(c) f(x)= 1+x^2/x-lnx Solution y = 1+x^2/x-lnx Interchange x and y and re-solve

Question

(c) f(x)= 1+x^2/x-lnx

Explanation / Answer

y = 1+x^2/x-lnx Interchange x and y and re-solve for y. x = 1+x^2/x-lnx ==> e^x = e^[2y + ln(y)] ==> e^x = e^(2y)*e^[ln(y)] ==> e^x = y*e^(2y) ==> 2*e^x = 2y*e^(2y) ==> W(2*e^x) = 2y W(2*e^x)/2 = y Therefore, the inverse is: f^-1(x) = W(2*e^x)/2 where W is the Lambert W-Function (check source to learn more). Check: Let x = 1 f(1) = 2(1) + ln(1) = 2 f^-1(2) = W(2*e^2)/2 = 2/2 = 1 Plugging in x = 1 gives y = 2 in the original function and plugging in x = 2 gives y = 1 in the inverse function. So that works! Let's try another value. Let x = 7 f(7) = 2(7) + ln(7) = 14 + ln(7) f^-1[14 + ln(7)] = W[2*e^(14 + ln 7)] = 7

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