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

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

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

inverse

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

- \frac{2 x - 1}{x ^{2}}, - - \frac{2 x - 1}{x ^{2}}

Solution steps

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

Replace

x

with

y

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

Solve for

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

- \frac{2 x - 1}{x ^{2}}, - - \frac{2 x - 1}{x ^{2}}

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