WEBVTT
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It's a problem of fifty eight of this tour.
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Calculus eighth edition Section two point five party proved that
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equation has at least one real route. Part B
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. Use your calculator to find and a roll of
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length point on one that contains a route. The
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function A dick question, given his Ellen X minus
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, equals three minus two x And what we do
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is we rearrange the function to have a form of
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effort X equals to zero. So moving the terms
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and the right to the left and we call all
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the terms on the left effects and finding the route
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means meaning meaning, meaning that we find an ex
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value that thanks this function equal to zero. Um
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, we will not exactly calculate this route, but
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we will first prove that it exists in party and
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then never down where it might be in part B
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of a party. Ah, we take a look
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at the function and we see that it's a log
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rhythmic function. Plus a linear function is a constant
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palli. The sum of all these functions since they
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are independently continuous provides us a function that is ah
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, totally continuous on its own domain. Allen of
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X has a certain domain. I've only zero not
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included to infinity s o. The domain that we're
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looking at is strictly X is greater than zero.
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Ah So with that in mind, we know that
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the functions continuous on this X is greater than zero
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domain and will only concern ourselves with that domain.
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So we will take a look at the extreme values
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for the function. First looking at exes, the
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function is an expert zero from the right next and
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do toothy ln function, this is approaching negative.
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Infinity is your princess from the right on this term
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just goes to zero and this is negative three.
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But this goes too negative affinity which are which dominates
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the function. And then as we approach infinity for
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axe the function ah also purchase Infinity is the natural
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argue function and the linear function does go towards infinity
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and so were we have shown is in the dark
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domain that we have of interest from zero not included
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to infinity. The function takes on all the values
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between native into the infinity because we know it's a
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continuous function on this domain and through the Internet about
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this here. It must have taken the value zero
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, which is between these two numbers. And so
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in that case, we could confirm that this function
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does have the route at least one from control zero
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to infinity. Now, in part B, we're
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trying to narrow down this interval. Since it is
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a very large interval, we want to narrow it
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down to an interval. That is length point one
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, you know, first, start off by some
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trial and error. Look, ATT. Some smaller
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numbers taking on values that say of us of one
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Ah, which in our case would be Alan of
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one zero plus two times one just to ministry.
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So this is negative one. And then after of
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two to this inn into a wall of from one
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to two. This will be Ellen of two plus
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two temperatures for minus three. Okay. And if
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we do Eleanor to your calculator, I add form
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. In this three, we get about one point
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sixteen and we see here that between one and two
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, the function changes values from a negative value to
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bother to value. So the function must have taken
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the value zero between this interval from one to was
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a lot more trial and error. We can narrow
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down the interval two very small length of point one
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. And if we take a look at the value
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of the function at one point three four and one
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point three five after much Charl, they're We have
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narrowed down to this small Inderal, where at one
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one point three four on the function takes on the
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value of about native zero point out to seven.
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And for one point two five, it's actually going
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to be a very small number because the value is
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very close to one point three five O. R
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. Zero is very close to one point three five
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on this day that it takes on this point Oh
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, Oh oh One oh five. So we know
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that zero is somewhere between these two because the function
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changes from a negative value to positive value. So
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with a lot of Charles, Harry can use a
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calculator to determine this exact interval, and you could
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even expand further, adding more decimals. Teo narrow
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down exactly what the value of the route is.
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But for the the purpose of this problem, we
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have just found the interval of link point on one
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. That disk contained this route