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Math Help - Find the equation of the tangent line to the curve ..

  1. #1
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    Find the equation of the tangent line to the curve ..

    A) y = e^(-x) @ x = -1

    dy/dx = (-1)(1)(e^(-x))
    = -e^(-x)

    f'(-1) = -0.368
    f(-1) = 0.368

    y =mx+b
    0.368 = (-0.38)(-1) + b
    0=b

    y=-0.368x

    B) y = 2 - e^(-x) @ x=1

    dy/dx = 0 +(-1)(-e^(-x))
    = e^(-x)

    f'(1) = -0.368
    f(1) = 1.632

    y=mx+b
    1.632 = (-0.38)(1)+b
    2 = b

    y= -0.368x + 2

    Is this correct? I feel funny about it.
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  2. #2
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    1. m=-e \Rightarrow y-e=-e(x+1)\Rightarrow y=-ex

    2. m=e^{-1}\Rightarrow y-(2-e^{-1})=e^{-1}(x-1)\Rightarrow y=e^{-1}x+2-2e^{-1}
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  3. #3
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    Quote Originally Posted by advancedfunctions2010 View Post
    A) y = e^(-x) @ x = -1

    dy/dx = (-1)(1)(e^(-x))
    = -e^(-x)

    f'(-1) = -0.368
    f(-1) = 0.368 , no, f(-1)=e, and f'(-1)=-e

    y =mx+b
    0.368 = (-0.38)(-1) + b
    0=b

    y=-0.368x

    B) y = 2 - e^(-x) @ x=1

    dy/dx = 0 +(-1)(-e^(-x))
    = e^(-x)

    f'(1) = -0.368 : This should be e^(-x) for x=1, so it's positive.
    f(1) = 1.632 : This is correct.

    y=mx+b
    1.632 = (-0.38)(1)+b
    2 = b

    y= -0.368x + 2

    Is this correct? I feel funny about it.

    For (A):

    \displaystyle e^{-x} at x=-1 is: \displaystyle e^{-(-1)}=e^1=e\approx 2.718281828

    Otherwise, what you did was right. (You should be able to check these with a graphing calculator.)

    For (B):

    \displaystyle e^{-(1)}=e^{-1}\approx 0.36788

    Again your method was correct.
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