How do you find the inverse of $f (x) = \frac{100}{1 + 2^{- x}}$ $f(x)= (100)/(1+2^-x)$ ?

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Answer 1 · 2015-12-17T16:11:47

$y = - {log}_{2} (\frac{100 - x}{x})$ $y=-log_2((100-x)/x)$

Rewrite as

$y = \frac{100}{1 + 2^{- x}}$ $y=100/(1+2^-x)$

Flip the $x$ $x$ and $y$ $y$ and solve for $y$ $y$ .

$x = \frac{100}{1 + 2^{- y}}$ $x=100/(1+2^-y)$

$x (1 + 2^{- y}) = 100$ $x(1+2^-y)=100$

$1 + 2^{- y} = \frac{100}{x}$ $1+2^-y=100/x$

$2^{- y} = \frac{100}{x} - 1$ $2^-y=100/x-1$

$2^{- y} = \frac{100 - x}{x}$ $2^-y=(100-x)/x$

$- y = {log}_{2} (\frac{100 - x}{x})$ $-y=log_2((100-x)/x)$

$y = - {log}_{2} (\frac{100 - x}{x})$ $y=-log_2((100-x)/x)$

The graphs should be reflections of themselves over the line $y = x$ $y=x$ .

How do you find the inverse of f(x)= (100)/(1+2^-x)f(x)=1001+2−xf(x)= (100)/(1+2^-x)?