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

#1**+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 ????

CPhill
Oct 23, 2018

#3**+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 !!!!

CPhill
Oct 23, 2018

#4**+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**+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.... ]

CPhill
Oct 23, 2018

#6**0 **

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**+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. 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**+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 !!!!

CPhill
Oct 23, 2018

#9**+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**+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 !!!

CPhill
Oct 23, 2018