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ive got a question on how to get a fixed point. on the equation for.

[tex]\frac{1}{8}X^2+\frac{11}{8}X+\frac{1}{2}[/tex]

do you find the two factors to get the fixed points. or run the equation though a quadratic formula to get the fixed points.

i have an = which is x^2-3X-4=0 but i dont know how the fraction 11/8X =3X in the equation.

A fixed point for a function, f, is a value of x such that f(x)= x.

Here

[tex]f(x)= \frac{1}{8}X^2+\frac{11}{8}X+\frac{1}{2}= x[/tex]

The easiest way to handle the fractions is to multiply the entire equation by 8:

[tex]x^2+ 11x+ 1= 8x[/tex]

or

[tex]x^2+ 3x+ 1= 0[/tex]

[tex]\frac{1}{8}X^2+\frac{11}{8}X+\frac{1}{2}[/tex]

do you find the two factors to get the fixed points. or run the equation though a quadratic formula to get the fixed points.

i have an = which is x^2-3X-4=0 but i dont know how the fraction 11/8X =3X in the equation.

## Homework Statement

## Homework Equations

## The Attempt at a Solution

A fixed point for a function, f, is a value of x such that f(x)= x.

Here

[tex]f(x)= \frac{1}{8}X^2+\frac{11}{8}X+\frac{1}{2}= x[/tex]

The easiest way to handle the fractions is to multiply the entire equation by 8:

[tex]x^2+ 11x+ 1= 8x[/tex]

or

[tex]x^2+ 3x+ 1= 0[/tex]

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