What is the inverse of f(x)= 1/(x+1)+1/x ?

The inverse of f(x)= 1/(x+1)+1/x is (-x+2+sqrt(x^2+4))/(2x),(-x+2-sqrt(x^2+4))/(2x)

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inverse of f(x)= 1/(x+1)+1/x

inverse

f (x) = \frac{1}{x + 1} + \frac{1}{x}

\frac{- x + 2 + x ^{2} + 4}{2 x}, \frac{- x + 2 - x ^{2} + 4}{2 x}

Solution steps

y = \frac{1}{x + 1} + \frac{1}{x}

Replace

x

with

y

x = \frac{1}{y + 1} + \frac{1}{y}

Solve for

y, x = \frac{1}{y + 1} + \frac{1}{y}

\frac{- x + 2 + x ^{2} + 4}{2 x}, \frac{- x + 2 - x ^{2} + 4}{2 x}

lo g_{10} (x - 4)

0.083

f (x) = 2 + \frac{1}{( 5 - x ) ^{2}}

f (x) = 8^{x} - 4

f (x) = \frac{8 x + 1}{16 x ^{2} - 4}

What is the inverse of f(x)= 1/(x+1)+1/x ?
The inverse of f(x)= 1/(x+1)+1/x is (-x+2+sqrt(x^2+4))/(2x),(-x+2-sqrt(x^2+4))/(2x)