Hi, I gotta show that f(x)=4x^3/(x^2+1) has an inverse and find f(-1)' (2) I cant seem to find y, and f(-1)(x). After I find that I guess I derivate and put in the derivate=2, but I need some help to get started off
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I don't know how to do it... but it seems very complicated http://www.wolframalpha.com/input/?i=x+%3D+\dfrac{4y^3}{y^2+%2B+1} Especially is this part:
Originally Posted by fredrikhoeg Hi, I gotta show that f(x)=4x^3/(x^2+1) has an inverse and find f(-1)' (2) I cant seem to find y, and f(-1)(x). After I find that I guess I derivate and put in the derivate=2, but I need some help to get started off One does not need the exact inverse. $\displaystyle \dfrac{{df^{ - 1} }}{{dx}}(2) = \dfrac{1}{{f'(f^{ - 1} (2))}}$
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