Please help me with a few questions I don't understand from my homework. Thank you!

Question

1. If a>0, show that the function f(x)=(x^5)+ax-1 has precisely one real root.

 

2. Find the critical points and apply the second derivative test (or state that it fails).

(a) f(x)=(sin^2)x+cosx, 0<x<pi

(b) f(x)=(e^-x)-(4e^-2x)

Answer 1

1. f(x)=x⁵+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)=5x⁴+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)= sin²x + cos x, 0<x<pi

f'(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. :)