# function

• October 25th 2007, 08:53 AM
DivideBy0
function
Given that $f(x+3)=x^3-3$, find $f(x)$

Doing it in reverse is odd...
• October 25th 2007, 09:23 AM
Plato
$f(x) = f\left( {(x - 3) + 3} \right) = (x - 3)^3 - 3$
• October 25th 2007, 09:49 AM
Soroban
Hello, DivideBy0!

Quote:

Given that $f(x+3)\:=\:x^3-3$, find $f(x)$

Doing it in reverse is odd... . . . . yes, it is

Then I baby-talked my way through it . . .

We have a functon $f$ such that $x+3$ is "processed" and becomes $x^3-3$

. . . $x + 3\quad\rightarrow\quad \boxed{\begin{array}{ccc} \\ \;\;f \;\; \\ \\\end{array}} \quad\rightarrow\quad x^3-3$

Then the steps are quite clear:
. . Subtract 3, then cube, then subtract 3.

Therefore: . $f(x) \;=\;(x-3)^3 - 3$