# fundamental theorem

• Nov 24th 2006, 06:33 AM
gracy
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)

For diagram please see the attach
• Nov 24th 2006, 06:44 AM
ThePerfectHacker
Quote:

Originally Posted by gracy
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,
$f'(1)=\sqrt{1-1^2}=0$
Quote:

(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.