What is the inverse of f(x)=(y^2+3)/(3y^2) ?

The inverse of f(x)=(y^2+3)/(3y^2) is sqrt(3/(3y-1)),-sqrt(3/(3y-1))

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inverse of f(x)=(y^2+3)/(3y^2)

inverse

f (x) = \frac{y ^{2} + 3}{3 y ^{2}}

\frac{3}{3 y - 1}, - \frac{3}{3 y - 1}

Solution steps

x = \frac{y ^{2} + 3}{3 y ^{2}}

Replace

y

with

x

y = \frac{x ^{2} + 3}{3 x ^{2}}

Solve for

x, y = \frac{x ^{2} + 3}{3 x ^{2}}

\frac{3}{3 y - 1}, - \frac{3}{3 y - 1}

x^{3} + 4 x + 5

f (x) = \frac{x ^{2} - x + 6}{x ^{2} - x - 20}

- x^{2} - 3

f (x) = 17 - x

3

What is the inverse of f(x)=(y^2+3)/(3y^2) ?
The inverse of f(x)=(y^2+3)/(3y^2) is sqrt(3/(3y-1)),-sqrt(3/(3y-1))