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Find the locations of the absolute extrema of the function on the given interval:f(x)20x + 8; [0,6]The absolule minimum occurs atx= The absolute maximum occurs alx=...

Question

Find the locations of the absolute extrema of the function on the given interval:f(x)20x + 8; [0,6]The absolule minimum occurs atx= The absolute maximum occurs alx= (Type exact answers.

Find the locations of the absolute extrema of the function on the given interval: f(x) 20x + 8; [0,6] The absolule minimum occurs atx= The absolute maximum occurs alx= (Type exact answers.



Answers

Find the absolute maximum and minimum values of $f$ on the given closed interval, and state where those values occur. $f(x)=8 x-x^{2} ;[0,6]$

For this problem we've been given a function G of x equals two X squared minus eight X. And we have a closed interval that we are examining from 0 to 6. And our goal is to find all of the absolute extreme values of the function, the absolute minimums and maximums. Well, those kinds of points occur at one of two locations. Either it's going to happen at the end points of the interval. In this case we have to look at both zero and six. They also can occur at critical numbers. Now remember critical numbers happen in one of two places as well. Either or the derivative of our function equals zero or it's undefined. So while we've been given our endpoints, we need to look for critical numbers. So we'll need to go back to our function and find the derivative. Our function is a fairly straightforward polynomial. So we'll need to use the power rule two X square take the derivative, We bring that to down that becomes four X. And then negative. Adex just becomes negative eight. So there is our derivative four X -8. I can put any value of X in there and will always have some value. So it's never going to be undefined. But there is a place where it equals zero. So let's take our derivative Set it equal to zero. And that gives us a value of X equals two. So that's three points. We need to look at our two endpoints of zero and six and are critical point of X equaling two. Now, when we go to actually evaluate these to see which one is our minimum and which ones are absolute maximum. Remember that we're going back to the original function, we use the derivative to find the critical number. But the function that we're given is where we go to actually plug these numbers in to find out which one is our minimum and which ones that are maximum. So we'll plug in our numbers GF- zero. Well, if I let x equal zero, that just becomes zero. X equaling six. Well fat becomes 36 times two. That's going to be 72. Six times eight is 48 which gives us a value of 24. And then if I plug in to, that's going to give me 8 -16 which is going to be -8. So out of these, The biggest one is 24. So this one is going to give us our maximum at x equaling six And the smallest one is the -8. This is our minimum when x equals two. So three critical points to examine those are our minimums and maximums.

The function efforts. It is given that to access Carmona's Tex last six. Okay, so we will get after sex will, because to four x money Soviet and the interval it is given that 02 infinity. Okay, so for the critical number have the sex is considered a hero. So for six months, there will be zero x will be too. So no, we will find the value at critical number and then points. So if our hero will be six, I hope you will be f of two and will be two into four minus of air into two plus six. That is eight. That is minus 16 plus six. That is eight minus 10. That is minus two. So it's up to will be minus two. Okay, about two billion managed to and I have serious six. So we can say that. Uh huh. So we can see that minimum half of two will be cause to manage to This is the answer. Okay, So I hope you understood. Thank you.

Eight x is equal to x a 1/3 from interval for neither one which is all sandwich access disaster equal to eight were also finder outside extreme houses, which is our absolute max forms and absolutely moms. So start off with we can try and find our quick values if there are any. So we have. Let's see, each prime of X is equal to actually write to this exit a 1/3 power. He's a par roll so we have all seen 1/3 extending choosers power The 1/3 accident Mega 2/3 power quote zero So what if I besides I wonder to get us accident Every 2/3 powers you go to zero when we get the X is a domain er first our exit people Soc There's just no excuse. No critical valleys cause domain air So we know that our ups a max of the minimums are on this interval are honors are cycling. So we have Let's see a tornado one. He was three to the negative one, which will get us negative one. Next, we have eight of eight to go to ST A C, which will get us to so we have our absolute long made the one on her absolute maximum to next year. Asta craftiest. So this monograph looks like cr. That's the minimum. Is that no one in our absolute next month? It's a chew, and so this is our answer.

Here we have the function F of X is equal to X plus four over X on the closed and draw from negative eight to negative one. So here first we go ahead and differentiate and find the derivative thought we have prime of X. We differentiate term by term. When we get while you're good over X is just one and driven for over X. Well, thesis is four times extra negative one. So we get negative four times extra negative two which is minus four over X squared. OK, so these are derivative one minus for over X squared and now defend the critical values. We take a negative and we set it equal to zero. So we have one minus four over X squared equal to zero. Well, that would imply that just X squared minus four is equal to zero, which implies that X squared is equal to four means that X is equal to positive or negative to. So we have two critical guy who's here positive or negative too. So then, well, we list out all cripple values and endpoint to, but notice here that positive two is not in our interview, right? So Therefore we can throw out positive to and say the only critical value here is negative two. So then we have our endpoints negative eight. Then we have a critical value of negative two and the other end point of native one. And then we go ahead and evaluate. Are given function effort backs at those values. Right now we have f off negative two that's gonna be native to plus for all the native to right, which is just negative. Two minus two, which is negative for so f of negative two is equal to negative four and then f of negative one. Well, that's equal to negative one plus four. All negative one Texas native one minus four which is negative five. And then f of negative eight f of negative eight Well again, just equal to negative eight plus four over negative eight, which is equal to negative eight minus one half, which is equal to while negative 17 halves so f of negative eight is equal to negative 17 half. So therefore, have they given intervals between negative eight and won. The absolute maximum well is the largest value that we found I evaluating at all the click of values and endpoints so the absolute maximum is going to be negative. Four. And what does that occur? What occurs when X is equal to negative two. So the absolute max is negative, for that occurs at ex being equal to you in negative too. And then for the absolute minimum off. The absolute minimum is the smallest follow we found, which was negative. 17 halves thing. Absolute minimum is negative 17 haves. And that occurs when X is equal to, um, negative eight. Right? So we have absolute minimum here and occurs ex being equal to negative eight. Alright, take what?


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