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anonymous

  • 4 years ago

Find the equation of the line tangent to the graph of x^3 - 4x^2 + 5x with the least slope. What can I do on this one?

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  1. anonymous
    • 4 years ago
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    have you studied calculus?

  2. anonymous
    • 4 years ago
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    yes. i think i should take the first derivative of this function and minimize it on the interval \[\left( -\infty ,\infty \right)\] i found that at x=4/3, it's slope will be least. i don't know what to do next.

  3. anonymous
    • 4 years ago
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    well the equation of the tangent at x = 4/3 is value of the the function at x=4/3 just plug in this value of into f(x)

  4. anonymous
    • 4 years ago
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    differentiate 3x^2 - 8x + 5 = 0 for critical points (3x -5)(x - 1) = 0 x = 5/3 or x = 1 there are two turning points second derivative = 6x-8 so x = 5/3 gives minimum and x=1 gives maximum

  5. anonymous
    • 4 years ago
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    there are two tangents with slope zero at x=1 and at x=5/3

  6. anonymous
    • 4 years ago
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    ah. but when you get the critical points of the first derivative, you are minimizing the value of the original function and not its slope. am i making sense?

  7. anonymous
    • 4 years ago
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    yes the critical points give you local minimum and local maximum of the original function or it could be a point of inflection. The tangents to the graph at these max and min points will have 0 slope. The question is a bit confusing it asks for the tangent (singular) with least slope but there are two of these. i'll try and draw it

  8. anonymous
    • 4 years ago
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    |dw:1327319828466:dw|

  9. anonymous
    • 4 years ago
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    excuse the bad drawing!

  10. anonymous
    • 4 years ago
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    i think we can get a single tangent line by just minimizing the second first derivative. love the drawing.

  11. anonymous
    • 4 years ago
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    lol

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