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# What is the inverse

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What is the inverse of f(x)=−3x+5 ?

f ^−1(x)=−3x+5

f ^−1(x)=1/3x−53

f ^−1(x)=−1/3x+53

f ^−1(x)=3x−5

Second question

What is the inverse of the function?

g(x)=−4/3x+2

g−1(x)=4/3x−2

g−1(x)=−4/3x−2

g−1(x)=−3/4x+32

g−1(x)=3/4x+23

Oct 23, 2018
edited by jjennylove  Oct 23, 2018

#1
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y  =  -3x  + 5      first....we want to get x by itself

y - 5  =  3x        divide both sides by 3

[ y - 5 ] / 3  = x

"Swap"  x and y and for, y write   f-1(x)

[ x - 5 ] / 3  =  y

[x - 5 ] / 3  =  f-1(x)   =  (1/3)x  - 5/3

Based on these steps.....can you do the second one, Jenny ????   Oct 23, 2018
#2
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based off your steps the second one would be -3/4x+3/2 ? Am I correct?

jjennylove  Oct 23, 2018
#3
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Correct  !!!!

Note....here's how you can check your solution

Put  a value  in the first equation...I'll choose  3...  and we get - (4/3) * 3  + 2  =  -4+ 2  =  -2

So...the point ( 3, -2)  is on the graph

The inverse "reverses" the   coordinates....so the point (-2, 3) should be on the inverse graph

So...putting -2 into your inverse and we get

-(3/4) (-2)  +  3/2  =

6/4 + 3/2  =

3/2 + 3/2  =

3  !!!!!

So   ( -2, 3)  shows that your inverse solution is correct  !!!!

Good Job  !!!!   CPhill  Oct 23, 2018
#4
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Thank You! I have another question , same like this one if I could check with you ?

What is the inverse of the function?

f(x)=1/3x−5

f−1(x)=3x+15

f−1(x)=−1/3x+5/3

f−1(x)=−1/3x+5

f−1(x)=−3x+5/3

I think the answer would be A , f−1(x)=3x+15 ?

jjennylove  Oct 23, 2018
#5
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Put 15  into the original function and we get 0  back

So

(15, 0)  is on the original function

So....0  should return 15 in your proposed inverse....so

3(0)  +  15  =  15  ....correct  !!!

Note...this isn't foolproof....but if given a list of proposed inverses...it should test out

There is a "foolproof" method  [ if you are interested.... ]   CPhill  Oct 23, 2018
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P. S.   ....I'm glad to see that you are trying to work toward the answers instead of just  accepting them blindly...!!!   CPhill  Oct 23, 2018
#7
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I am most definitely going to use your feedback for any future problems and make sure to always check my problems afterward! I appreciate the tips and tricks, it has and will help me. Also, yes I really do enjoy learning and I do want to make sure I fully understand everything just so for my own knowledge as well as to just learn new things. I really appreciate you helping me on this, you explained everything clearly that made me understand so it was super easy , and now I can use this knowledge you gave me for any other problems! Thank you.

I would like to learn the foolproof method if you dont mind explaining,I think it would really help me.

jjennylove  Oct 23, 2018
#8
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Here's the foolproof method :

Using

(1/3)x  -  5      and    3x + 15

Put the second function into the first for each x and we have

(1/3) (3x + 15)  - 5

x +  5  -  5

x

Now...put the first function  nto the second for each x  and we have

3 [ (1/3)x -  5 ]  + 15

x - 15  +  15

x

If we get an x in each case.....the functions are inverses  !!!!   Oct 23, 2018
edited by CPhill  Oct 23, 2018
#9
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I think I prefer the first method but its nice seeing 2 different ways . I will make notes of these! Thank you for showing me both.

jjennylove  Oct 23, 2018
#10
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OK....but be careful with this method..it isn't foolproof in all cases

For instance

y  = 3x       and y   = (1/4)x

Putting  0   into the first returns 0

So  (0,0)  is on the first graph

So....putting the "returned" value of 0  into the second graph gives us back the original value of 0

So...these would appear to be inverses....but they are not

Using the foolproof method

3 [ (1/4)x ]  =    (3/4) x

This isn't   "x"......so....we can stop here....they are NOT inverses  !!!   CPhill  Oct 23, 2018