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MarcLeclair

  • 3 years ago

Find the point on the graph of √(x+1) that lies closest to point ( 4,0).

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  1. galanh
    • 3 years ago
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    The distance between a point on the curve and the point (0,4) is: \[d= \sqrt{(x-4)**2+x+1}=\sqrt{x**2 -7x+17}\] \[d=\sqrt{(x-3.5)**2 +4.75}\] d is minimum when x=3.5 and y=sqrt(3.5+1)=sqrt(4.5)=2.12

  2. MarcLeclair
    • 3 years ago
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    Is that the formula for the distance of a curve to a point, the one they teach in grade 10 or something?

  3. MarcLeclair
    • 3 years ago
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    we dont want the distance, we want the point closest, and the answer is wrong :/

  4. galanh
    • 3 years ago
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    the closed point is (3.5,2.12)

  5. MarcLeclair
    • 3 years ago
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    wait... is that a basic equation? it's for my calculus class and I didn't see you use any derivative or anything

  6. MarcLeclair
    • 3 years ago
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    @galanh its odd my solution sheet has a different way, I'm just trying to understand the method sorry :/

  7. galanh
    • 3 years ago
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    I used this graph

  8. MarcLeclair
    • 3 years ago
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  9. MarcLeclair
    • 3 years ago
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    thats the solution using derivative, but I don't understand why we would set the first derivative to 0 to find the critical points, just seems odd to me

  10. galanh
    • 3 years ago
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    Thanks for the solution with derivative.

  11. MarcLeclair
    • 3 years ago
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    do you understand that solution? I don't understand it that was why i asked the question in the first place

  12. galanh
    • 3 years ago
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    I am not good with derivatives! derivative of a function equal to zero, gives you the minimum of the function!

  13. MarcLeclair
    • 3 years ago
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    Errr thanks for the help :) i'll try to figure out the rest on my own haha

  14. galanh
    • 3 years ago
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    My pleasure!

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