Question

Transformation of Functions:

p(x) is the PRE-IMAGE of f(x)

1. Find the mapping notation
2. Find the points on 'p' for the given points on 'f'

Answer 1

\[\frac{f(x)-3}{2}=~p(\frac{x-1}{4})\]

Answer 2

What I think:
I found the inverse of the x-part and didn't take the inverse for the y-part:
\[T(x,y)=(4x+1,2y+3)\]

I assume this is the correct mapping notation of p to f right?

Answer 3

And so IF that's the mapping notation from p to f, then the inverse of that is from f to p, right?

And that's how I would find the points on p given the points on f right?

Answer 4

Have you been stuck on this question for 45 minutes?

Answer 5

Still need help at this question, @mhchen?

Answer 6

Yep. Confused on whether to use the mapping notation or the equation itself to find the pre-image points. @3mar

Answer 7

Where is the equation itself?

Answer 8

\[\frac{f(x)-3}{2}=~p(\frac{x-1}{4})\] @3mar

Answer 9

f(x) is the transformation of p(x)

Answer 10

And I want to know like if there's a point on f(x) that's (-5,1)

I want to know where the same point on p(x) is

Answer 11

the part I'm confused about is whether I need to use the inverse of not for the x-value

Answer 12

\[\frac{f(x)-3}{2}=~p(\frac{x-1}{4})\]
\[f(x)-3=~2*p(\frac{x-1}{4})\]
\[f(x)=~2*p(\frac{x-1}{4})+3\]

got that?

Answer 13

Yep

Answer 14

And I know the y-value uses that.

I wonder if I have to do anything special with the x-value?

Answer 15

Can you find the mapping notation using this?

Answer 16

\[T(x,y)=(4x+1,2y+3)\]

Answer 17

Question: How? The problem says T(p(x)) = f(x) , it doesn't say anything about x. How can you assume that T(x) = 4x+1?

Answer 18

Is it not that T (x, p(x)) = (x, f(x)) ??

Answer 19

Um, I did the inverse for the x-part, and left the y-part as it is?

Answer 20

I don't know!! what is the given point on f?

Answer 21

(-5,1)

Answer 22

And using my transformation, I was thinking

for what value of x would
4x+1 = -5 <-- using my mapping notation

which would be -6/4 right?


The other thing I'm not sure is if I am supposed to do:

for what value of x would
-5 = p((x-1)/4)

Answer 23

And using my transformation, I was thinking

for what value of x would
4x+1 = -5 <-- using my mapping notation

which would be -6/4 right?


The other thing I'm not sure is if I am supposed to do:

for what value of x would
\[-5 = p(\frac{x-1}{4})\]

which would be -19