So . Now simplify.
Find f(x) if f(x+1) = x^2-5x+3. Hint: let u =x+1 and find f(u). (ps do you think the author of this question is curseing at me? haha)
I got f(x)= x(x+2)-9x because f(x+1) = (x+1)(x+1+2)-9x=x^2-5x+3.
But this just seems fishy. like there can be other answers right? So am I right or am i missing something?
I kind of understand what your doing... if f(x+1) = x^2-5x+3, then f(x-1) should equal f(x). So your putting x-1 into thef(x) equation that has f( x+1) equation in it.
However.... (x-1)^2-5(x-1)+3 simplifys to x^2-7x+9.
So your saying f(x) = x^2-7x+9
But then how come f(x+1) = x^2-5x+10 ? shouldnt putting (x+1) threw f(x) = x^2-5x+3 like the original equation? thats why I dont get it.
No I dont understand. I thought that if f(x+1) equals the above equation, then if you put (x-1) threw it it might go back to f(x). Im pretty sure I was wrong thinking that. SO I dont really get what to do. (other then just make up any equation that when you put x+1 threw would equal the same as f(x+1)).
As for x(a) and x(a+1), I dont know how to answer that without having f(x) to plug them into. I guess you MIGHT mean the constant before x^2 for "a".
Please help.
I dont really see where to go because im thinking it says find f(u), u = x+1. So thats really saying find f(x+1) isnt it? and isnt f(x+1) the equation that they give me? So thats not really helping me. So im definitly missing something.
Here's another way to look at it.
Then you could re-examine the previous posts by "Prove It" and "HallsoIvy"
since the logic is more compact and faster.
We can try to write this in terms of (x+1) instead of x.
We can also write the "x" term as an (x+1) term...
These equations are identical,
but they are both
Hence, to write we can now substitute x for x+1
in the equation written in terms of (x+1).
Then you really need to go over "function notation" with your teacher.
No, I didn't. I replaced x with u- 1 which is equal to x.I thought that if f(x+1) equals the above equation, then if you put (x-1) threw it it might go back to f(x).
Im pretty sure I was wrong thinking that. SO I dont really get what to do. (other then just make up any equation that when you put x+1 threw would equal the same as f(x+1)).
As for x(a) and x(a+1), I dont know how to answer that without having f(x) to plug them into. I guess you MIGHT mean the constant before x^2 for "a".
Please help.
I dont really see where to go because im thinking it says find f(u), u = x+1. So thats really saying find f(x+1) isnt it? and isnt f(x+1) the equation that they give me? So thats not really helping me. So im definitly missing something.
Hello, frankinaround!
This is a variation of what Archie did . . .
With a little imagination, we can hammer the function into the desired form.
We have: .
Add and subtract
. .
. . . . . . . . .
Subtract and add 7:
. .
. . . . . . . . .
Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I just found yet another approach.
Let the original function be: .
Since , we have:
. .
This simplifies to:
. .
Equate coefficients: .
. . and solve the system: .
Therefore: .