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cuzzin Group Title

Find all critical numbers of f and classify the extreme values given x ∈ [0,12] and f(x)=x^2-22x+7.

  • one year ago
  • one year ago

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  1. cuzzin Group Title
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    f'(x)=2x-22 f'(x)=2(x-11) critical numbers: x=11

    • one year ago
  2. swissgirl Group Title
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    Well firstly you gotta find the derivative of f(x)

    • one year ago
  3. cuzzin Group Title
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    What's my next move here? I'm stuck.

    • one year ago
  4. cuzzin Group Title
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    Right, which is 2(x-11)

    • one year ago
  5. cuzzin Group Title
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    Oh, and problem correction. It is x^2-22x+7

    • one year ago
  6. cuzzin Group Title
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    So the critical numbers are found using the first derivative, right? I get x=11 as the only critical number.

    • one year ago
  7. cuzzin Group Title
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    And 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?

    • one year ago
  8. swissgirl Group Title
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    wait just trying to remember how I use to solve this give me a sec

    • one year ago
  9. swissgirl Group Title
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    Well when f'(x)=0 its either a max or a min correct?

    • one year ago
  10. cuzzin Group Title
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    Yes, I think so. You find the first derivative, set it equal to zero, and then solve to find the critical numbers.

    • one year ago
  11. swissgirl Group Title
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    So 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

    • one year ago
  12. cuzzin Group Title
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    I found, using a slope chart, that 11 is a local min.

    • one year ago
  13. swissgirl Group Title
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    in this case its a parabola so its simple to know the shape|dw:1351565881044:dw|

    • one year ago
  14. swissgirl Group Title
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    Not exactly sure what a slope chart is but ya

    • one year ago
  15. swissgirl Group Title
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    We must take into consideration our endpoints too

    • one year ago
  16. cuzzin Group Title
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    So does this answer choice sound right?: C. Critical no. 11; absolute max f(0) = 7; local and absolute min f(11) = -114.

    • one year ago
  17. swissgirl Group Title
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    Correct

    • one year ago
  18. JakeV8 Group Title
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    These problems get a little more interesting, tricky, whatever, when the critical point falls outside the allowed region of values for x.

    • one year ago
  19. swissgirl Group Title
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    For example @JakeV8 ?

    • one year ago
  20. cuzzin Group Title
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    I 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.

    • one year ago
  21. swissgirl Group Title
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    ok so let me draw a pic to illustrate

    • one year ago
  22. cuzzin Group Title
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    What am I plugging 11 into to get -114?

    • one year ago
  23. JakeV8 Group Title
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    If 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!!

    • one year ago
  24. swissgirl Group Title
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    |dw:1351566181002:dw|

    • one year ago
  25. swissgirl Group Title
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    We are only looking at the graph between 0 and 12

    • one year ago
  26. JakeV8 Group Title
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    yes, I shouldn't have jumped in a minute ago... was offering a hypothetical different problem, but that only confuses things.

    • one year ago
  27. JakeV8 Group Title
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    @cuzzin are you following this? It seemed like you were, then you said you were confused.

    • one year ago
  28. swissgirl Group Title
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    No what he is missing is that we are only looking at the area in within the boundry

    • one year ago
  29. cuzzin Group Title
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    I follow everything except how we arrived at the (0,7) and (11,-114) points.

    • one year ago
  30. swissgirl Group Title
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    We would like to find the max and the min within that boundry

    • one year ago
  31. swissgirl Group Title
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    So 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)^2-22(11)+7=-114 So our global minimum is at (11,-114)

    • one year ago
  32. swissgirl Group Title
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    Now 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)=0^2-22(0)+7=7 f(12)=12^2-22(12)+7=-113 Now as you seee (0,7) is the maximum point

    • one year ago
  33. cuzzin Group Title
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    Ah, ok then. That makes a lot of sense. Thanks for the help, I get it now.

    • one year ago
  34. cuzzin Group Title
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    Thanks for being so in-depth.

    • one year ago
  35. JakeV8 Group Title
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    @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.

    • one year ago
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