+0  
 
+1
33
10
avatar+186 

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

jjennylove  Oct 23, 2018
edited by jjennylove  Oct 23, 2018
 #1
avatar+90968 
+2

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 ????

 

cool cool cool

CPhill  Oct 23, 2018
 #2
avatar+186 
+1

based off your steps the second one would be -3/4x+3/2 ? Am I correct?

jjennylove  Oct 23, 2018
 #3
avatar+90968 
+2

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  !!!!

 

 

cool cool cool

CPhill  Oct 23, 2018
 #4
avatar+186 
+1

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
avatar+90968 
+2

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.... ]

 

cool cool cool

CPhill  Oct 23, 2018
 #6
avatar+90968 
0

P. S.   ....I'm glad to see that you are trying to work toward the answers instead of just  accepting them blindly...!!!

 

 

 

cool cool cool

CPhill  Oct 23, 2018
 #7
avatar+186 
+1

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.smiley 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
avatar+90968 
+2

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  !!!!

 

 

 

cool cool cool

CPhill  Oct 23, 2018
edited by CPhill  Oct 23, 2018
 #9
avatar+186 
+1

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
avatar+90968 
+2

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  !!!

 

 

 

cool cool cool

CPhill  Oct 23, 2018

36 Online Users

avatar
avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.