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anonymous
 3 years ago
Find all critical numbers of f and classify the extreme values given x ∈ [0,12] and f(x)=x^222x+7.
anonymous
 3 years ago
Find all critical numbers of f and classify the extreme values given x ∈ [0,12] and f(x)=x^222x+7.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f'(x)=2x22 f'(x)=2(x11) critical numbers: x=11

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well firstly you gotta find the derivative of f(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What's my next move here? I'm stuck.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right, which is 2(x11)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, and problem correction. It is x^222x+7

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So the critical numbers are found using the first derivative, right? I get x=11 as the only critical number.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And the second derivative test gives the local min/max of that point. The second derivative in this case is only 2. Is it better to use a slope chart in this case?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait just trying to remember how I use to solve this give me a sec

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well when f'(x)=0 its either a max or a min correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, I think so. You find the first derivative, set it equal to zero, and then solve to find the critical numbers.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So what I tend to do was just plug in a 11 back into the original equation and you will right away know if its the min or the max

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I found, using a slope chart, that 11 is a local min.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in this case its a parabola so its simple to know the shapedw:1351565881044:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not exactly sure what a slope chart is but ya

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We must take into consideration our endpoints too

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So does this answer choice sound right?: C. Critical no. 11; absolute max f(0) = 7; local and absolute min f(11) = 114.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0These problems get a little more interesting, tricky, whatever, when the critical point falls outside the allowed region of values for x.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For example @JakeV8 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I guess that's where I get stuck then. I get how to find absolute max, min, critical points and all of that. But I don't know how to arrive at the last bit of information.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so let me draw a pic to illustrate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What am I plugging 11 into to get 114?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you restrict x such that it doesn't include x = 11, then you can use the same curve, but there is no critical point in the region. So you end up with max and min points at the two ends of f(x). Wow, a picture would be much easier!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351566181002:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We are only looking at the graph between 0 and 12

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, I shouldn't have jumped in a minute ago... was offering a hypothetical different problem, but that only confuses things.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@cuzzin are you following this? It seemed like you were, then you said you were confused.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No what he is missing is that we are only looking at the area in within the boundry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I follow everything except how we arrived at the (0,7) and (11,114) points.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We would like to find the max and the min within that boundry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So how do we go about this? Well we know that the max or min occur at the point where f'(x)=0 IN this case we found out that the derivative equals 0 when x=11 And then we found out if it was a max or min which it was a min in this case. Now since its a parabola we know that the vertex is the global minimum point. Then we plugged 11 back into our original equation which was f(11)=(11)^222(11)+7=114 So our global minimum is at (11,114)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now we know another thing that the end point usually are a max or min. Imagine if we didnt have a boundry this parabola would just go on for forever and we would have no maximum point. It would just continue on to infinity. When we have a boundry it stop the parabola at some point so we end up having a maximum. So look back at my picture and you can see that the maximum is at the boundry. So what we do is, we plug in 0 and 12 back into our original equation and try to see which one is a maximum f(0)=0^222(0)+7=7 f(12)=12^222(12)+7=113 Now as you seee (0,7) is the maximum point

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, ok then. That makes a lot of sense. Thanks for the help, I get it now.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for being so indepth.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@cuzzin... so one last thought... think if the region for x had been 0 to 2. That region would not include the critical point at x = 11. So within that region, you could calculate the points f(0) and f(2)... these would be the max and min points over the region x = 0 to x = 2, but the region would NOT include the critical point or the global minimum. That wasn't important in this question, but you may see one like that sometime.
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