Cal 1 Proofs?

Question

Hmm i'm kinda stuck on some homework proofs and wonder if anyone can help me?

Using the applications of Extreme Value Theorem, Intermediate Value Theorem, Mean Value Theorem, & Rolle's Theorem detemine the following questions.

1) Suppose f, g are continuous on [a,b], f(a) < g(a), f(b) > g(b). Prove f(x) = g(x) for some x E (a,b)

2) Suppose f is continuous on [0,1], 0 ≤ f(x) ≤ 1 for x E [0,1]. Prove f(x) = x for some x E [0,1]

3) a) For f as in problem 2, prove f(x) = -x for some x E [0,1]

b) If g is continuous on [0,1], g(0) = 0, g(1) = 1. Prove f(x) = g(x) for some x E [0,1]

Answer 1

#1: Let h(x) = f(x) - g(x). Then h is a continuous function such that h(a) < 0 and h(b) > 0, so by the intermediate value theorem there exists a c in the interval (a, b) such that h(c)=0. But h(c) = 0 means f(c) = g(c), so there is a point in (a, b) where f(x) = g(x).

#3b (sic): f and g are continuous on (0, 1), f(0) ≥ g(0) = 0, f(1) ≤ g(1) = 1. If we have equality in either case then we are done. If not, then this meets the conditions of the functions in problem 1, so we may apply that result.

#2: Let g(x) = x. Apply the result of 3b.

#3a: This one must have a typo. f(x) is nonnegative over [0, 1], so the only way f(x) could equal -x is if f(0) = 0, and there's nothing guaranteeing that it does. Perhaps you mean f(x) = 1-x. In that case f(0) ≤ g(0) = 1, f(1) ≥ g(1) = 0, and so if h(x) = f(x) - g(x), h(0) ≤ 0, h(1) ≥ 0. If we have equality in either case then we are done. Else, we may invoke the intermediate value theorem much as we did in problem 1.

Answer 2

There are a couple of ways to do it I think you want d/dx(cos x) let d/dx(cos x) = t we know sin ^ 2 x + cos ^2 x = 1 differentiate both sides 2sinx d/dx(sinx) + (2 cos x ) t = 0 or 2 sinx cos x + 2t cos x =0 so t = - 2sinx cosx /(2cosx) = - sin x 2nd method let d/dx(cos x) = t we know exp(ix) = cos x + i sin x differentiate both sides i exp(ix) = t + i cos x devide both sides by i exp(ix) = -it + cos x = cos x -it equating imaginary part t = -sin x