What is the inverse of f(x)= 1/3 ln((x-4)/(x+1)) ?

The inverse of f(x)= 1/3 ln((x-4)/(x+1)) is (e^{3x}+4)/(1-e^{3x)}

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inverse of f(x)= 1/3 ln((x-4)/(x+1))

inverse

f (x) = \frac{1}{3} ln (\frac{x - 4}{x + 1})

\frac{e ^{3 x} + 4}{1 - e ^{3 x}}

Solution steps

y = \frac{1}{3} ln (\frac{x - 4}{x + 1})

Replace

x

with

y

x = \frac{1}{3} ln (\frac{y - 4}{y + 1})

Solve for

y, x = \frac{1}{3} ln (\frac{y - 4}{y + 1})

\frac{e ^{3 x} + 4}{1 - e ^{3 x}}

\frac{1 - x}{x}

y = 64 - (x - 2.9)^{2} - 3

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

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

(x + 5)^{2}, x \geq - 5

What is the inverse of f(x)= 1/3 ln((x-4)/(x+1)) ?
The inverse of f(x)= 1/3 ln((x-4)/(x+1)) is (e^{3x}+4)/(1-e^{3x)}