Question

If f(x) = (5x)/(x + 3), find f(x)^-1. Please help, I am stumped, I have a bit of it solved but I'm stuck at one part:

If we make f(x) = y, switch y with x and then solve for y we get:

x = (5y)/(y+3)
x(y + 3) = 5y

It's here I get stumped, because I can bring over the y and get (xy + 3x)/(y) = 5, but where do I go from here?

Answer 1

good.
\[\huge\rm \frac{xy+3x}{y}=5\]
is same as \[\huge\rm \frac{xy}{y} +\frac{3x}{y}=5\]
rewrite it as a separate fraction

Answer 2

\[\large\rm \frac{ x\cancel{y} }{\cancel{y} } +\frac{3x}{y}=5\]
\[x+\frac{3x}{y}=5\]
solve for y

Answer 3

Okay so now I'm at

\[\frac{ 3x }{ y }=5-x\]

How do I bring over the 3x? I don't think I can just multiply... Or can I?

Answer 4

well how would you get rid of the fraction ?

Answer 5

Hang on I think I got it

\[\left(\begin{matrix}1 \\ 3x\end{matrix}\right)\left(\begin{matrix}3x \\ y\end{matrix}\right) = 5-x\]

Which then would give me

\[y = 5-x \left(\begin{matrix}1 \\ 3x\end{matrix}\right)\]

Which simplifies to

\[y = \frac{ 5-x }{ 3x}\]

Is that correct?

Answer 6

\[\frac{ 1 }{ \cancel{3x} } \cdot \frac{\cancel{3x}}{y} =\frac{1}{y} \]
\[\frac{1}{y} \cancel{= }y\]

Answer 7

y/1= y
1/y is just 1/y

we can move y to the right side.. right?
\[\huge\rm \frac{ 3x }{ y }=5-x\]
\[\huge\rm y \cdot \frac{ 3x }{ y }= (5-x) \cdot y\]
\[\huge\rm \cancel{y} \cdot \frac{ 3x }{\cancel{ y} }= (5-x) \cdot y\]
\[\rm 3x= \color{Red}{y}(5-x)\]

Answer 8

OH Okay, I see now.

So then we can bring over 5 - x instead...

So that gives us \[\frac{ 3x }{ 5-x } = y\]

And then we flip everything back to their original variables meaning:

\[f(x)^-1 = \frac{ 3x }{ 5-x }\]

And if I graph that it is in fact the inverse, fantastic, thank you so so much!

Answer 9

yw o^_^o