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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{7}{( x - 8 ) ^{3}}

6 - x

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

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

\frac{6 x ^{2} + 35 x - 6}{4 x ^{2} + 23 x - 6}

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