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SanjanaP

  • 10 months ago

Help with Increasing and Decreasing Derivatives Please!!

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  1. myininaya
    • 10 months ago
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    what is the question exactly?

  2. myininaya
    • 10 months ago
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    If f'>0, then f is increasing If f'<0, then f is decreasing

  3. SanjanaP
    • 10 months ago
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    Oh...sorry I thought I had the attachment on

  4. SanjanaP
    • 10 months ago
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    I neeed help with C and D

  5. SanjanaP
    • 10 months ago
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    @myininaya

  6. myininaya
    • 10 months ago
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    so are you having problems with the part highlighted in yellow?

  7. SanjanaP
    • 10 months ago
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    Basically...

  8. myininaya
    • 10 months ago
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    ok then so you know that g' will tell us if g is decreasing or increasing right?

  9. SanjanaP
    • 10 months ago
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    Yeah

  10. myininaya
    • 10 months ago
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    so find g' given that: \[g(x)=\int\limits_{0}^{x} f(t) dt \] don't look at the picture yet just tell what g' is

  11. SanjanaP
    • 10 months ago
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    f(x)+0.5

  12. SanjanaP
    • 10 months ago
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    F(x)+c right?

  13. myininaya
    • 10 months ago
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    no i'm sorry that isn't correct ok let's look at this... I will rewrite it a little... \[g(x)=F(x)-F(0)\] where F'=f Now differentiate to find g'

  14. SanjanaP
    • 10 months ago
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    I'm sorry..so what is g'?

  15. myininaya
    • 10 months ago
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    you have to differentiate g(x)=F(x)-F(0) to find g' can you do that?

  16. SanjanaP
    • 10 months ago
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    I'm not sure what differentiate means anymore?

  17. myininaya
    • 10 months ago
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    to find derivative

  18. SanjanaP
    • 10 months ago
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    It's f'(x)-f'(0)

  19. SanjanaP
    • 10 months ago
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    f'(x)(x) right?

  20. myininaya
    • 10 months ago
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    I will help you out some more. derivative of g is g' derivative of F is f (this was given above when I said F'=f) derivative of a constant is 0 everything i said in this little post right here will need to be used

  21. SanjanaP
    • 10 months ago
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    oh....so the derivative of F(x) is f(x). Is that what you mean?

  22. myininaya
    • 10 months ago
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    That is what I said

  23. myininaya
    • 10 months ago
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    so do you know F(0) is a constant?

  24. SanjanaP
    • 10 months ago
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    its 0.5

  25. myininaya
    • 10 months ago
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    I think you are thinking of f(0) not F(0) f is given not F

  26. SanjanaP
    • 10 months ago
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    yea?

  27. myininaya
    • 10 months ago
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    Anyways F(0) is a constant. Because F is a just a function of x any if you plug in a number for x then you will definitely receive a constant ---------------------------------------------------- example: Say F(x)=cos(x) well F(0) is definitely a constant because F(0) is 1 and 1 never changes (it is and will always remain 1)

  28. myininaya
    • 10 months ago
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    So going back to \[g(x)=\int\limits_{0}^{x} f(t) dt \\ g(x)=F(x)-F(0)\] can you differentiate g now?

  29. myininaya
    • 10 months ago
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    what that means is you will have to differentiate both sides (not just one side)

  30. SanjanaP
    • 10 months ago
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    Can you just tell me where it is increasing and concave up...so I'll try to figure it out?

  31. myininaya
    • 10 months ago
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    Try to use what I said earlier... derivative of g is g' derivative of F' is f derivative of a constant is 0 you can do this

  32. SanjanaP
    • 10 months ago
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    I would prefer the answer because I have to write the explanation anyway.

  33. SanjanaP
    • 10 months ago
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    Okay...so you g(x)= F(x)-F(0)

  34. SanjanaP
    • 10 months ago
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    that means g'(x) = f(x)-0?

  35. myininaya
    • 10 months ago
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    I'm not going to give just the answer. Sorry. But right g'=f so that means the picture given is g'

  36. myininaya
    • 10 months ago
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    and you know if g'>0, then g is increasing and you know if g'<0, then g is decreasing

  37. myininaya
    • 10 months ago
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    where on the picture that is given is g' above the x axis (because that is where g is increasing) and when g' is below the x-axis that is where g is decreasing

  38. SanjanaP
    • 10 months ago
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    what about concave up?

  39. SanjanaP
    • 10 months ago
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    |dw:1407794546041:dw|

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