# Intermediate Value Theorem

• Feb 28th 2010, 03:15 PM
ExCaLiBuR
Intermediate Value Theorem
Maybe its just me, but I feel like this question is worded wrong in some way.

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

the cubed root of x=1-x, from 0 to 1.

I really don't even know where to start. My notes tell me what to do relatively well, but they also tell me that i need a constant in which to solve for. Help please?
• Feb 28th 2010, 03:17 PM
Drexel28
Quote:

Originally Posted by ExCaLiBuR
Maybe its just me, but I feel like this question is worded wrong in some way.

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

the cubed root of x=1-x, from 0 to 1.

I really don't even know where to start. My notes tell me what to do relatively well, but they also tell me that i need a constant in which to solve for. Help please?

I would assume this is asking show that $\displaystyle f(x)=x+\sqrt[3]{x}-1=0$ for some $\displaystyle x\in[0,1]$. It doesn't strike you as odd that $\displaystyle f(0)<0,f(1)>0$?
• Feb 28th 2010, 03:21 PM
ExCaLiBuR
Quote:

Originally Posted by Drexel28
I would assume this is asking show that $\displaystyle f(x)=x+\sqrt[3]{x}-1=0$ for some $\displaystyle x\in[0,1]$. It doesn't strike you as odd that $\displaystyle f(0)<0,f(1)>0$?

I understand that, but where does that fit into showing that there is a root of the equation? I'm sorry if I'm being extremely difficult, but I'm new to this
• Feb 28th 2010, 03:24 PM
Drexel28
Quote:

Originally Posted by ExCaLiBuR
I understand that, but where does that fit into showing that there is a root of the equation? I'm sorry if I'm being extremely difficult, but I'm new to this

I mean this in the most non-condescending way possible. Write out the hypothesis and conclusion the the IVT. I bet half way through it will hit you.
• Feb 28th 2010, 03:26 PM
Plato
Quote:

Originally Posted by ExCaLiBuR
I understand that, but where does that fit into showing that there is a root of the equation? I'm sorry if I'm being extremely difficult, but I'm new to this

In fact it is very clear. If $\displaystyle f(c)=0$ then $\displaystyle c$ is a root of the equation $\displaystyle f(x)=0$.
What do you not understand about that?
• Feb 28th 2010, 03:31 PM
ExCaLiBuR
Quote:

Originally Posted by Drexel28
I mean this in the most non-condescending way possible. Write out the hypothesis and conclusion the the IVT. I bet half way through it will hit you.

i really have no idea what you are talking about haha. I've tried to graph it out and my professor said that there is a number c in (0,1) such that f(c) = k (on the graph). The question confuses me as to where the c and k come into play.
• Feb 28th 2010, 03:35 PM
ExCaLiBuR
Quote:

Originally Posted by Plato
In fact it is very clear. If $\displaystyle f(c)=0$ then $\displaystyle c$ is a root of the equation $\displaystyle f(x)=0$.
What do you not understand about that?

That does make sense, I just don't understand where you got f(c)=0 from? I mean- where do you get the root of the equation, are there more than one answers?
• Feb 28th 2010, 03:52 PM
Plato
Quote:

Originally Posted by ExCaLiBuR
That does make sense, I just don't understand where you got f(c)=0 from? I mean- where do you get the root of the equation, are there more than one answers?

The key to this is that the graph of a continuous function cannot have a ‘hole’ in it.
If $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and $\displaystyle C$ is between $\displaystyle f(a)~\&~f(b)$ then the graph cannot ‘skip’ over $\displaystyle C$.
So there is a $\displaystyle c,~a\le c\le b$ such that $\displaystyle f(c)=C$. No skipping.