Question

PLEASE HELP!!!

Given the function f(x) = 6(x+2) − 3, solve for the inverse function when x = 21.

Answer 1

@sangya21 @tkhunny

Answer 2

\[\large f(x) = 6(x + 2) - 3\]

Let's rewrite this as

\[\large y = 6(x + 2) - 3\]

To find the inverse of a function...we switch the 'x' and the 'y' and then solve again for 'y'

\[\large y = 6(x + 2) - 3\]

turns into

\[\large x = 6(y + 2) - 3\]

Now how would we solve that for 'y'?

Answer 3

Find inverse of f(x) first
put f(x) = y
y = 6(x+2) - 3
thus inverse will be
\[ \frac{ y-3 }{ 6 } -2 = x\]
Now x = 21
then find y

Answer 4

sorry its y + 3

Answer 5

wait, why isn't it (y+2)/6 -3 ?

Answer 6

I'm just curious

Answer 7

We have to find inverse of x. so simply putting y in position of x wont give us that.

we assume that our function = y
thus
y = 6(x+2) - 3
now whats x in terms of y?
x = y+3/6 -2 -------- (1)
thats our inverse function BUT its not the answer yet.
our f(x) = y
thus f(x)^-1 = y
now we change position of x and y
thus getting,
y = x+3/6 - 2


sorry missed last part in above comments

Answer 8

okay thanks :) so y = 125, right?

Answer 9

\[y = \frac{ 14 }{ 6 } \]

Answer 10

how did you get that?
I did:

21 = (y+3)/6 -2
126 = y + 1
125 = y

Answer 11

\[y = \frac{ x+3 }{ 6 } - 2\]
\[y = \frac{ 21+3 }{ 6 } - 2\]
\[y = \frac{ 26 }{ 6 } - 2\]
\[y = \frac{ 26 - 12 }{ 6 } \]

Answer 12

I mentioned in explanation that I forgot mentioning the last part. Sorry

Answer 13

http://www.purplemath.com/modules/invrsfcn3.htm

i hope this helps

Answer 14

just plug in 21 for f(x)

Answer 15

instead of going through all complex steps

Answer 16

when I just plugged in 21 for f(x), I got 2, not 14/6. Did I do something wrong?

Answer 17

^You did it correctly @RosieF There was a typo in the response up there

Answer 18

okay :) thank you everyone that helped :)

Answer 19

inverse of a linear function. For a linear function f(x), find \({\rm f}^{-1}(a)\)

\(\large\color{black}{ \displaystyle f(x)=mx+b }\)
\(\large\color{black}{ \displaystyle y=mx+b }\)
\(\large\color{black}{ \displaystyle x=my+b }\)
\(\large\color{black}{ \displaystyle x-b=my }\)
\(\large\color{black}{ \displaystyle y=(x-b )/m }\)
\(\large\color{black}{ \displaystyle y=(a-b )/m }\)
\(\large\color{black}{ \displaystyle f^{-1}(a)=(a-b )/m }\)

or, going with a trick:
\(\large\color{black}{ \displaystyle f(x)=mx+b }\)
\(\large\color{black}{ \displaystyle a=mx+b }\)
\(\large\color{black}{ \displaystyle a-b=mx }\)
\(\large\color{black}{ \displaystyle (a-b)/m=x }\) (don't forget x is the inverse function)
\(\large\color{black}{ \displaystyle (a-b)/m={\rm f}^{-1}(a) }\)

Answer 20

so you can see it is all the same thing

Answer 21

thank you for explaining how the trick works @SolomonZelman