# Thread: Real analysis - proving the number of solutions

1. ## 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

2. 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:

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

or:

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

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

Now:

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

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

RonL

3. ## thank u

thank u so much, that was really helpful!
cheers