Use the intermediate value theorem to show that each polynomial f...

Question

Use the intermediate value theorem to show that each polynomial function has a real zero between the numbers given. See Example 5. ƒ(x)=2x^4-4x^2+4x-8; 1 and 2

Accepted Answer

Hey, everyone in this problem, we're asked to express that the given function has a real zero between the X values one and three using the intermediate value zero. And the function we're given is F O X is equal to nine X to the exponent four minus three, X squared plus three, X minus 10. We're given four answer choices A through D and they each show the value of the function F at the 40.1 and at the 0.3. And we're gonna come back to those as we work through this problem. The first thing I want to think about with the intermediate value theorem is satisfying the conditions of the theorem. So we have our interval. OK? We have our X values one and three. And so the interval we're interested in is from 1 to 3. Yeah. Now our function F FX is a polynomial. So we know that it is continuous everywhere. OK? So F of X is gonna be continuous on the interval we're interested in from 1 to 3. OK? That closed interval from 1 to 3, this tells us that there exists a value of C such that the minimum oh one, F 53 is less than F F C which is less than the maximum of those two F at one or F at three. OK. And this is sort of tell what the intermediate value theorem tells us. OK. So we get F is a polynomial that tells us it's continuous everywhere. OK. So that means it's gonna be continuous on our close interval from 1 to 3. Since it is continuous on this closed interval. The intermediate value theorem tells us that there is some value of C where the function evaluated that, that value K lies between the minimum and the maximum of the function evaluated at either end point in our inter. OK. So we can see that this depends on the value of F of one and F of three. Same way that in the answer choices, we're looking at the value of F of one and F three. So let's go ahead and calculate each of those. So F evaluated at one, we're gonna take our function, we're gonna substitute in one for X. So we have nine multiplied by one to the exponent four minus three, multiplied by one squared plus three, multiplied by one minus 10. OK? We can simplify this, this is gonna be nine minus three plus three minus 10. Then we get that F one is equal to negative one. OK? So F one is a negative value. And what about F at three? Well, F at three, we're gonna do the same thing. Nine multiplied by three to the exponent four minus three, multiplied by three squared plus three, multiplied by three minus 10. This is gonna be equal to 729 minus 27 course nine minus 10, which gives us an fa three value of 701. Now, the question is asking us how to express that there is a zero, right? And the important part here is that one of our end values in this case, F of one is equal to a negative value. OK? This is less than zero and the other F F three is equal to a positive value. OK? 701 is greater than zero. So one of them is equal to a positive, one of them is equal to a negative value. The intermediate value theorem guarantees that there is going to be some F F C. OK? That is equal to zero. OK. That tells us that there is a zero that again is because one is positive one is negative, that function is continuous between those two points. So it must cross zero at some point. So we're looking at our answer choices, OK? We found that F of one is equal to negative one. So we can eliminate option C and D because it has an equal to positive one and that value is negative. And then as of three, we found was equal to 700. And one which is greater than zero. And so the correct answer here is option A. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with College Algebra?

Use the intermediate value theorem to show that each polynomial function has a real zero between the numbers given. See Example 5. ƒ(x)=2x^4-4x^2+4x-8; 1 and 2

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