• Oct 13th 2007, 09:41 AM
rlarach
:confused::confused:

Let f(x) be a polynomial function such that, for all real x.

f(x^2 +2) = x^4 + 10x^2 + 4.

Evaluate f(x^2 - 2).

• Oct 13th 2007, 09:50 AM
topsquark
Quote:

Originally Posted by rlarach
:confused::confused:

Let f(x) be a polynomial function such that, for all real x.

f(x^2 +2) = x^4 + 10x^2 + 4.

Evaluate f(x^2 - 2).

Well, the "brute force" method would be:
$f(x^2 + 2) = x^4 + 10x^2 + 4$

Let $y = x^2 + 2$.

Solve for x:
$x^2 = y - 2$

$x = \pm \sqrt{y - 2}$

Ignoring the difficulty of the $\pm$ let's blithely continue:
$f(y) = ( \pm \sqrt{y - 2} )^4 + 10 ( \pm \sqrt{y - 2} ) + 4$

$f(y) = (y - 2)^2 + 10(y - 2) + 4$

Now let $y = x^2 - 2$.

$f(x^2 - 2) = (x^2 - 2 - 2)^2 + 10(x^2 - 2 - 2) + 4$

$f(x^2 - 2) = (x^2 - 4)^2 + 10(x^2 - 4) + 4$

which you can expand, or not, as you choose.

-Dan
• Oct 13th 2007, 09:57 AM
CaptainBlack
Quote:

Originally Posted by rlarach
:confused::confused:

Let f(x) be a polynomial function such that, for all real x.

f(x^2 +2) = x^4 + 10x^2 + 4.

Evaluate f(x^2 - 2).

$
f(x)=x^2+6x-12
$

RonL
• Oct 13th 2007, 11:16 AM
rlarach
How Do You Get X^2 + 6x - 12
• Oct 13th 2007, 01:11 PM
CaptainBlack
Quote:

Originally Posted by rlarach
How Do You Get X^2 + 6x - 12

coefficient 1, so is of the form:

f(x)=x^2+bx+c.

Then:

f(x^2+2)=(x^2+2)^2+b(x^2+2)+c = x^4+10x^2+4

Now expend the middle term and equate coefficients.

RonL