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MTALHAHASSAN2

  • one year ago

find the point on the curve f (x)=(x^2-1)^2 for which the tangent line is horizontal

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  1. IrishBoy123
    • one year ago
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    \(f = (x^2-1)^2\) you learning calculus?

  2. MTALHAHASSAN2
    • one year ago
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    yes

  3. IrishBoy123
    • one year ago
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    so you know to differentiate f(x) and set \(f'(x) = 0\) ??

  4. MTALHAHASSAN2
    • one year ago
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    yeah i do

  5. MTALHAHASSAN2
    • one year ago
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    wait let be show you then

  6. IrishBoy123
    • one year ago
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    cool!

  7. MTALHAHASSAN2
    • one year ago
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    |dw:1443871570969:dw|

  8. MTALHAHASSAN2
    • one year ago
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    right?

  9. IrishBoy123
    • one year ago
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    nah \(f = (x^2-1)^2\) \(f' = 2(x^2-1)(2x)=\dots\) if that is confusing, let \(u = x^2 -1 \) and use \(\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\)

  10. MTALHAHASSAN2
    • one year ago
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    then is it be 4x(x^2-1)

  11. IrishBoy123
    • one year ago
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    yes let's do it long way too \(f = (x^2-1)^2 = x^4 - 2 x^2 + 1\) \(f' = 4x^3 - 4x = 4x(x^2-1)\) \(\checkmark\)

  12. IrishBoy123
    • one year ago
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    |dw:1443861295810:dw|

  13. MTALHAHASSAN2
    • one year ago
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    oh ok got you

  14. MTALHAHASSAN2
    • one year ago
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    thnx a lot so the 4x(x^2-1) be the tangent line is

  15. IrishBoy123
    • one year ago
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    \(f' = 0 \implies 4x(x^2-1) = 0\) solve for x, you will get 3 turning points

  16. IrishBoy123
    • one year ago
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    |dw:1443862179850:dw|

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