What is the inverse of f(x)=1+2*sin(1-2/(x+1)) ?

The inverse of f(x)=1+2*sin(1-2/(x+1)) is (-arcsin((x-1)/2)-1)/(arcsin((x-1)/2)-1)

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inverse of f(x)=1+2*sin(1-2/(x+1))

inverse

f (x) = 1 + 2 \cdot sin (1 - \frac{2}{x + 1})

\frac{- arcsin ( \frac{x - 1}{2} ) - 1}{arcsin ( \frac{x - 1}{2} ) - 1}

Solution steps

Solve for

sin (1 - \frac{2}{x + 1}), y = 1 + 2 \cdot sin (1 - \frac{2}{x + 1})

sin (1 - \frac{2}{x + 1}) = \frac{y - 1}{2}

sin (a) = b \Rightarrow a = arcsin (b)

1 - \frac{2}{x + 1} = arcsin (\frac{y - 1}{2})

Solve for

x, 1 - \frac{2}{x + 1} = arcsin (\frac{y - 1}{2})

\frac{- arcsin ( \frac{y - 1}{2} ) - 1}{arcsin ( \frac{y - 1}{2} ) - 1}

Substitute

y = x

\frac{- arcsin ( \frac{x - 1}{2} ) - 1}{arcsin ( \frac{x - 1}{2} ) - 1}

f (x) = 3 lo g_{2} (x - 1) + 2

f (x) = (x - 1) g (x) = x - 1

\frac{d}{d} \frac{5}{x}

f (x) = 6 + 2 x - 6

h (x) = (x)^{- 1} = x + 5

What is the inverse of f(x)=1+2*sin(1-2/(x+1)) ?
The inverse of f(x)=1+2*sin(1-2/(x+1)) is (-arcsin((x-1)/2)-1)/(arcsin((x-1)/2)-1)