+1995 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
+129852 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 ????
+1995 based off your steps the second one would be -3/4x+3/2 ? Am I correct?
+129852 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 !!!!
+1995 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 ?
+129852 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.... ]
+129852 P. S. ....I'm glad to see that you are trying to work toward the answers instead of just accepting them blindly...!!!
+1995 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.
+129852 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 !!!!
+1995 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.
+129852 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 !!!
