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anonymous

  • one year ago

Suppose that f has a continuous second derivative for all x, and that f(0) = 1, f ' (0) = 2, and f''(0) = 0. A. Does f have an inflection point at x = 0? Explain your answer. B. Let g'(x) = (3x^2 + 2)f(x) + (x^3 + 2x + 5)f'(x). The point (0,5) is on the graph of g. Write the equation of the tangent line to g at this point. C. Use your tangent line to approximate g(0. 3). D. Find g''(0).

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  1. anonymous
    • one year ago
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    I feel like A is no because f'(0) would have to = 0 for it to be a critical point? Is that correct?

  2. dan815
    • one year ago
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    no it can be an inflection point even if its not a critical point

  3. dan815
    • one year ago
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    |dw:1435145399302:dw|

  4. dan815
    • one year ago
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    inflection points are where the concavity flips, and critical point is where slope is 0

  5. dan815
    • one year ago
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    and roots are where the function is 0

  6. Michele_Laino
    • one year ago
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    I thin|dw:1435145496095:dw|k that x=0 is an inflection point with inclined tangent, like this:

  7. anonymous
    • one year ago
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    And concavity is usually determined by the second derivative? So what does it mean that this second derivative is always 0?

  8. Michele_Laino
    • one year ago
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    oops.. ...I think that x=0 is an inflection point with inclined tangent, like this...

  9. Michele_Laino
    • one year ago
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    the second derivative is zero only at x=0 so we can not say that there exist a complete neighborhood in which f ''(x) >0 and we can not say that there exists a complete neighborhood in which f'' (x) <0

  10. anonymous
    • one year ago
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    Oh, I totally misread the question! sorry

  11. anonymous
    • one year ago
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    So how should I be testing for it being an inflection point?

  12. Michele_Laino
    • one year ago
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    since f ' (2) is positive, then we can find a complete neighborhood of the point x=2 such that the function f is an increasing function

  13. Michele_Laino
    • one year ago
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    so, being f ''(2)=0 in that complete neighborhood there are points for which f ''(x) >0 and points for which f ''(x) <0

  14. anonymous
    • one year ago
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    Which would make it an inflection point because there are values in either direction at f''(x)? Or am I jumping ahead here?

  15. Michele_Laino
    • one year ago
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    |dw:1435146282570:dw|

  16. anonymous
    • one year ago
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    I see! That makes sense

  17. Michele_Laino
    • one year ago
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    you are right, we have an inflection point since there are values in either direction of f''(x), inside a complete neighborhood of x=2

  18. anonymous
    • one year ago
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    Great. So the derivative of g(x) is really throwing me off in the next part. But I just plug in f(0) and f'(0) and (0,5), right? then use that as a slope for a y-y1=m(x-x1) formula for the tangent line?

  19. Michele_Laino
    • one year ago
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    yes! that's right you should get this:

  20. Michele_Laino
    • one year ago
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    \[y - 5 = 2\left( {x - 5} \right)\]

  21. anonymous
    • one year ago
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    great! So then to find g(.3) using a tangent line approximation, do i plug in .3=x to that tangent line equation?

  22. Michele_Laino
    • one year ago
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    oops...wrong equation. Here is the right equation \[y - 5 = 2x\]

  23. Michele_Laino
    • one year ago
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    we can write this: \[g\left( {0.3} \right) - g\left( 0 \right) \cong g'\left( 0 \right)\left( {0.3 - 0} \right)\]

  24. anonymous
    • one year ago
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    where does the g(0) come from?

  25. Michele_Laino
    • one year ago
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    since by thefinition of first derivative we can write: \[g\left( x \right) - g\left( {{x_0}} \right) = g'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + o\left( {x - {x_0}} \right)\]

  26. Michele_Laino
    • one year ago
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    where: \[o\left( {x - {x_0}} \right)\] is a quantity which goes to zero as x approaches to x_0

  27. anonymous
    • one year ago
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    okay

  28. Michele_Laino
    • one year ago
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    so substituting your data we can write: \[g\left( {0.3} \right) = 5 + 0.6 = 5.6\]

  29. anonymous
    • one year ago
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    Awesome! And then for g''(x)... does this look correct: 6x^2+6x+4 so that g''(0)=4

  30. Michele_Laino
    • one year ago
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    Starting for m your formula for g '(x), after simple computation, I got this:

  31. Michele_Laino
    • one year ago
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    \[g''\left( x \right) = 6xf\left( x \right) + \left( {6{x^2} + 4} \right)f'\left( x \right) + \left( {{x^3} + 2x + 5} \right)f''\left( x \right)\]

  32. Michele_Laino
    • one year ago
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    so, substituting x=0, we can write: \[g''\left( 0 \right) = 4\]

  33. Michele_Laino
    • one year ago
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    oops.. starting from your formula...

  34. anonymous
    • one year ago
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    Thank you so much! This has been a great help!!

  35. Michele_Laino
    • one year ago
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    :)

  36. Michele_Laino
    • one year ago
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    please wait, I have made an error, we have: g' (0) = 2+10=12 so the equation for tangent line is: \[y - 5 = 12x\]

  37. Michele_Laino
    • one year ago
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    sorry for my error of computation

  38. Michele_Laino
    • one year ago
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    I have to be grateful to @Loser66 , who kindly has warned me. Thanks !! @Loser66

  39. anonymous
    • one year ago
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    so g(.3) now is 8.6, yes?

  40. Michele_Laino
    • one year ago
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    yes! we have: \[g\left( {0.3} \right) = 5 + 12 \times 0.3 = 8.6\]

  41. anonymous
    • one year ago
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    Thank you!!

  42. Michele_Laino
    • one year ago
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    :)

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