# Find Function expression

• May 1st 2010, 08:34 AM
Monster32432421
[Solved]Find Function expression
The function f is defined by f(x) = x^3-x for -1=< x < 0 and f(x+1) = f(x) for all real x.
Find an expression for f(x) for 0=< x < 1

Dont worry I solved it...
• May 1st 2010, 10:02 AM
Soroban
Hello, Monster32432421!

You solved it? . . . Good for you!

It's a strange problem . . .

Quote:

The function $\displaystyle f(x)$ is defined by: .$\displaystyle f(x) \:=\: x^3-x\;\text{ for }\text{-}1 \leq x < 0$

and $\displaystyle f(x+1) \:=\: f(x)\;\text{ for all real }x.$

Find an expression for $\displaystyle f(x)\:\text{ for }0 \leq x < 1$

$\displaystyle f(x) \:=\:x^3-x$ has this graph:

Code:

                  |               *  |           *    * |          *         *      *|                   |     - - * - - - - * - - - - * - -       -1        |        1                   |*      *       *          | *    *                   |  *                   |

For the domain $\displaystyle [\text{-}1,0)$, we have this graph:

Code:

                  |               *  |           *    * |         *      *|                   |     - - * - - - - * - - - -       -1        |0                   |

Since $\displaystyle f(x+1) = f(x),$ the graph on $\displaystyle [0,1)$ is identical to the graph on $\displaystyle [\text{-}1,0)$

The graph looks like this:

Code:

                  |               *  |    *           *    * |  *    *         *      *|*      *                   |     - - * - - - - * - - - - * - -       -1        |0        1                   |

So the graph of: $\displaystyle f_1(x) \:=\:x^3-x$ . is moved one unit to the right.

We have: .$\displaystyle f_2(x) \;=\;(x-1)^3 - (x-1) \:=\:x^3 - 3x^2 + 2x$

Therefore: .$\displaystyle f(x) \;=\;\begin{Bmatrix}x^3-x && \text{-}1 \leq x < 1 \\ x^3-3x^2+2x && 0 \leq x < 1 \end{Bmatrix}$