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

The inverse of f(x)=((1-3x))/(x+1) is (1-x)/(x+3)

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inverse of f(x)=((1-3x))/(x+1)

inverse

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

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

Solution steps

y = \frac{1 - 3 x}{x + 1}

Replace

x

with

y

x = \frac{1 - 3 y}{y + 1}

Solve for

y, x = \frac{1 - 3 y}{y + 1}

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

\frac{5}{3 x + 2}

\frac{3 x - 2}{5}

f (x) = 9 x^{- \frac{1}{2}} + 16 x^{\frac{1}{2}} - 24

f (x) = \frac{6 x}{5 x - 4}

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

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