Given #f(x) = (2x)/(x+3)#, how do you find #f^(-1) (4)#?

Question

Given #f(x) = (2x)/(x+3)#,  how do you find #f^(-1) (4)#?

Cora Alexander · Accepted Answer

I found: #f^-1(4)=-6# I start rearranging the original function #y=f(x)=(2x)/(x+3)# solving it for #x# to get: #(x+3)*y=2x# #xy+3y-2x=0# #xy-2x=-3y# #x(y-2)=-3y# and: #x=f(y)=(-3y)/(y-2)# or changing #x# with #y# you get: #f^-1(x)=(-3x)/(x-2)# and #f^-1(4)=(-3*4)/(4-2)=-12/2=-6#

Caleb Alexander · Answer

undefined To find $ f^{-1}(4) $, first, set $ f(x) = 4 $ and solve for $ x $.

$$ f(x) = \frac{2x}{x + 3} = 4 $$

Then, solve this equation for $ x $ to find $ f^{-1}(4) $.

$$ \frac{2x}{x + 3} = 4 $$

$$ 2x = 4(x + 3) $$

$$ 2x = 4x + 12 $$

$$ -2x = 12 $$

$$ x = -6 $$

Therefore, $ f^{-1}(4) = -6 $.

Cora Alexander · Answer

undefined To find $ f^{-1}(4) $ for the function $ f(x) = \frac{2x}{x+3} $, we first need to find the inverse function $ f^{-1}(x) $.

1. Start by replacing $ f(x) $ with $ y $: $ y = \frac{2x}{x+3} $.
2. Swap the roles of $ x $ and $ y $: $ x = \frac{2y}{y+3} $.
3. Solve this equation for $ y $ to find $ f^{-1}(x) $.

$$ x = \frac{2y}{y+3} $$

$$ x(y+3) = 2y $$

$$ xy + 3x = 2y $$

$$ 3x = 2y - xy $$

$$ 3x = y(2-x) $$

$$ y = \frac{3x}{2-x} $$

Now that we have $ f^{-1}(x) = \frac{3x}{2-x} $, we can find $ f^{-1}(4) $ by substituting $ x = 4 $ into the inverse function:

$$ f^{-1}(4) = \frac{3(4)}{2-4} $$

$$ f^{-1}(4) = \frac{12}{-2} $$

$$ f^{-1}(4) = -6 $$

Therefore, $ f^{-1}(4) = -6 $.