✔ 最佳答案
Well, you know (a) !
(a) Intermediate Value Theorem. Let f(x) be a function which is continuous on the closed interval [a, b]. Suppose that d is a real number between f(a) and f(b); then there exists c in [a, b] such that f(c) = d.
Rolle's Theorem: Suppose that y = f(x) is continuous at every point of the closed interval[a,b] and differentiable at every point of its interior (a,b).If f(a) = f(b), then there is at least one number c between a and b at which f'(c) = 0.
(b) f(x) = exp(cosx) - 4sin(x^2)
f'(x) = -sinx[exp(cosx)] - 8xcos(x^2) < 0 in [0,√(π/2)]
So, f(x) is strictly decreasing in [0,√(π/2)]
f(0) = e and f[√(π/2)] = exp[cos√(π/2)] - 4 < 0
So, f(x) = 0 have a unique solution in [0,√(π/2)]
(c) f(0) , f (√π), f(√(2π)), .... f(√(mπ)) > 0
f((√π/2)), f((√3π/2)),...f((√(2m + 1)π/2)), < 0
So, there should be at least one root in each interval
[√(mπ) ,√((2m + 1)π/2)] and so there are infinite roots of f(x)
If the no. of roots of f'(x) = 0 is finite. Then f'(x) should be greater than or smaller than zero after two specific value x = a and b. This means that f(x) is a strictly increaing or decreasing function in (-∞,a) and (b,∞) and the no. of roots should be finite which contradict the previous result. So, the no. of roots of f'(x) = 0 is infinite.
(d) f(x)
= exp(cosx) - 4sin(x^2)
< e - 4sin(x^2) since cosx <= 1
< e + 4 since sinx >= -1
2011-04-24 12:46:14 補充：
f(√(mπ)) > 0
Second term is 0 while exp(x) > 0 for any value of x
