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March 28, 2017

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consider g(x)= {a sin x + b, if x ≤ 2pi} {x^2 - pi x + 2, if x > 2pi}
A. Find the values of a a b such that g(x) is a differentiable function.
B. Write the equation of the tangent line to g(x) at x = 2pi.
C. Use the tangent line equation from part B to write an approximation for the value of g(6). Do not simplify.

  • calculus - ,

    since each part of g is differentiable, we need the slopes to match up at x=2π, so that dg/dx is defined everywhere.

    dg/dx =
    a cos x for x <= 2π
    2x-π for x > 2π

    so, we need

    a cos(2π) = 2(2π)-π
    a = 3π

    g(x) is thus
    3π sinx + b
    x^2 - πx + 2

    and we need g to be continuous, so needs must

    3π*0 + b = (2π)^2 - π(2π) + 2
    b = 2π^2 + 2

    g(x) =
    3π sinx + 2π^2 + 2
    x^2 - πx + 2

    g(2π) = 2π^2+2
    g'(2π) = 3π

    so we want the line through (2π,2π^2+2) with slope 3π:

    y-(2π^2+2) = 3π(x-2π)

    when x=6,

    y-(2π^2+2) = 3π(6-2π)
    y = 18π-6π^2 + 2π^2+2
    = 18π-4π^2+2 = 19.070

    In fact, g(6) = 3π sin(6) + 2π^2 + 2 = 19.106

  • calculus - ,

    sayyyyyyyyyyyy that

    f(x) = a sinx + b
    h(x) = x^2 - πx + 2

    find h(2pi) and set it equal to f(2pi). also find h'(2pi) and set it equal to f'(2pi). together that should give you enough info to find a and b, and then the rest of the problem tests other calc concepts.

  • calculus - ,

    Part a:
    g(x) = { asinx + b, for x ≤ 2π
    .........{ x² - πx + 2, for x > 2π

    g must be continuous such that
    lim x→2π⁻ g(x) = lim x→2π⁺ g(x)

    Lefthand limit:
    lim x→2π⁻ (asinx + b) = asin(2π) + b = 0 + b = b

    Right hand limit:
    lim x→2π⁺ (x² - πx + 2) =
    (2π)² - π(2π) + 2
    4π² - 2π² + 2
    2π² + 2

    b = 2π² + 2

    Also, for the derivative to exist at all values of x you must also check the limit of the derivative to see that the derivative approaches the same value from both sides:
    ..........{ acosx, for x < 2π
    g'(x) = { ?, for x = 2π
    ..........{ 2x - π, for x > 2π

    lim x→2π⁻ g'(x) =
    lim x→2π⁻ (acosx) = acos(2π) = a

    lim x→2π⁺ g'(x) =
    lim x→2π⁺ (2x - π) = 2(2π) - π = 3π

    a = 3π

    So,
    g(x) = { 3πsinx + 2π² + 2, for x ≤ 2π
    .........{ x² - πx + 2, for x > 2π,

    and g'(2π) = 3π

    Part b:
    g(2π) = b = 2π² + 2
    This gives you the point (2π, 2π² + 2)

    y - (2π² + 2) = 3π(x - 2π)
    y - 2π² - 2 = 2πx - 6π²
    y = 2πx - 4π² + 2

    Part c:
    g(6) ≈ g(2π) + g'(π)(6 - 2π)
    g(6) ≈ (2π² + 2) + 3π(6 - 2π)
    g(6) ≈ 2π² + 2 + 18π - 6π²
    g(6) ≈ -4π² + 18π + 2
    g(6) ≈ 19.07

    Using a calculator, I evaluated g(6) and got 19.11

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