Calculus

If the local linear approximation of f(x)=4x+e^2x at x=1 is used to find the approximation for f(1.1), then the % error of this approximation is
Between 0% and 4%
Between 5% and 10%
Between 11% and 15%
Greater than 15%

  1. 👍 0
  2. 👎 0
  3. 👁 955
  1. df/dx = f'(x) = 4+ 2x e^2x

    f(1) = 4 + e^2 = 4 + 7.389 = 11.389
    f'(1) = 4 + 2 e^2 = 4 + 14.778 = 18.778
    f(1.1) = f(1) + 0.1 f'(1)
    = 11.389 + 1.878 = 13.27

    real = 4.4 + 9.02 = 13.42

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  2. f(x)=4x+e^2x
    f ' (x) = 4 + 2e^(2x)
    f ' (1) = 4 + 2e^2
    f(1) = 4 + e^2

    linear equation: y - (4 + e^2) = (4+2e^2)(x-1)
    y - 4 - e^2 = 4x + (2e^2)x - 4 - 2e^2
    y = 4x + 2e^2 x - e^2 OR y ≂ 18.778x - 7.389

    when x = 1.1, y ≂ 18.778(1.1) - 7.389 = appr 13.267
    from original function:
    f(1.1) = 4(1.1) + 2e^(2.2) = appr 13.65

    see what you can do with that.

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    2. 👎 1
  3. go with Damon's

    I messed up in my 2nd last line
    should have been:
    f(1.1) = 4(1.1) + e^(2.2) = appr 13.425

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    2. 👎 1

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