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anonymous

  • one year ago

Help would be much appreciated ^.^ -Find the absolute and local maximum and minimum values of f(t) = 2/t + 10 , 0 < t ≤ 4

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  1. myininaya
    • one year ago
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    Well we know we need to find f'(t) is f(t)=2/t+10 or f(t)=2/(t+10)?

  2. anonymous
    • one year ago
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    the 1st one :)

  3. myininaya
    • one year ago
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    just so i can know to check your derivative

  4. myininaya
    • one year ago
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    ok cool

  5. myininaya
    • one year ago
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    \[f(t)=\frac{2}{t}+10 \\ f(t)=2t^{-1}+10 \\\] do you know how to find f'?

  6. anonymous
    • one year ago
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    Yes! It's \[f'(t)= \frac{ -2 }{ t^2 }\] am I right?

  7. myininaya
    • one year ago
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    yes and f' is not zero ever but f' dne at t=0 but we f(0) also doesn't exist at t=0 we actually have a vertical asymptote |dw:1434748364786:dw| so we don't have to worry about critical numbers but this just shows we also don't have any max of any type but w do have a min and it is an absolute min (you know since local min can't occur at endpoints)

  8. myininaya
    • one year ago
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    so you just need to consider the endpoints of your interval given

  9. myininaya
    • one year ago
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    tell me if you can find the absolute min?

  10. anonymous
    • one year ago
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    Ohhh I understand. How can I find the absolute max. and min. ?

  11. myininaya
    • one year ago
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    there is a vertical aysmptote in the pic and as x approaches it from the right the y values keep getting bigger and bigger there is no max you only need to find the absolute min and I gave you a big hint for that it occurs at the endpoint for this question

  12. myininaya
    • one year ago
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    |dw:1434748634077:dw|

  13. anonymous
    • one year ago
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    Oh yeah! So I plug in 4 into the original function and that will be my abs. min. ?

  14. myininaya
    • one year ago
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    yes

  15. anonymous
    • one year ago
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    So that means it is 10.5 ?

  16. myininaya
    • one year ago
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    yes

  17. myininaya
    • one year ago
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    no calculus was actually needed on this problem

  18. anonymous
    • one year ago
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    If I did not known what the graph looked like would I still be able to find it?

  19. myininaya
    • one year ago
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    Yes.

  20. myininaya
    • one year ago
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    I said no calculus was needed because g(x)=1/x is a pretty basic graph learned in algebra. f(x)=2/x+10 is just a translation of that graph. Anyways f(x)=2/x+10 , 0<x<=4 We already know this as a vertical asymptote at x=0 And we know in the interval (0,4] the y values are approaching positive infinity since as the x's move closer to 0 since \[\lim_{x \rightarrow 0^+}(\frac{2}{x}+10)=\infty \] Means there is no max at all. You could find f' to see if there are any critical numbers on (0,4) but we did that and there are none. So f will have no local min. But you can also say since \[f'(t)=\frac{-2}{t^2}<0 \text{ means } f \text{ is decreasing }\] You have definitely ensure f(4) is a absolute min

  21. anonymous
    • one year ago
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    @myininaya thank you so much!!!

  22. myininaya
    • one year ago
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    Means there is no abs max at all.* Since there were no critical numbers then there are no relative max/min

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