Andre W. answered • 10/29/13

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1. f(x)=x

^{5}+ax-1, a>0, has AT LEAST one root, since it is continuous and f(0)=-1<0 and f(1)=a. By the intermediate value theorem, f(x) must assume every value between -1 and a>0, including 0, so it has a root between 0 and 1.f(x) has AT MOST one root, since it is differentiable and f'(x)=5x

^{4}+a, which is always positive, meaning f(x) is strictly monotonic (increasing). Monotonic functions have at most one root.Therefore, f(x) has EXACTLY one root.

2. (a) f(x)= sin

^{2}x + cos x, 0<x<pif'(x) = 2sin x cos x - sin x = sin x (2cos x -1)

Set the derivative equal to zero to find the critical points:

sin x (2cos x -1) =0

sin x≠0 on 0<x<pi, which leaves us with

2cos x -1 =0, cos x = 1/2, x= pi/3

The only critical point in the interval 0<x<pi is at x = pi/3

To determine its nature look at the 2nd derivative:

f''(x) = 2cos² x -2sin² x - cos x

At x=pi/3,

f''(pi/3) = -2sin²(pi/3) <0

By the 2nd derivative test, there is a maximum at x=pi/3.

Its value is f(pi/3) = sin²(pi/3) = 3/4.

I'll leave the rest of your problems to others. :)

Leigh A.

10/30/13