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Math Help - fundamental theorem

  1. #1
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    fundamental theorem

    Let f(x)=integral(-1 to x) sqrt(1-t^2) dt for all x E |-1,1|
    (1)Use fundamental theorem of calculus to evaluate F'(x) for all x |-1,1|
    (2)explain why f(x) is invertiable on |-1,1|
    (3)Let G(y) denote inverse of F(x).state domain and range of g(y) and evaluate G(pi/4)
    (4)evaluate G'(pi/4)
    (Hint:use inverse function theorem to express the derivative ofG(y) in terms of derivative of F(x)

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  2. #2
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    Quote Originally Posted by gracy View Post
    Let f(x)=integral(-1 to x) sqrt(1-t^2) dt for all x E |-1,1|
    (1)Use fundamental theorem of calculus to evaluate F'(x) for all x |-1,1|
    Okay you have,
    f(x)=\int_{-1}^x \sqrt{1-t^2}dt
    Thus, (I would like a different lower bound point inside the interval rather than on its endpoint, but anyway)
    f'(x)=\sqrt{1-x^2}
    Thus,
    f'(1)=\sqrt{1-1^2}=0
    (2)explain why f(x) is invertiable on |-1,1|
    The function is continous and increasing. Thus it is a bijective map. Thus, there exists an inverse function on this interval.
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