Intermediate Value Theorem

Intermediate Value Theorem

Postby Guest » Mon Nov 22, 2021 7:12 am

A solution of the polynomial
[tex]f(x)= x^5 - 6x^4 - 12x^3 + 32x^2 + 27x - 42[/tex]
lies between x = 1.7 and x = 1.8. Use the Intermediate Value Theorem to isolate the x value of this solution to within 0.01.

I have no clue where to start this.
Guest
 

Re: Intermediate Value Theorem

Postby Guest » Tue Nov 23, 2021 11:51 am

When I did this before, in more detail than be for I kept being told that there were "too many smilies". I did not think I had put any "smilies" in!
Apparently every time I had an 8 and ) next to each other that was interpretd as a "smilie"! Here is what I had intended with spaces between the 8 and ):

First, a rather picky technical point: a polynomial does not have a "solution"! An equation does. A polynomial, p(x), might have a "zero" which is the solution to the equation, p(x)= 0. I assume you are asking about a solution to the equation [tex]f(x)= x^5- 6x^4- 12x^3+ 32x^2+ 27x- 42= 0[/tex]. Is that correct or is it equal to some number other than 0?

You titled this "Intermediate Value Theorem" and the instructions you copied say "use the intermediate value theorem". Do you know what that is?

The "Intermediate Value Theorem" says that is f is a continuous function, f(a) is negative, and f(b) is positive then there exist x between a and b such that f(x)= 0.

You are told that this polynomial has a zero (not "solution") between 1.7 and 1.8. You can check that:
[tex]f(1.7)= 1.7^5- 6(1.7^4)- 12(1.7^3)+ 32(1.7^2)+ 27(1.7)- 42= 1.50997[/tex]
and
[tex]f(1.8)= 1.8^5- 6(1.8^4)- 12(1.8^3)+ 32(1.8^2)+ 27(1.8 )- 42= -3.79392[/tex] .

Yes, f(1.8 ) is negative and f(1.7) is positive so there exist some x between 1.7 and 1.8 where f(x)= 0. Where between? We have no idea! The "Intermediate Value Theorem" only says it is somewhere in between. It could be anywhere between. So TRY a number! A lot of people, as long as there is no clue where, like to try "half way between". Half way between 1.7 and 1.8 is (1.7+ 1.8 )/2= 3.5/2= 1.75 (no surprize). What is f(1.75)? Is IT 0?

[tex]f(1.75)= 1.75^5- 6(1.75^4)- 12(1.75^3)+ 32(1.75^2)+ 27(1.75)- 42= -0.9228[/tex]
No, that's not 0. But since it is not 0 it must be either positive of negative so the opposite sign to one of the previous values! f(1.75) is negative while f(1.7) was positive. There must be a zero somewhere between 1.7 and 1.75. Where? No idea! Might as well try half way between. Half way between 1.7 and 1.75 is 1.725.

Calculate f(1.725). If it is 0 you are done! If it is positive you know there is a zero between 1.725 and 1.75. Try half way between and continue. If it is negative you know there is a zero between 1.725 and 1.7. Try half way between and continue.

The problem says "isolate the x value of this solution to within 0.01". Since the "exact" solution is always between two consecutive "partial solutions" once you get two consecutive "partial solutions" that are no more than 0.01 apart, either one is sufficiently accurate.
Guest
 

Re: Intermediate Value Theorem

Postby Guest » Wed Nov 24, 2021 9:38 am

I got this,.. thanks.
Guest
 


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