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nickersia Group Title

Tangent at the point of inflection question.

  • 8 months ago
  • 8 months ago

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  1. nickersia Group Title
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    http://prntscr.com/356je6

    • 8 months ago
  2. nickersia Group Title
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    Part d, help anyone? Thanks :)

    • 8 months ago
  3. mathmale Group Title
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    Would you please start the necessary work, and share with us what you have done. also, please explain what it is that you need to know. Let's build upon what you do already know.

    • 8 months ago
  4. nickersia Group Title
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    Everything is on the picture, I've done parts a,b and c, and I don't know how to do part d

    • 8 months ago
  5. mathmale Group Title
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    Your work is very neat. I apologize for not having scrolled down to see what you'd already done. Let me look at Part D. What do you think you need to know to answer Part D correctly?

    • 8 months ago
  6. nickersia Group Title
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    \[y-y _{1}=m(x-x _{1})\] \[y- (\frac{ 2 }{ 4 }-1+\ln \frac{ 2 }{ 4 }) = \frac{ x _{a} -2}{ x _{a} ^{2} }(x-4)\] where Xa is 4 because point B is (4,0.193)

    • 8 months ago
  7. nickersia Group Title
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    y1 (this thing in the bracket) is equal to 0.193 which I got as a coordinate for B, and the slope of the tangent should be equal to derivative of y at point B

    • 8 months ago
  8. mathmale Group Title
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    I'm going to assume that your point of inflection, (4,0.193), is correct. then \[x _{0}=4,y _{0}=0.193\] and you need only substitute these into the point-slope formula for the equation of a straight line. What is your assumed value for the slope, m?

    • 8 months ago
  9. nickersia Group Title
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    \[m = \frac{ x-2 }{ x ^{2} }\] where x is 4, so it gives me 0.125

    • 8 months ago
  10. nickersia Group Title
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    Because \[\frac{ dy }{ dx } = m\]

    • 8 months ago
  11. mathmale Group Title
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    Using your formula for the slope, and calculating the slope that way, I get the same result calculating the slope differently. I, too, get m=0.125, which is the same as 1/8. I'd suggest you substitute these numerical results into the point slope form. Point of tangency is (4,0.193), as before, and slope is either 1/8 or 0.125.

    • 8 months ago
  12. mathmale Group Title
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    \[y-0.193 = (?)*(x - ?)\]

    • 8 months ago
  13. nincompoop Group Title
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    the second derivative must equal zero to be an inflection point

    • 8 months ago
  14. mathmale Group Title
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    @nincompoop, are you saying that the inflection point is not (4,0.193)?

    • 8 months ago
  15. nincompoop Group Title
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    I didn't look at his solution, but that is what to bear in mind when looking for the inflection point.

    • 8 months ago
  16. nincompoop Group Title
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    http://www.math.sc.edu/~diestelr/4.2Notes.pdf

    • 8 months ago
  17. nickersia Group Title
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    I got it!

    • 8 months ago
  18. mathmale Group Title
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    @nincompoop: If you haven't looked at Nick's solution, then commenting as you have is irrelevant and distracting. If you had found a mistake, then I would have wanted to hear from you.

    • 8 months ago
  19. nickersia Group Title
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    \[y - (-\frac{ 1 }{2 }+\ln2) = \frac{ 1 }{ 8 }(x-4)\] . . . \[x-8y+8(\ln2-1)=0\] The problem was that I didn't changed the slope with 1/8 at the beginning so it got more and unnecessary complicated.

    • 8 months ago
  20. mathmale Group Title
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    @nickersia: Congrats! Again, I admire the neatness and precision of your work.

    • 8 months ago
  21. nickersia Group Title
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    @nincompoop yes, I used that in part b) Thank you @mathmale ! :)

    • 8 months ago
  22. mathmale Group Title
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    Thanks for the medal and for becoming a fan of mine! Hope to work with you again soon.

    • 8 months ago
  23. nickersia Group Title
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    You're welcome, me too! Good luck :)

    • 8 months ago
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