Finley,HowGan

The problem is "If f(x)=sinx+2x+1 and g is the inverse of f, what is the value of g'(1) (the derivative of g at x=1)?"

I have no clue how to inverse this. What is the solution?

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3 months ago 3

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  1. Captain Matticus, La

    y = sin(x) + 2x + 1
    x = sin(y) + 2y + 1
    x = sin(g(x)) + 2 * g(x) + 1

    Derive implicitly and solve for g'(x)

    1 = g'(x) * cos(g(x)) + 2 * g'(x)
    1 = g'(x) * (2 + cos(g(x))
    g'(x) = 1 / (2 + cos(g(x))
    g'(1) = 1 / (2 + cos(g(1))

    x = sin(g(x)) + 2 * g(x) + 1
    1 = sin(g(1)) + 2 * g(1) + 1
    0 = sin(g(1)) + 2 * g(1)
    -2 * g(1) = sin(g(1))

    This only happens when g(1) = 0

    g'(1) = 1 / (2 + cos(g(1))
    g'(1) = 1 / (2 + cos(0))
    g'(1) = 1 / (2 + 1)
    g'(1) = 1/3

    EDIT:

    Just for fun, let's see how this corresponds to f'(0)?

    f(x) = sin(x) + 2x + 1
    f'(x) = cos(x) + 2
    f'(0) = 1 + 2
    f'(0) = 3

    So, it appears that for inverse functions f(x) and g(x), f'(x) * g'(x) = 1. It'd be worth investigating in order to determine if that holds true all the time.

  2. cidyah

    f(x) = sin x + 2x + 1
    f^-1(x) = g(x)
    Let f^-1(1) = c
    f(c) = 1
    g(1) = c

    sin c + 2c + 1 = 1
    sin c + 2c = 0
    c= 0 is a solution
    g(1) = 0

    g'(1) = 1/ f'( g(1) )
    g'(1) = 1 / f'(0)
    f'(x) = cos x + 2
    f'(0) = cos (0) + 2
    f'(0) = 3

    g'(1) = 1/3

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