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nickersia

  • 2 years ago

Tangent at the point of inflection question.

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  1. nickersia
    • 2 years ago
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    http://prntscr.com/356je6

  2. nickersia
    • 2 years ago
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    Part d, help anyone? Thanks :)

  3. mathmale
    • 2 years ago
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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.

  4. nickersia
    • 2 years ago
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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

  5. mathmale
    • 2 years ago
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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?

  6. nickersia
    • 2 years ago
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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)

  7. nickersia
    • 2 years ago
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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. mathmale
    • 2 years ago
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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?

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

  10. nickersia
    • 2 years ago
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    Because \[\frac{ dy }{ dx } = m\]

  11. mathmale
    • 2 years ago
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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.

  12. mathmale
    • 2 years ago
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    \[y-0.193 = (?)*(x - ?)\]

  13. nincompoop
    • 2 years ago
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    the second derivative must equal zero to be an inflection point

  14. mathmale
    • 2 years ago
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    @nincompoop, are you saying that the inflection point is not (4,0.193)?

  15. nincompoop
    • 2 years ago
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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.

  16. nincompoop
    • 2 years ago
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    http://www.math.sc.edu/~diestelr/4.2Notes.pdf

  17. nickersia
    • 2 years ago
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    I got it!

  18. mathmale
    • 2 years ago
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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.

  19. nickersia
    • 2 years ago
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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.

  20. mathmale
    • 2 years ago
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    @nickersia: Congrats! Again, I admire the neatness and precision of your work.

  21. nickersia
    • 2 years ago
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    @nincompoop yes, I used that in part b) Thank you @mathmale ! :)

  22. mathmale
    • 2 years ago
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    Thanks for the medal and for becoming a fan of mine! Hope to work with you again soon.

  23. nickersia
    • 2 years ago
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    You're welcome, me too! Good luck :)

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