💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Prove, without graphing, that the graph of the function has at least two $x$-intercepts in the specified interval. $y = x^2 - 3 + 1/x$, $(0, 2)$

Check back soon!

Limits

Derivatives

### Discussion

You must be signed in to discuss.
##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp

### Video Transcript

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