Inverse functions... yay!

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Xclueless

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Jan 29, 2011
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Right now we're studying inverse functions in my Algebra 2 class. There's one problem i'm stuck on, and I can't find the answer.

f(x)=3-(x/4)

(That's 3 minus x over 4)

To find an inverse, I was told to change x and y. So the function would become x=3-(y/4) and then solve for y. So I started by multiplying both sides by 4, to get rid of the fraction.
4x=3-y

Then I added y.
4x+y=3

Subtracted 4x to get y alone
y=4x-3.

I don't think this is right. Help would be appreciated. Thank you :)
 
Xclueless said:
Right now we're studying inverse functions in my Algebra 2 class. There's one problem i'm stuck on, and I can't find the answer.

f(x)=3-(x/4)

(That's 3 minus x over 4)

To find an inverse, I was told to change x and y. So the function would become x=3-(y/4) and then solve for y. So I started by multiplying both sides by 4, to get rid of the fraction.
4x=3-y

Then I added y.
4x+y=3

Subtracted 4x to get y alone
y=4x-3.

I don't think this is right. Help would be appreciated. Thank you :)
Click to expand...

It would be almost correct if the problem was:

\(\displaystyle y \ = \ \frac{3-x}{4}\)

which should be posted as:

y = (3-x)/4

The problem you posted:

\(\displaystyle y \ = \ 3 \ - \ \frac{x}{4}\)

switch 'x' and 'y'

\(\displaystyle x \ = \ 3 \ - \ \frac{y}{4}\)

multiply by '4'

\(\displaystyle 4x \ = \ 12 \ - \ y\)

\(\displaystyle y \ = \ 12 \ - \ 4x\)
 
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