# Stuck on inverting a function

• Jul 11th 2012, 03:21 PM
AZach
Stuck on inverting a function
$\displaystyle f(x)=0.5+\frac{5}{0.2x}$ $\displaystyle ,x\neq0$

I'm having some difficulty on finding the inverse function for this problem.

I follow the first method of switching the 'x' and 'y' variables and then subtracting the $\displaystyle 0.5$ from both sides, $\displaystyle x-0.5=\frac{5}{0.2y}$. Here's where my intuition breaks down. I know the objective is to solve for 'y'. Thus, multiply '0.2y' on both sides? $\displaystyle 0.2y(x-0.5)=5$.

Should I divide by $\displaystyle (x-0.5)$? That would give me $\displaystyle 0.2y=\frac{5}{x-0.5}$

And then divide $\displaystyle 0.2$ from both sides and I end up with $\displaystyle y=\frac{1}{x-0.5}$

I've tried working some inputs and the function is definitely not the inverse function, so I know my math is wrong. If somebody could toss me a tip I would appreciate it.
• Jul 11th 2012, 03:45 PM
skeeter
Re: Stuck on inverting a function
Quote:

Originally Posted by AZach
$\displaystyle f(x)=0.5+\frac{5}{0.2x}$ $\displaystyle ,x\neq0$

I'm having some difficulty on finding the inverse function for this problem.

I follow the first method of switching the 'x' and 'y' variables and then subtracting the $\displaystyle 0.5$ from both sides, $\displaystyle x-0.5=\frac{5}{0.2y}$. Here's where my intuition breaks down. I know the objective is to solve for 'y'. Thus, multiply '0.2y' on both sides? $\displaystyle 0.2y(x-0.5)=5$.

Should I divide by $\displaystyle (x-0.5)$? That would give me $\displaystyle 0.2y=\frac{5}{x-0.5}$

And then divide $\displaystyle 0.2$ from both sides and I end up with $\displaystyle y=\frac{1}{x-0.5}$

5 divided by 0.2 is 25, not 1

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