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onegirl

  • 2 years ago

Find an equation of the tangent line to y = f(x) at x = a. f(x) = sqrt(x^2 + 16), x = 13

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  1. SithsAndGiggles
    • 2 years ago
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    \[f(x)=\sqrt{x^2+16}\Rightarrow f(13)=\sqrt{185}\\ f'(x)=\frac{2x}{\sqrt{x^2+16}}\Rightarrow f'(13)=\frac{26}{\sqrt{185}}\] The tangent line is given by the point-slope formula using the point (13, √185) and the slope 26/√185: \[y-\sqrt{185}=\frac{26}{\sqrt{185}}(x-13)\]

  2. onegirl
    • 2 years ago
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    so thats the equation? @SithAndGiggles

  3. onegirl
    • 2 years ago
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    @SithsAndGiggles

  4. onegirl
    • 2 years ago
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    @myko

  5. myko
    • 2 years ago
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    the derivative is wrong. The nominator should be just x

  6. onegirl
    • 2 years ago
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    okay u mean in 2x?

  7. myko
    • 2 years ago
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    yes

  8. onegirl
    • 2 years ago
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    ok

  9. myko
    • 2 years ago
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    x/sqrt(x^2+16)

  10. onegirl
    • 2 years ago
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    okay

  11. onegirl
    • 2 years ago
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    so it wouldn't be 26

  12. myko
    • 2 years ago
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    rest looks right, just change 26 for 13 and that's it

  13. onegirl
    • 2 years ago
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    okay thx

  14. myko
    • 2 years ago
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    yw

  15. SithsAndGiggles
    • 2 years ago
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    Thanks for catching the mistake

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