# Homework Help: Numerical analisys question

1. May 29, 2010

### nhrock3

2. May 29, 2010

### HallsofIvy

To get $g_1$, they simply added x to both sides of $x^3- x- 5= 0$ to get $x^3- 5= x$.

To get $g_2$, they added x+ 5 to both sides to get $x^3= x+ 5$ and took the cube root of both sides.

To get $g_3$, they subtracted 5 from both sides to get $x^3- x= 5$, then factored, $x(x^2- 1)= 5$ and, finally, divided both sides by $x^2- 1$ to get $x= 5/(x^2- 1)$.

3. May 29, 2010

### nhrock3

ok i solved a similar equation:
$$x^3+2x^2+4-x=0$$
$$g1=x^3+2x^2+4$$

why for g1 L=0.5
how to find L for other g
what is the meaning of L?

Last edited: May 29, 2010