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

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

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inverse of f(x)= 1/(3x^2+1)

inverse

f (x) = \frac{1}{3 x ^{2} + 1}

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

Solution steps

y = \frac{1}{3 x ^{2} + 1}

Replace

x

with

y

x = \frac{1}{3 y ^{2} + 1}

Solve for

y, x = \frac{1}{3 y ^{2} + 1}

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

f (x) = \frac{3}{x ^{2} + 1}

f (x) = 6 + 3 + x

F (s) = \frac{3 - s}{s ^{2} - 4 s + 7}

f (x) = \frac{10 x - 9}{7 x - 1}

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