# Thread: Roots of a function

1. ## Roots of a function

I help some help please.

Basically if i had a graph of e.g.

y=-(x-3)^2+1

I know that my graph would intersect the x-axis at x=2 and x=4.

However if I didn't know the equation of my curve, but knew the shape of the graph was like y=-(x-3)^2+1, and it crossed the x-axis at 2 points.

How could I prove that it had 2 roots?

I think I may have something to do with continuity, but not sure.

Any help is appreciated.

2. I'm not sure I understand the problem. Crossing the x-axis at 2 points means the same thing as having 2 roots, doesn't it?

If you have a continuous function f(x), and it is positive for $\displaystyle x=x_0$, negative for $\displaystyle x=x_1$, and then positive again for $\displaystyle x=x_2\text{ (with }x_0<x_1<x_2\text{)}$, then you know that it has at least 2 roots.

If you can tell me more, I might be able to help more. If the problem is from a book or problem set, it usually helps to copy the whole problem word-for-word. It also helps to know what you're studying.

Post again in this thread and I'll answer as soon as I can.

- Hollywood

3. Basically I have the function v(u) = -u + 1/RoInU + C where C is a constant of integration.

I need to determine relevant solutions so that v(u)=0

I already know that v(1)=0

I need to sketch the curves with 0.1<=u<=1 for an illustrative range of values of Ro.

Out of these curves could also be traced out during an epidemic.

4. So you have a function $\displaystyle v(u)=-u+\frac{1}{R_0I_nu}+C$ and you need to find where v(u)=0.

You can use the boundary condition v(1)=0 to determine C:

$\displaystyle v(1)=-1+\frac{1}{R_0I_n}+C=0\text{, so }C=1-\frac{1}{R_0I_n}$

Substituting back in to the equation gives:

$\displaystyle v(u)=-u+1+\frac{1}{R_0I_n}\left(\frac{1}{u}-1\right)$

And the roots are given by:

$\displaystyle v(u)=-u+1+\frac{1}{R_0I_n}\left(\frac{1}{u}-1\right)=0$

$\displaystyle 1-u+\frac{1}{R_0I_n}\left(\frac{1-u}{u}\right)=0$

We already know about the root at u=1, so we can cancel 1-u, giving:

$\displaystyle 1+\frac{1}{R_0I_n}\left(\frac{1}{u}\right)=0$

$\displaystyle u=-\frac{1}{R_0I_n}$ as the second root.

I don't know if the combination $\displaystyle R_0I_n$ can be negative. If it has to be positive, and assuming u is always positive, then there is only the one root at u=1. If it can be negative, there is a second root.

I hope that was helpful. Post again in this thread if you're still having trouble.

- Hollywood