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QUESTION 17f(x) =6x3 2x+4on the interval [0,4] is The absolute minimum for the function400 3-172and submit: Click Save All Ansters save all answers Click Save and S...

Question

QUESTION 17f(x) =6x3 2x+4on the interval [0,4] is The absolute minimum for the function400 3-172and submit: Click Save All Ansters save all answers Click Save and Submit t0 Soue

QUESTION 17 f(x) =6x3 2x+4on the interval [0,4] is The absolute minimum for the function 400 3 -172 and submit: Click Save All Ansters save all answers Click Save and Submit t0 Soue



Answers

In Exercises $17-36,$ locate the absolute extrema of the function on the closed interval.
$$
f(x)=x^{3}-12 x,[0,4]
$$

Hey, it's clear. So when you married here, So we're gonna find our absolute maximum in. So first, we're gonna check the values off the function at the end points. So from zero is equal to zero. An effort for is equal to around 1.348 And this is after we simplify for minus, um, to inverse Tena four. Next, we're gonna find our derivative equal to X graham. Honest one over X square plus one. We're gonna check this, Um, make this equal to zero. So after we do that, we get X is equal to one a negative one. So off of one is equal to one minus to turn in verse of one. This is equal to negative 10.57 We're not gonna check negative one, since it's not in the interval. So you see that we are about absolute Max X equals four and absolute Men X is equal to one

So here we get this function and we need to find the absolute extreme off or f basically so help to obtain the absolute extreme A for some function. Well, given a close interval, what we need to do is first calculate. The critical points and the critical point are obtained by equating the derivative of the function to zero. So let's see what the durability of dysfunction so f prime of X is equal to three quarters which clearly is different from zero. So this implies that there is no critical point. But this is not all we need to ever wait on the endings of the interval. That means taking, evaluating the function at zero and at four. So for that we need to take the values of F at the ending points in this case zero and four. And in case that we got critical points, we're going to adhere the critical points. But we don't have critical points that we just need to evaluate the function on the ending point of the interval. This at the end, is equal to zero, corresponds to to an F evaluated that four is equal to five. So from this These are the only references that we got for this function. So F four corresponds to an absolute maximum. An f evaluated at zero corresponds to an absolute minimum. So this is the direct procedure. However, in some cases you can do this by some graphical analysis of the function. And what happened is that you can observe the dysfunction is more than a straight line. So is in a straight line to find from zero 24 hear from 0 to 4 we, the ending points included on the function and clearly in the middle of the interval there is no maximum or minimum. So the absolute maximum and the absolute meaning for this kind of functions are going to be the ending points.

Well, given this function defined pads for X squared minus one, half X tuned. And we only care about the interval from zero to age. So when it gets time to it, we are going to check these endpoints as well as the critical numbers for the absolute extreme. So what we need to do is find the derivative and we're going to use the power room or we multiply the exponent by the coefficient. Subtract one from the exponents. Yeah, three minutes, one is two and we have to set that equal to zero. So if I were doing this problem, other what I would do is factor out just an X. Keep it simple and you'd have eight minus three halves X. So this ex for a critical number could be zero but it was already one of our endpoints. So we're going to check that anyway. And then a minus three has X equals zero. So I would probably just uh had three has x 20 multiply the two over and then divided by three, you know, to get rid of each piece. Um So my critical numbers would be 16 3rd to check out. So what we need to check for the absolute extreme is first of all er and points 0.5 of eight but we also have to check out f of 16 3rd. Well this first one would be really easy because if I plug in zero for all those values, I'm going to get zero. Um Otherwise I need to go back to the original function, plug in those values for X columns are looking at for X squared minus one, half execute and I need to plug 16 3rd. Um um I don't know if you have a professor that would prefer to see the fraction or the rounded answer, but then you also have to check with eight so confidence when I did that, I got zero. So what I just found out by testing these values, whether you write down the 1024 over 27 or this 37 9 to 6, that's an absolute max. And then these other values because of the smallest between the two of them, that would be your absolute mint. Now the phrase is the absolute min value is zero and then you can get the X. Values and it occurs at at X equals zero or eight. Um but we mostly care about the men value of zero and then max values this one.

It's clear. So when you right here. So here we have our function. So we're gonna first find the function value, um, at the both ends of that interval. So when it's equal to negative, one goes to zero, and when it's two, it is 27. To find the critical values, we're gonna get our derivative and sulfur X by making equal to zero. So this is our derivatives equal to six specs. Times x square minus one square. Make this equal to zero. We're gonna use our zero product will. So we have six X is equal to zero. So we know for our first case, access equal two's Ciro. And for our second case, we have X square minus one square is equal to zero. You have access equal two plus and minus one. Now it's time to calculate our critical volume. So one of his negative X is equal to negative one. We have cereal F 00 up of one is zero and up of two. It was 27. So we see that there's an absolute max at 27 and absolute men at zero


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