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

The inverse of f(x)=arccos((4-x)/(3+x)) is -(3cos(x)-4)/(1+cos(x))

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

inverse

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

- \frac{3 cos ( x ) - 4}{1 + cos ( x )}

Solution steps

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

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

Solve for

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

- \frac{3 cos ( y ) - 4}{1 + cos ( y )}

Substitute

y = x

- \frac{3 cos ( x ) - 4}{1 + cos ( x )}

\frac{6}{8 - x}

\frac{x}{3} - 2

f (x) = x^{\frac{- 1}{0.06}}

f (x) = e^{8 - 5 l n (x)}

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

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