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

Doing it in reverse is odd...

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- Oct 25th 2007, 08:53 AMDivideBy0function
Given that $\displaystyle f(x+3)=x^3-3$, find $\displaystyle f(x)$

Doing it in reverse is odd... - Oct 25th 2007, 09:23 AMPlato
$\displaystyle f(x) = f\left( {(x - 3) + 3} \right) = (x - 3)^3 - 3$

- Oct 25th 2007, 09:49 AMSoroban
Hello, DivideBy0!

Quote:

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

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

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

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

Then the steps are quite clear:

. . Subtract 3, then cube, then subtract 3.

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