For which of the following intervals is the function f(x)=cos(πx)+√3sin(πx) decreasing throughout the interval? (1 point) Responses (0,1) ( 0 , 1 ) (1/3,4/3) 1 3 , 4 3 (1/2,3/2) 1 2 , 3 2 (2/3,5/3) 2 3 , 5 3 (1,2) ( 1 , 2 )
To determine where the function is decreasing, we need to find the intervals where the derivative of the function is negative. Taking the derivative of f(x), we have:
f'(x) = -πsin(πx) + π√3cos(πx)
To find the intervals where f(x) is decreasing, we need to solve the inequality f'(x) < 0.
Let's look at each option and determine if it satisfies the inequality:
(A) (0,1):
f'(x) = -πsin(πx) + π√3cos(πx)
At x=0.5, f'(0.5) = -πsin(0.5π) + π√3cos(0.5π) = -π(1) + π√3(0) = -π < 0
Since the derivative is negative for the entire interval (0,1), the function is decreasing throughout this interval.
(B) (1/3,4/3):
f'(x) = -πsin(πx) + π√3cos(πx)
At x=1, f'(1) = -πsin(π) + π√3cos(π) = 0 + π√3(-1) = -π√3 < 0
Since the derivative is negative at x=1, the function is decreasing to the left of 1. However, at x=4/3, f'(4/3) = -πsin(4/3π) + π√3cos(4/3π) = -π(√3/2) + π(1/2) = -π(1/2 - √3/2) = π(√3/2 - 1/2) > 0. The derivative is positive at x=4/3, so the function is not decreasing throughout (1/3,4/3).
(C) (1/2,3/2):
f'(x) = -πsin(πx) + π√3cos(πx)
At x=1/2, f'(1/2) = -πsin(0.5π) + π√3cos(0.5π) = -π(√3/2) + π(0) = -π√3/2 < 0
Since the derivative is negative at x=1/2, the function is decreasing to the left of 1/2. However, at x=3/2, f'(3/2) = -πsin(1.5π) + π√3cos(1.5π) = -π(-√3/2) + π(0) = π√3/2 > 0. The derivative is positive at x=3/2, so the function is not decreasing throughout (1/2,3/2).
(D) (2/3,5/3):
f'(x) = -πsin(πx) + π√3cos(πx)
At x=2/3, f'(2/3) = -πsin(2/3π) + π√3cos(2/3π) = -π(√3/2) + π(1/2) = -π(1/2 - √3/2) = π(√3/2 - 1/2) > 0
Since the derivative is positive at x=2/3, the function is not decreasing to the left of 2/3.
(E) (1,2):
f'(x) = -πsin(πx) + π√3cos(πx)
At x=1, f'(1) = -πsin(π) + π√3cos(π) = 0 + π√3(-1) = -π√3 < 0
Since the derivative is negative at x=1, the function is decreasing to the left of 1. However, at x=2, f'(2) = -πsin(2π) + π√3cos(2π) = 0 + π√3(1) = π√3 > 0. The derivative is positive at x=2, so the function is not decreasing throughout (1,2).
The correct interval where the function f(x) = cos(πx) + √3sin(πx) is decreasing throughout is (0,1) or option A.