Find f^-1 in similar form for each of the following functions.
Question no. 1 ↓
f:x→3x over x-3,x not equal to 3
Question no. 2 ↓
f:x→2x+4 over 2x-1,x not equal to 1over2

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1. I will do the 2nd, you do the 1st in the same way

f(x) = (2x+4)/(2x-1)
or
y = (2x+4)/(2x-1)
inverse by switching x's and y's
x = (2y+4)/(2y-1)
simplify and solve this for y ....
2xy - x = 2y+4
2xy - 2y = x+4
y(2x - 2) = x+4
y = (x+4)/(2x-2)

f^-1 (x) = (x+4)/(2x-2) , x ≠ 1

I usually test my answer, let x = 3 , (I get no fractions)
f(3) = 10/5 = 2
f^-1 (2) = 6/2 = 3

(this does not "prove" that my answer is right, but if I did not get 3 as my return answer, by equation would have been wrong. My test "strongly suggests" that I am right)

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posted by Reiny
2. Reiny,thanks a lot! You are right...can you explain why the answer x is not equal to 1 ?

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posted by mandy
3. What would happen if you sub x = 1 into our new denominator?
Can we divide by zero ?

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posted by Reiny
4. Ohh...ok..thank you!

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posted by mandy
5. No.1
y=3x/x-3
x=3y/y-3
xy-3y=3x
y(x-3)=3x
y=3x/x-3
f^-1(x)=3x/x-3,x≠3
Am i correct ?:D

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posted by mandy
6. yes, but in this form of typing you have to use brackets.

f^-1 (x) = 3x/(x-3) or else the 3x is divided by x only

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posted by Reiny

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