) = 0), and λ\lambda is the coefficient in front of the square root term that makes the formula work.

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