What is the inverse of f(x)=(arccos(1-x)-pi) ?

The inverse of f(x)=(arccos(1-x)-pi) is -cos(x+pi)+1

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inverse of f(x)=(arccos(1-x)-pi)

inverse

f (x) = (arccos (1 - x) - π)

- cos (x + π) + 1

Solution steps

Solve for

arccos (1 - x), y = arccos (1 - x) - π

arccos (1 - x) = y + π

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

1 - x = cos (y + π)

Solve for

x, 1 - x = cos (y + π)

- cos (y + π) + 1

Substitute

y = x

- cos (x + π) + 1

1 - 5 x

lo g_{10} (x - 11) + 6

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

f (x) = (- 4, 5), (2, 1) (4, 9)

f (x) = (x - 7)^{3} + 8

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