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Given the graph of f below.refkFind all critical numbers ofFind the absolute maxi. and absolute minimum value of on the interval [2,5]...

Question

Given the graph of f below.refkFind all critical numbers ofFind the absolute maxi. and absolute minimum value of on the interval [2,5]

Given the graph of f below. refk Find all critical numbers of Find the absolute maxi. and absolute minimum value of on the interval [2,5]



Answers

Refer to the graph of $y=f(x)$ shown here. Find the absolute minimum and the absolute maximum over the indicated interval. $$[2,5]$$

For this given problem, we want to refer to the graph and find the absolute maximum and minimum over the interval from 2-5. Over the course of this interval, we see that the graph um rises, hits a peak and then comes down. So we see that at two um we have the value of eight and at five we have the value of seven. So we see that that's actually gonna be the lowest point on the graph. So what we're gonna do is we're gonna have our absolute minimum, that's going to be five or seven, and that's when X equals five in the absolute maximum is going to be that local maximum, the peak that we see, that's when X equals three and f of X is going to equal nine. So that's our final answer.

Question here is if a box is equal to seven minus four x on the closed interval from negative 2 to 5. So to find the absolute max and men, well, first, we're gonna go ahead and differentiate our function. So we find the derivatives that would be f prime of acts. Well, we just have a linear function. So the derivative here is just equal to while negative four. So those are derivative. And then, um, to find any critical values we would set are derivative equal to zero. But since our delivers a constant right, negative four is never equal to zero. So therefore, we have no critical values here. So then, well, we just list out any, um, and he couldn't go values and our endpoints. But here we only have our end points. We just check negative two and five. So we do right? Negative two and five. Supposed to do f off and go back to our original function and evaluate our function at the end points and any critical values. But here again, no critical values. So first we're just evaluating at the end points. So we do f off negative, too. Well, that's gonna be equal to just seven minus four times negative too. That's seven plus eight, which is 15. Okay, so f of negative two is 15 and then we evaluate the function at five. So f five that's gonna be equal to seven minus four time 57 minus 20 which is negative. 13. So f of five is equal to negative 13. So therefore, we have a maximum, um, value of 15. So the absolute max right is the Onley, Max. The absolute max would be, um, 15. And that would occur when X is equal to negative to that occurs at X being equal to negative two. And then our minimum is while the Onley minimum fist the absolute minimum. And that would be I'm negative. 13. And that would occur when access equal to the other end point five. So at X equal to five. All right, take care

Okay, so? Well, here we have our function. F of X is equal to negative five. And then we're on the closed interval from negative one toe one. And that's to find the while the absolute maximum it. So if you think about this, I mean the function here f of X is equal to negative five. This is a constant function. This is just a horizontal line, right? So Okay, so we're gonna have a maximum men. Well, how we would go about solving this, Normally, it would take the derivative of our function. So we have our function. F of X is equal to negative five. While the derivative of prime of X, it's just equal to zero, right? The derivative of a constant is just zero. So are derivative is equal to zero. And then for, um, are critical values. We would say we would set the function equal to zero. So where is zero equal to zero? Well, everywhere. Right. So we could say that there are no critical. I mean, really, there are no click of values, or I guess you could say maybe that everything is critical value. Um, maybe since but there are no critical values. Um, So then what we do is just list the endpoints and say, Well, OK, eso over the given interval, right? You just evaluate the function at the two endpoints. But if you have value, if you evaluate ffx at any value, right, if you put the endpoints it if you do f of negative one whatever. Negative one, that's negative. Five. What is f of one? Well, that's five, right? What is what is left of any value on this interval or even not on the interval? Right. The value is I mean, negative five, right? The value is always negative. Five because of a constant function. So therefore, what is the absolute max? Well, negative five, right? The only valid stakes is negative. Five. So the absolute max is negative. Five. Um And what is that? Occur? Or it occurs on any value. So you could say, you know, negative one. Um, less than or equal to x less than or equal to one, right. Any value of X on the clothes interval. The value of the function is going to be negative. Five. So the absolute max isn't gonna find and the absolute men is also negative. Five. Right? Because we have a constant function. So, um so yeah, right. Um, don't over think that one, right. The value of dysfunction is always negative. Five. So therefore, the absolute max. Absolute men is negative. Five. That's the only rally dysfunction takes. All right, take care. Oh!

Now f of X is X to the fifth power minus 25 3rd X cubed plus 20 X minus one. And the interval is from negative three 22 The derivative is five X. The fourth power minus 25 X squared plus 20. And we need to set that equal to zero. Going to divide by five. And that's gonna give me X. To the fourth power minus five. X squared plus four. Equal zero. Now I'm going to let you equals X squared. So it's going to be u squared minus five. You plus four equals zero. Now I can factor it you minus four times U minus one equals zero. Now I'll substitute back in X squared minus four, X squared minus one equals zero. So that means that we've got four critical points negative one negative two, one and two. We also have the endpoints. The endpoints are at negative three and two. So we've got to check the end point as well. Um So I need to check f of negative three F of negative two F of negative one F of one and F of two. Well it'd be easier just to graph this thing. Um And so I think that's what I'm gonna do. Uh So I'm going to dez most and I'm typing in X to the fifth power but it's 25 3rd X to the third power. It was 20 X. Ryan is one now zooming way out here. Okay, from now I can see the values of these points at negative two negative 6.3333 Um At negative one. Negative 13.7 a positive one 11.7 approximately. And at positive two 4.3 approximately. F of negative three. So if I put a negative three in for X, this is going to be the the starting point. Put in a negative three for X. I'm trying to do that right now. Then I get negative 79 so it looks like oops. The minimum is negative 79. The maximum was found at F of one, which was 11.7 approximately.


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