Int. Value Thm: use to show root exists on 1/4 < x < 1

[Sophie]

I have to use the intermediate Value theorem to show there is a root of

x6 + 2x - (1/x) In the interval (1/4 , 1)

Which I have done as follows

f(1/4) = -3.484
f(1) = 2

Since -3.484 < 0 < 2 there is a number such that f(c) = 0 by the intermediate value theorem.

My first problem is I must show that the hypotheses of theorem are satisfied before you can apply the theorem and I am not sure whether this is what I have shown?

My second problem is I have to find this root. I know it is close to 0.12766, through the use of graphing. Can this be solved using factoring?

Thanks, Sophie[/list]

 

[stapel]

x6 + 2x - (1/x)

Does the above mean "x^6 + 2x - 1/x", or something else?

I am not sure whether this is what I have shown?

What are the hypotheses, the "if's", in the IVT, as stated in your text?

My second problem is I have to find this root.

Why? The exercise, as posted, only says to prove that a root exists. Is there another part of this exercise, as yet unposted, which says that you need to actually find the zero?

Thank you.

Eliz.

 

[Sophie]

Oops! The function was supposed to be "x^3 + 2x - (1/x)" in (1/4 , 1)

Why? The exercise, as posted, only says to prove that a root exists. Is there another part of this exercise, as yet unposted, which says that you need to actually find the zero?

Yes there is a second part to find this root to two decimal places, which I have only managed through the use of a graph.

Thanks, Sophie

 

[stapel]

What are the hypotheses, the "if's", in the IVT, as stated in your text?

...there is a second part to find this root to two decimal places....

For future reference, it is generally best to include the entire question with which you are requesting help. Thank you.

What techniques does your book provide for finding approximate solutions? Obviously, you aren't expected to find the exact solution ("to two decimal places"). Back in algebra, you of course did numerical approximations, where you found approximate zeroes by using more and more precise x-values. Are you not supposed to use that method here?

Thank you.

Eliz.

 

[Sophie]

Thanks

f(x) must be continouse

and I used the IVT to locate a root that lies between (.643, 0.645)

Showing f(.64) aproximatly = 0.

Thanks

 

[stapel]

f(x) must be continouse

I will guess that you mean "continuous". And I will guess that the above is the only "if" for the IVT, as given in your text, though it is extremely odd that your book says nothing about intervals, endpoints, or the signs of functional values.

But since all you are required to show, in order to apply the IVT, is that the function is continuous, then you would want to show (prove, state, whatever) something about the continuity of f(x).

I used the IVT to locate a root....

As mentioned previously, the IVT only says that a root exists. It does nothing to locate that root. So I will guess that you are referring to something else, perhaps an algorithm in your book.

Unfortunately, without clear communication from you, we are left with only our guesses. Sorry.

Eliz.