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

Question

Sadie Alexander · Accepted Answer

undefined To find the inverse of the function $ f(x) = \frac{100}{1+2^{-x}} $, you can follow these steps:

1. Replace $ f(x) $ with $ y $ to rewrite the function as an equation: $ y = \frac{100}{1+2^{-x}} $.
2. Swap the roles of $ x $ and $ y $ in the equation.
3. Solve for $ y $ to find the inverse function $ f^{-1}(x) $.

Here are the calculations:

1. Start with $ y = \frac{100}{1+2^{-x}} $.

2. Swap $ x $ and $ y $: $ x = \frac{100}{1+2^{-y}} $.

3. Solve for $ y $:
   $$ x(1+2^{-y}) = 100 $$
   $$ 1+2^{-y} = \frac{100}{x} $$
   $$ 2^{-y} = \frac{100}{x} - 1 $$
   $$ 2^{-y} = \frac{100-x}{x} $$
   $$ -y = \log_2\left(\frac{100-x}{x}\right) $$
   $$ y = -\log_2\left(\frac{100-x}{x}\right) $$

Therefore, the inverse of the function $ f(x) = \frac{100}{1+2^{-x}} $ is $ f^{-1}(x) = -\log_2\left(\frac{100-x}{x}\right) $.

Theodore Amidon · Answer

#y=-log_2((100-x)/x)#

Rewrite as

#y=100/(1+2^-x)#

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

#x=100/(1+2^-y)#

#x(1+2^-y)=100#

#1+2^-y=100/x#

#2^-y=100/x-1#

#2^-y=(100-x)/x#

#-y=log_2((100-x)/x)#

#y=-log_2((100-x)/x)#

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