Question

If 4f(x)+f(3-x)=x^2 what is f(x)?

Answer 1

I think you solve by switching the x to (3-x) and the 3-x to x if that makes sense. I don't know where to go from there.

Answer 2

What level math are you doing?

Answer 3

College Algebra.

Answer 4

it looks to be that x = 3-x if anything, its worth a shot :)

2x = 3; x = 3/2

4(3/2) + 3/2 ?= 9/4

12/2 + 3/2 = 15/2 ..... well that dint work out lol

Answer 5

its hard to tell without more information ... at least for me it is

Answer 6

I have noooooo idea. Lol. I googled and found a similar problem.
It might have something to do with synthetic division

Answer 7

Please share!

Answer 8

i cant make sense of it at the moment :)

Answer 9

well.. i don't know the url. I googled it the other day =T but this might help. http://au.answers.yahoo.com/question/index?qid=20110522002904AATkbUG

Answer 10

I had this problem the other day for an assignment, 5f(x)+f(3-x)=x^2 and the right answer was supposed to come out to f(x)=1/6x^2+1/4x-3/8 .

Answer 11

I'm just working the example problem trying to figure out how they got their answer.
If I figure it out I'll give you a yell!

Answer 12

PLEASE DO! lol

Answer 13

I ran out of time on the assignment :/ But the answer is supposed to be 1/5x^2 + 2/5x - 3/5.

Answer 14

do you have any of the steps? think this has to do with finding the inverse yes?

Answer 15

or are you supposed to know that
\[f(x)=ax^2+bx+c\] and solve for a, b and c?

Answer 16

Unfortunately, no. It wasn't a problem that was solved in class.

Answer 17

ok was the first term
\[\frac{1}{5}x^2\]?

Answer 18

From the example I found online it looks like a combination of that formula above f(x)=ax^2+bx+c and synthetic division.

Answer 19

if that is the right first term we can do this

Answer 20

It was!

Answer 21

whew. ok ready? here we go!

Answer 22

i guess you have to know that
\[f(x)\] is going to be a quadratic of some sort. that is
\[f(x)=ax^2+bx+c\]

Answer 23

we just don't know what a, b and c are and that is what we have to find.

Answer 24

so we compute
\[4f(x)=4ax^2+4bx=4c\] easy enough. now what is
\[f(3-x)\]?
\[f(3-x)=a(3-x)^2+b(3-x)+c\] and so
\[4f(x)+f(3-x)=4ax^2+4bx+c+a(3-x)^2+b(3-x)+c\]

Answer 25

sorry typo on first line. it should be
\[4f(x)=4ax^2+4bx+4c\]

Answer 26

and we know that all that muck is equal to
\[x^2\] because that is what we are told

Answer 27

now the fun job is to multiply out. you get
\[4ax^2+4bx+4c+9a-6ax+ax^2 +3b-bx+c=x^2\]

Answer 28

now we collect like terms. for the
\[x^2\] term we have
\[4ax^2+ax^2=5ax^2\] and we know that this is equal to \[x^2\] which means \[5a=1\] so
\[a=\frac{1}{5}\]

Answer 29

did i lose you yet?

Answer 30

I'm following you, but why is c not c(3-x) as well?

Answer 31

oh because c is a constant. if i replace x by 3-x is
\[ax^2+bx+c\] i get
\[a(3-x)^2+b(3-x)+c\]

Answer 32

hope it is more or less clear what i did in the second part. i was not multiplying by 3-x i was replacing x by 3-x

Answer 33

Oh. That makes more sense. The first part was multiplication and the second part was filling in for x then?

Answer 34

yes the first part was multiply by 4 because it was
\[4f(x)\] the second part was replacing x by 3-x because it was
\[f(3-x)\]

Answer 35

so now at least we know that
\[a=\frac{1}{5}\] by equating coefficients. you have on one side
\[5ax^2\] and on the other side you have
\[x^2\] so as i said
\[5a=1\]
\[a=\frac{1}{5}\]

Answer 36

now we do the same for the "x" term. there are no x's on the right because you just have
\[x^2\]

Answer 37

on the left i have to scroll up and see what we get

Answer 38

yes on the left we have for the "x" term
\[4bx-6ax-bx\]
\[3bx-6ax\]
\[(3b-6a)x\]

Answer 39

and we know that
\[3b-6a=0\] because we should have no x term

Answer 40

\[3b=6a\]
\[3b=\frac{6}{5}\]
\[b=\frac{2}{5}\] this because we already knew that
\[a=\frac{1}{5}\]

Answer 41

is that what we should have for b?

Answer 42

and finally we do the same thing with the constant. the constant on the right is 0. the constant on the left is ... let me scroll up.

Answer 43

It's right. I'm a little bit stuck on how you figured out that 4ax^2 +ax^2=5ax^2

Answer 44

oh i can explain that. we call that "combining like terms" yes?

Answer 45

and the 3b-6a part.

Answer 46

4 somethings and 1 something is 5 somethings.

Answer 47

I get that.

Answer 48

yes?
\[4x+x=5x\]
\[4ax+ax=5ax\]
\[4ax^2+ax^2=5ax^2\] tex

Answer 49

k?

Answer 50

now lets go all the way back to
\[4ax^2+4bx+4c+9a−6ax+ax^2+3b−bx+c=x^2\]

Answer 51

and rewrite the left and side in standard form. in other words combine like terms

Answer 52

Ohh. Okay I think I get it. I was confused about what the 9a-6ax+ax^2+3b-bx+c what about.

Answer 53

\[4ax^2+ax^2+4bx-6ax-bx+4c+9a+3b+c=x^2\]

Answer 54

\[5ax^2+(3-6a)x+(5c+9a+3b)=x^2\]

Answer 55

damned typo!

Answer 56

\[5ax^2+(3b-6a)x+(5c+9a+3b)=x^2\]

Answer 57

this tells us that
\[5a=1, 3b-6a=0,5c+9a+3b=0\]

Answer 58

is that step clear?

Answer 59

because thats is what you get when you combine like terms on the left, and on the right you have
\[1x^2+0x+0\]

Answer 60

so if that part is ok it is gravy from here on in. first solve for a, then solve for b, then solve for c

Answer 61

as you can see setting this up is a pain. solving not so bad

Answer 62

wait, why exactly is that 0? :/

Answer 63

ok you are given that
\[4f(x)+f(3-x)=x^2\] yes?

Answer 64

and we just rewrote the left hand side but the right hand side is still just
\[x^2\]

Answer 65

so whatever "x" terms we have on the left, on the right there are none

Answer 66

on the left side we collect like terms and get
\[(3a-6b)x\] but on the right side we have
\[0x\]

Answer 67

no x term on the right,
\[3a-6b\]of them on the left.

Answer 68

that tells us that
\[3a-6b=0\]

Answer 69

analogous to how we knew that
\[5a=1\]
on the left we have
\[5ax^2\] on the right we have
\[x^2\] so we know \[5a=1\]

Answer 70

on the left we have
\[(3a-6b)x\] on the right we have
\[0x\]
on the left the "constant" is
\[9a+3b+5c\] on the right the "constant" is 0

Answer 71

that's pretty clear now

Answer 72

ok good. so we so\
lve one at a time. first we found
\[a=\frac{1}{5}\] then we know that
\[3b-6a=0\]
\[3b=6a\]
\[3b=\frac{6}{5}\]
\[b=\frac{2}{5}\]

Answer 73

and finally we solve
\[5c+9a+3b=0\] and now this is too much arithmetic for me but the idea is that we know this is
\[5c+\frac{9}{5}+\frac{6}{5}=0\] etc

Answer 74

might as well finish...
\[5c+\frac{15}{5}=0\]
\[5c+3=0\]
\[5c=-3\]
\[c=-\frac{3}{5}\]

Answer 75

I got it. :D

Answer 76

Thank you so much!