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IsTim

Find the equation of the tangent to the curve at the given x value. y=sqrt(x^2+3),x=1

  • one year ago
  • one year ago

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  1. IsTim
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    |dw:1335241905273:dw| That was my final answer.

    • one year ago
  2. IsTim
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    The correct answer os x-2y+3=0

    • one year ago
  3. satellite73
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    find the derivative, replace x by 1 to get your slope, then use point slope formula

    • one year ago
  4. IsTim
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    I tried to, but I reached a point where my denominator equals 0.

    • one year ago
  5. IsTim
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    unless I can factor "h" out of my answer?

    • one year ago
  6. satellite73
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    \[f(x)=\sqrt{x^2+3}\] \[f'(x)=\frac{x}{\sqrt{x^2+3}}\] \[f'(1)=\frac{1}{2}\] is your slope, and your point is \((1,f(1))\) i.e. \((1,2)\)

    • one year ago
  7. IsTim
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    We're not using f′(x) yet.

    • one year ago
  8. satellite73
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    really? then what are you supposed to do ?

    • one year ago
  9. IsTim
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    Well, we're either using substitution, factoring, or the conjugate method.

    • one year ago
  10. IsTim
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    Oops, I think I forgot to properly do the Conjugate Method (Working off the Calculus Phobe Videos at Calculis Help)

    • one year ago
  11. satellite73
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    oh ok, in that case start with \[\frac{\sqrt{1+h+3}-2}{h}\] or rather \[\frac{\sqrt{4+h}-2}{h}\] and multiply top and bottom by conjugate

    • one year ago
  12. satellite73
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    they still have calculus phobe videos? i thought they were gone

    • one year ago
  13. IsTim
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    My teacher showed me in class.

    • one year ago
  14. satellite73
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    ok that was wrong, i forgot about the square start with \[\frac{\sqrt{(1+h)^2+3}-2}{h}\]

    • one year ago
  15. satellite73
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    or with \[\frac{\sqrt{1+2h+h^2+3}-2}{h}=\frac{\sqrt{4+2h+h^2}-2}{h}\] and then multiply by the conjugate

    • one year ago
  16. IsTim
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    @satellite73 Umm, my final answer is 1/2x+1.5, which isn't the correct answer...

    • one year ago
  17. satellite73
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    well slope is \[\frac{1}{2}\] and the point is \((1,2)\) so lets see \[y-2=\frac{1}{2}(x-1)\] \[y=\frac{1}{2}x+\frac{3}{2}\] looks good to me

    • one year ago
  18. IsTim
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    They're saying its x-2y+3=0

    • one year ago
  19. satellite73
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    mutiply by 2 and put everything on one side of the equal sign and you get the same thing

    • one year ago
  20. IsTim
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    Just wondering, how would I know to multiply by 2 on a test?

    • one year ago
  21. IsTim
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    @satellite73 Just wondering, how would I know to multiply by 2 on a test?

    • one year ago
  22. satellite73
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    no idea. i guess it says "standard form" or something? it matters not

    • one year ago
  23. IsTim
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    So I guess the teacher wouldn't deduct me for marks on a test for something like that right?

    • one year ago
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