What is the inverse of f(x)= 1/(2x-1)-2,2<= x<= 9 ?

The inverse of f(x)= 1/(2x-1)-2,2<= x<= 9 is (x+3)/(2(2+x))

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inverse of f(x)= 1/(2x-1)-2,2<= x<= 9

inverse

f (x) = \frac{1}{2 x - 1} - 2, 2 \leq x \leq 9

\frac{x + 3}{2 ( 2 + x )}

Solution steps

y = \frac{1}{2 x - 1} - 2

Replace

x

with

y

x = \frac{1}{2 y - 1} - 2

Solve for

y, x = \frac{1}{2 y - 1} - 2

\frac{x + 3}{2 ( 2 + x )}

f (x) = \frac{( x + 6 )}{( x + 13 )}

f (x) = \frac{6 + 2 x}{6 - 2 x}

f (x) = \frac{x + 9}{1 - 2 x}

y = - x - 6

y = x^{2} + 6 x + 5

What is the inverse of f(x)= 1/(2x-1)-2,2<= x<= 9 ?
The inverse of f(x)= 1/(2x-1)-2,2<= x<= 9 is (x+3)/(2(2+x))