Asked by Lindsay
f(x) =x-2/x-3. I calculated f^-1 (x)to be 3x-2/y-1. How do I use f^-1 (x) to evaluate (x-2/x-3)=4.
P.S/ f^-1 means 'f' inverse.
P.S/ f^-1 means 'f' inverse.
Answers
Answered by
Reiny
First of all, I am sure you meant
f(x) = (x-2)/(x-3) , those brackets are essential!
or
y = (x-2)/(x-3)
step1 in forming the inverse:
switch the x and y variables
---> x = (y-2)/(y-3)
step2: solve this new equation for y
xy - 3x = y-2
xy - y = 3x - 2
y(x-1) = 3x-2
y = (3x-2)/(x-1) , notice the brackets again ?
(you had y-1 in the denominator , I hope it was just a typo )
f^-1 (x) = (3x-2)/(x-1)
I don't understand what you mean by
"How do I use f^-1 (x) to evaluate (x-2/x-3)=4"
did you mean evaluate
f( (x-2)/(x-3) ) ??
or
are you solving (x-2)/(x-3) = 4 ??
f(x) = (x-2)/(x-3) , those brackets are essential!
or
y = (x-2)/(x-3)
step1 in forming the inverse:
switch the x and y variables
---> x = (y-2)/(y-3)
step2: solve this new equation for y
xy - 3x = y-2
xy - y = 3x - 2
y(x-1) = 3x-2
y = (3x-2)/(x-1) , notice the brackets again ?
(you had y-1 in the denominator , I hope it was just a typo )
f^-1 (x) = (3x-2)/(x-1)
I don't understand what you mean by
"How do I use f^-1 (x) to evaluate (x-2/x-3)=4"
did you mean evaluate
f( (x-2)/(x-3) ) ??
or
are you solving (x-2)/(x-3) = 4 ??
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