- #1

- 14

- 0

## Homework Statement

http://postimg.org/image/48919pl1x/

## Homework Equations

f(f(x))= (x^4)-4(x^2)+2,

## The Attempt at a Solution

[f(x)]^2=

(x^4)-4(x^2)+2=

[(x-2)^2]-(4x^2)+2

i cant seem to square root (x^4)-4(x^2)+2...

- Thread starter GreenTea09
- Start date

- #1

- 14

- 0

http://postimg.org/image/48919pl1x/

f(f(x))= (x^4)-4(x^2)+2,

[f(x)]^2=

(x^4)-4(x^2)+2=

[(x-2)^2]-(4x^2)+2

i cant seem to square root (x^4)-4(x^2)+2...

- #2

- 3,812

- 92

The question states that f(x) is a polynomial. Use this.## Homework Statement

http://postimg.org/image/48919pl1x/

## Homework Equations

f(f(x))= (x^4)-4(x^2)+2,

## The Attempt at a Solution

[f(x)]^2=

(x^4)-4(x^2)+2=

[(x-2)^2]-(4x^2)+2

i cant seem to square root (x^4)-4(x^2)+2...

http://en.wikipedia.org/wiki/Polynomial

- #3

- 14

- 0

let f(x)=a(x^2)+bx+c

(a(x^2)+bx+c)*(a(x^2)+bx+c) =(x^4)-4(x^2)+2

comparing the coefficient of x^2

2c+b^2=-4 ---eqn(2)

comparing the coefficient of constant,

i get a c value of sqrt2 and -sqrt2

which i try to fit into eqn(2) but dosent work out..

- #4

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

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f(f(x)) must also be evaluated on the bf(x)+c terms!!

- #5

Mark44

Mentor

- 34,687

- 6,394

The image link is broken.## Homework Statement

http://postimg.org/image/48919pl1x/

No one else pointed this out, so I will. f(f(x)) is not the same as [f(x)]## Homework Equations

f(f(x))= (x^4)-4(x^2)+2,

## The Attempt at a Solution

[f(x)]^2=

That's not relevant here, since you are not being asked to solve for f(x) in the equation [f(x)](x^4)-4(x^2)+2=

[(x-2)^2]-(4x^2)+2

i cant seem to square root (x^4)-4(x^2)+2...

What you're working with is a composite of f with itself -- (f o f)(x) --, not the square of f(x).

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