# Real analysis - proving the number of solutions

• Aug 4th 2008, 05:52 AM
kevin28
Real analysis - proving the number of solutions
Hey everyone,

Basically this kinda question comes up every year on the Real Analysis exam, but i really don't know how to do it.
I'm stuck on the last part of the question with
f(x) = 1-x.

PLEASE if anyone can explain to me how you go about doing it...let me know.

thank you
• Aug 4th 2008, 08:08 AM
CaptainBlack
Quote:

Originally Posted by kevin28
Hey everyone,

Basically this kinda question comes up every year on the Real Analysis exam, but i really don't know how to do it.
I'm stuck on the last part of the question with
f(x) = 1-x.

PLEASE if anyone can explain to me how you go about doing it...let me know.

thank you

For the last part you are looking for the solutions to:

$\displaystyle x^3-6x^2+12x-8=1-x$

or:

$\displaystyle x^3-6x^2+13x-9=0$

Decartes rule of signs tells us this has $\displaystyle 1$ or $\displaystyle 3$ positive real roots. As the function changes sign between $\displaystyle x=2$ and $\displaystyle x=3$ there is at least one root in this interval.

Now:

$\displaystyle \frac{d}{dx}x^3-6x^2+13x-9=3x^2-12x+13$

Examining the discriminant of the quadratic we find this has no real zeros, and as this is positive for large $\displaystyle x$, the cubic is everywhere an increasing function, and so can have at most one zero, which we have already located between $\displaystyle 2$ and $\displaystyle 3$.

RonL
• Aug 6th 2008, 02:58 AM
kevin28
thank u
thank u so much, that was really helpful!
cheers