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Prove, without graphing, that the graph of the function has at least two $ x $-intercepts in the specified interval.

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

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Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

04:40

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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this problem Number sixty two of the Stuart Calculus, a eighth edition section two point five prove without graphing that the graph of the function has at least two ex intercepts the specified interval. Why equals X squared minus three plus one over X and intervals from zero to two? Um, we approaches problem as the fact that the ex intercepts for this graph are associated with the roots of the function and our function here is X squared minus three plus one. Rex. So we want to know where this function is equal to zero, um, and the way that we can approach to seize into money value theorem that states that, um, a function continuous function takes all the values between within an interval between the function evaluated at each point in the interval. Um, and we can demonstrate that by evaluating this function for a few choice values. So our first choice value won't be the first term in initial and the interval, the interval eyes from zero to two. But notice that there is a discontinuity at this end point of zero because of this division by there. So instead, we're going to choose a smaller and number close to zero such a zero point one. And this function is equal to approximately seven point. No one had zero point one. We're gonna choose value between zero and two TV with the function just to see what we get. That's just one. In this case, we warn Mars three's NATO to plus one is another one. And then finally we'LL take the Endpoint two and evaluate the function, and we'LL see that will get a bark similarly one point. So what this means is that dysfunction throughout this interval takes on the values seven point one point one NATO, one at one and one point five at two. And it must take all of the values between these numbers, since this function is definitely continuous throughout this interval, except at X equals zero. But everywhere else X squared minus three plus one. Rex is a continuous function on the central, and if that is true, then from going from positive seven to negative one, there is a point in between these two points where the function is equal to zero and that is one root or one x intercept. And between one and two, the values are on eight of one point five and the function also takes on zero. Because again, according to intermediate body serum, the continuous function must take all the values between ah one. Enough of to in this case. So there is another route or another X intercept within this sub interval. So overall, for the interval between zero m two for this function, there is at least two ex intercepts two routes for this function and we have shown that using the intermediate values

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