Equation of the Tangent Line...2

Discussion in 'Calculus' started by nycmathguy, Mar 25, 2022.

  1. nycmathguy

    nycmathguy

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    Find the equation of the tangent line to the curve
    y = x^3 - x + 5 at the point P (1, 5).

    Let me see.

    d/dx [x^3 - x + 5] = 2x^2 - 1.

    Let x = 1

    2(1)^ - 1 = 2 - 1 = 1

    The slope is 1.

    y = mx + b

    Let x = 1, m = 1, and y = 5.

    5 = (1)(1) + b

    5 = 1 + b

    5 - 1 = b

    4 = b

    y = mx + b

    y = (1)x + 4

    y = x + 4

    Yes?
     
    nycmathguy, Mar 25, 2022
    #1
  2. nycmathguy

    MathLover1

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    correct
     
    MathLover1, Mar 25, 2022
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Very good. I think Calculus 1 will be fun. However, related rates is a totally different ball game.
     
    nycmathguy, Mar 26, 2022
    #3
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