1. ## Is my answer right? Function problem

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?

2. $\displaystyle f(x+1) = x^2 - 5x + 3$

So $\displaystyle f(x) = f(x + 1 - 1) = (x - 1)^2 - 5(x - 1) + 3$. Now simplify.

3. Originally Posted by Prove It
$\displaystyle f(x+1) = x^2 - 5x + 3$

So $\displaystyle f(x) = f(x + 1 - 1) = (x - 1)^2 - 5(x - 1) + 3$. Now simplify.
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.

4. You were given the hint "let u =x+1 and find f(u)". Doing that, x= u-1 and $f(u)= (u- 1)-^2- 5(u- 1)+ 3= u^2- 2u+ 1- 5u+ 5+ 3= u^2- 7u+ 9$ which is the same as $f(x)= x^2- 7x+ 9$.

Do you understand what "f(x)= x^2- 7x+ 9" means?

What would f(a) equal? What about f(a+1)? What would f(x+ 1) equal?

5. Originally Posted by HallsofIvy
You were given the hint "let u =x+1 and find f(u)". Doing that, x= u-1 and $f(u)= (u- 1)-^2- 5(u- 1)+ 3= u^2- 2u+ 1- 5u+ 5+ 3= u^2- 7u+ 9$ which is the same as $f(x)= x^2- 7x+ 9$.

Do you understand what "f(x)= x^2- 7x+ 9" means?

What would f(a) equal? What about f(a+1)? What would f(x+ 1) equal?
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".

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.

6. Originally Posted by frankinaround
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".

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.

$f(x+1)=x^2-5x+3$

We can try to write this in terms of (x+1) instead of x.

$(x+1)^2=(x+1)(x+1)=x^2+2x+1$

$\Rightarrow\ x^2=(x+1)^2-2x-1$

We can also write the "x" term as an (x+1) term...

$x^2-5x+3=(x+1)^2-2x-1-5x+3=(x+1)^2-7x+2$

$-7x+2=-7(x+1)+7+2$

$\Rightarrow\ x^2-5x+3=(x+1)^2-7(x+1)+9$

These equations are identical,

but they are both $f(x+1)$

Hence, to write $f(x),$ we can now substitute x for x+1
in the equation written in terms of (x+1).

7. Originally Posted by frankinaround
No I dont understand.
Then you really need to go over "function notation" with your teacher.

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).
No, I didn't. I replaced x with u- 1 which is equal to 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".

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.

8. Hello, frankinaround!

This is a variation of what Archie did . . .

$\text{Find }f(x)\,\text{ if }f(x+1) \:=\: x^2-5x+3.$

With a little imagination, we can hammer the function into the desired form.

We have: . $f(x+1) \;=\;x^2 - 5x + 3$

Add and subtract $(2x+1)$

. . $f(x+1) \;=\;x^2 {\bf + 2x + 1} - 5x + 3 \:{\bf -\,2x - 1}$

. . . . . . . . . $\;=\;(x+1)^2 - 7x + 2$

. . $f(x+1) \;=\;(x+1)^2 -7x {\bf- 7} + 2 \:{\bf+\: 7}$

. . . . . . . . . $=\;(x+1)^2 - 7(x+1) + 9$

Therefore: . $f(x) \;=\;x^2 - 7x + 9$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I just found yet another approach.

Let the original function be: . $f(x) \;=\;Ax^2 + Bx + C$

Since $f(x+1) \:=\:x^2 - 5x + 3$, we have:

. . $A(x+1)^2 + B(x+1) + C \;=\;x^2 - 5x + 3$

This simplifies to:

. . $Ax^2 + (2A+B)x + (A+B+C) \;=\;x^2-5x+3$

Equate coefficients: . $\begin{Bmatrix}A \;=\;1 \\ 2A + B \;=\;\text{-}5 \\ A+B+C \;=\;3 \end{Bmatrix}$

. . and solve the system: . $A = 1,\;B = \text{-}7,\;C = 9$

Therefore: . $f(x) \;=\;x^2 - 7x + 9$