Applications of Differentiation

Question

Consider the function f(x)= cos(x) - (1/2)x.  This function has two critical numbers  A<B in [0,2pi]. Give the following:    

A=

B=

f''(A)=

f''(B)=

Answer 1

f^'(x) = -sinx - 1/2 = 0

sinx = - 1/2

x = 7π/6 , 11π/6 ; A = 7π/6 , B = 11π/6

f^"(x) = -cosx

f^"(7π/6) = -√3/2 , f^"(11π/6) = √3/2

Answer 2

f'(x) = –sin(x) – 1/2 = 0 at x = arcsin(–1/2) = –π/6.

Since sin(x) = sin(π – x) for all x, A = 7π/6 is critical too.

Since sin(x) = sin(2π + x) for all x, B = 11π/6 is critical too.

f''(x) = –cos(x) = –√(1 – sin²(x)) for |x – π| > π/2 and √(1 – sin²(x)) for |x – π| < π/2.

f''(A) = √(1 – 1/2²) = √3/2.

f''(B) = –√(1 – 1/2²) = –√3/2.

Graph: https://www.desmos.com/calculator/amhej7txk5