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

The inverse of f(x)=arccos^3((3x+4)/(x-3)) is (-3cos(\sqrt[3]{x})-4)/(3-cos(\sqrt[3]{x))}

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inverse of f(x)=arccos^3((3x+4)/(x-3))

inverse

f (x) = arccos^{3} (\frac{3 x + 4}{x - 3})

\frac{- 3 cos ( 3 x ) - 4}{3 - cos ( 3 x )}

Solution steps

Solve for

arccos (\frac{3 x + 4}{x - 3}), y = arccos^{3} (\frac{3 x + 4}{x - 3})

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

arccos (a) = b \Rightarrow a = cos (b)

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

Solve for

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

\frac{- 3 cos ( 3 y ) - 4}{3 - cos ( 3 y )}

Substitute

y = x

\frac{- 3 cos ( 3 x ) - 4}{3 - cos ( 3 x )}

f (x) = (x - 5)^{3} - 6

(x - 16)^{2}

f (x) = x^{2} - 4 x, x \geq 2

f (x) = 3 ln (5 - x) - 2

\frac{1}{s ^{4}}

What is the inverse of f(x)=arccos^3((3x+4)/(x-3)) ?
The inverse of f(x)=arccos^3((3x+4)/(x-3)) is (-3cos(\sqrt[3]{x})-4)/(3-cos(\sqrt[3]{x))}