Intermediate Value Theorem

**ExCaLiBuR** · February 28th 2010, 03:15 PM

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?

**Drexel28** · February 28th 2010, 03:17 PM

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 $f(x)=x+\sqrt[3]{x}-1=0$ for some $x\in[0,1]$ . It doesn't strike you as odd that $f(0)<0,f(1)>0$ ?

**ExCaLiBuR** · February 28th 2010, 03:21 PM

Originally Posted by Drexel28

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

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

**Drexel28** · February 28th 2010, 03:24 PM

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.

**Plato** · February 28th 2010, 03:26 PM

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 $f(c)=0$ then $c$ is a root of the equation $f(x)=0$ .
What do you not understand about that?

**ExCaLiBuR** · February 28th 2010, 03:31 PM

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.

**ExCaLiBuR** · February 28th 2010, 03:35 PM

Originally Posted by Plato

In fact it is very clear. If $f(c)=0$ then $c$ is a root of the equation $f(x)=0$ .
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?

**Plato** · February 28th 2010, 03:52 PM

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 $f$ is continuous on $[a,b]$ and $C$ is between $f(a)~\&~f(b)$ then the graph cannot ‘skip’ over $C$ .
So there is a $c,~a\le c\le b$ such that $f(c)=C$ . No skipping.

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