How to solve this problem

[Tenor]

At which points on the graph of inverse of [math]f(x)= \frac 1 {x^2+1} +\sqrt[3]{1-2x}[/math] [math]x \ge 0[/math] the tangents of f(x) and its inverse are perpendicular?

 

[blamocur]

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[mmm4444bot]

Hello. Where do you need help, in this exercise? Please share how far you got, so that people can get a sense of which parts you already understand. For example, have you learned the concept of inverse functions and how to find them?
[imath]\;[/imath]

 

[skeeter]

At which points on the graph of inverse of [math]f(x)= \frac 1 {x^2+1} +\sqrt[3]{1-2x}[/math] [math]x \ge 0[/math] the tangents of f(x) and its inverse are perpendicular?

Finding an explicit expression for [imath]f^{-1}(x)[/imath] algebraically is going to be quite difficult. The best that I could do to find a perpendicular pair of tangent lines was to look at the graphs of the function and its inverse. Given the restriction that [imath]x \ge 0[/imath], I can see one possible pair.