What is the inverse of f(x)= 1/pi arctan(x-3)+0.5pi ?

The inverse of f(x)= 1/pi arctan(x-3)+0.5pi is tan((pi(2x-pi))/2)+3

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inverse of f(x)= 1/pi arctan(x-3)+0.5pi

inverse

f (x) = \frac{1}{π} arctan (x - 3) + 0.5 π

tan (\frac{π ( 2 x - π )}{2}) + 3

Solution steps

Solve for

arctan (x - 3), y = \frac{1}{π} arctan (x - 3) + 0.5 π

arctan (x - 3) = \frac{π ( 2 y - π )}{2}

arctan (a) = b \Rightarrow a = tan (b)

x - 3 = tan (\frac{π ( 2 y - π )}{2})

Solve for

x, x - 3 = tan (\frac{π ( 2 y - π )}{2})

tan (\frac{π ( 2 y - π )}{2}) + 3

Substitute

y = x

tan (\frac{π ( 2 x - π )}{2}) + 3

\frac{x + 1}{7}

f (x) = ln (1 - x - 2)

f (x) = 25 x + 10

g (x) = - 2 (x + 3)^{3}

f (x) = 3 (x + 4)^{2} - 2, x \leq - 4

What is the inverse of f(x)= 1/pi arctan(x-3)+0.5pi ?
The inverse of f(x)= 1/pi arctan(x-3)+0.5pi is tan((pi(2x-pi))/2)+3