What is the inverse of f(x)= 1/x+1/(x^2) ?

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

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inverse of f(x)= 1/x+1/(x^2)

inverse

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

\frac{1 + 1 + 4 x}{2 x}, \frac{1 - 1 + 4 x}{2 x}

Solution steps

y = \frac{1}{x} + \frac{1}{x ^{2}}

Replace

x

with

y

x = \frac{1}{y} + \frac{1}{y ^{2}}

Solve for

y, x = \frac{1}{y} + \frac{1}{y ^{2}}

\frac{1 + 1 + 4 x}{2 x}, \frac{1 - 1 + 4 x}{2 x}

f (x) = ln (x + 6)

\frac{2 x}{x ^{2} + 1}

f (x) = \frac{x ^{3}}{81 - x ^{2}}

1 + x^{2}

(2, - 3), (4, 5)

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