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# Find the absolute maximum and absolute minimum values of $f$ on the given interval.$f(x) = \frac{x}{x^2 - x + 1}$, $[0, 3]$

## Absolute minimum value is 0 which occurs at $x=0$Absolute maximum value is 1 which occurs at $x=1$

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All right, let's go ahead and answer this question. We have the function. Why equals two X over X squared minus X plus one X is defined to be between zero and three inclusive. And it's asking you what's the what are the absolute extreme. Okay, so whatever you solve these kind of problems, there are two things you need to keep in mind. First, always check the endpoints. Okay, because the extreme value theory, um basically says that if there if a function has an extreme value, it will always happen at the end point or somewhere in between. Okay, So if the end points are included in the in the analysis, we always check, So we're going to plug in zero and three to figure out whether it's going to be an extreme. Okay, The second thing you want to keep in mind is to check the critical points. So we're going to look at the derivative, let it equal to zero. And if you want to be very precise, we do want to check the second derivative to see if it's going to be a local maximum or local minima or by observing the first derivative and check if it's actually going to change the signs. But I am going to admit, omit that detailed process of that. Okay, so I'm going to assume that when the derivative is equal to zero, it actually is a critical point rather than a saddle. Okay, so let's go through the calculation here. First, I am going to call this F and plug in. Zero f of zero is very simple. The numerator is zero. The denominator is not zero. So it'll end up being zero next f off three. A quick calculation tells you that the top is three. Bottom is nine minus 366 plus one is seven. So it's 37 So at least I know that zero is not an absolute maximum seven third probably. Okay, Now let's check F prime. This is a question so we can use the quotient rule. So it is going to be ex prime times the denominator. Of course, Ex prime is one. So you will have X squared minus X plus one minus the denominator prime. So it is going to be two x minus one times the numerator, which is X now the denominator. Of course you're gonna have to square it, so I'm just gonna But I'm just gonna call it triangle. And the reason is I know that I'm gonna let this equal to zero. And if you observe that the denominator is an irreducible quadratic meaning that the B squared minus four A. C is a positive number, uh, negative number. The denominator is never going to be negative. So I don't have to worry about the case where there's a division by zero or anything like that's all. I'm gonna omit it because the numerator simplifies to one minus X squared. And if I multiply the denominator on both sides, it disappears anyway, so I don't have to worry about that. Okay, so you can see that X is plus or minus one. But one is the only number that belongs in this limit. So we know that one is the only point that we need to check. Let's figure out what effort of one is equal to the gnome writers one and the denominator X squared minus one is simply zero. So the remaining one makes this equal to one. And one is, of course, larger than 3/7. Okay, so from this analysis, we figured out that this guy, it is going to be the local minimum or the absolute minimum. Excuse me. The bottom one here is going to be a an absolute maximum. And 7 30 is just a point at the end point. If you use a graphing utility, you will actually see that this is a type of graph that looks something like this and 0 to 3 is actually a region from here. So there. And this is going to be one. This is a zero, and this height right here is the seven third that we actually observed. And this is how I answer this problem.

University of California, Berkeley

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