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

$ x^4 + x - 3 = 0 $, $ (1, 2) $

$f(x)=x^{4}+x-3$ is continuous on the interval $[1,2], f(1)=-1,$ and $f(2)=15 .$ since $-1<0<15,$ there is a number $c$in (1,2) such that $f(c)=0$ by the Intermediate Value Theorem. Thus, there is a root of the equation $x^{4}+x-3=0$ in theinterval (1,2)

03:17

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Campbell University

Harvey Mudd College

Baylor University

Boston College

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All right. Um question asks you to use the intermediate value serum to show that there is a route Um of that polynomial in the interval from 1 to 2. Um Okay, so let's just look uh this is defined at the end points even though it's not included in the interval. So let's just look at what the value is. Well, even call it um Let's just call it F of x equals X. The four minus X plus x minus three. Let's make sure we coffee things down right? Um F of one since it is to find that one is one plus one minus three is negative one. And then f of two. Yeah. Is 2 to 4. 16 plus two minus three is positive 15. Okay, so let's just look on a graph where that is, so one comma negative one is there? And to calm a very big positive 15 is there? And uh this function is a polynomial. So it's continuous everywhere. In particular its continuous on that interval. So we've got to get from negative 12 positive 15. As we walk from 1 to 2. And there is no way that we can do that continuously without passing through zero. So the intermediate value theorem says uh there exists an X in the interval such that F of x equals zero. In fact we can we can choose any value between negative one and 15 and there would have to be a value where F of X equals that number. Um but for sure it works for zero

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