Given f(x)=sin(x)-2cos(x) on the interval [0,2pi].
a) Determine where the function is increasing and decreasing.
b) Determine where the function is concave up and concave down.
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I know:
f′(x)=cos(x)+2sin(x)
f″(x)=-sin(x)+2cos(x)
Other than that^^ I have no clue how to get the intervals where the function is increasing/decreasing and where it's concave up/concave down.
Go back and review he subject.
The function is increasing where f' > 0
The curve is concave up where f'' > 0.
Now just check 'em out.
To determine where a function is increasing and decreasing, we need to find the critical points and analyze the sign of the derivative.
1. Finding the critical points:
Critical points occur when the derivative equals zero or is undefined. In this case, we need to solve for the values of x where f'(x) = 0.
f'(x) = cos(x) + 2sin(x) = 0
To solve this equation, rearrange it as:
sin(x) = -cos(x) / 2
Now, use the identity sin^2(x) + cos^2(x) = 1 to get:
sin^2(x) + (cos^2(x) / 4) = 1
4sin^2(x) + cos^2(x) = 4
Substituting sin^2(x) = 1 - cos^2(x), we get:
4 - 3cos^2(x) = 4
3cos^2(x) = 0
cos^2(x) = 0
Solving for cos(x), we find two critical points:
cos(x) = 0 --> x = π/2, 3π/2
2. Analyzing the sign of the derivative:
To determine where the function is increasing or decreasing, we need to analyze the sign of the derivative in different intervals.
Interval 1: (0, π/2)
Pick a value x = 0 within this interval and substitute it into the derivative:
f'(0) = cos(0) + 2sin(0) = 1
Since f'(0) > 0, the function is increasing in this interval.
Interval 2: (π/2, 3π/2)
Pick a value x = π within this interval and substitute it into the derivative:
f'(π) = cos(π) + 2sin(π) = -1
Since f'(π) < 0, the function is decreasing in this interval.
Interval 3: (3π/2, 2π)
Pick a value x = 2π within this interval and substitute it into the derivative:
f'(2π) = cos(2π) + 2sin(2π) = 1
Since f'(2π) > 0, the function is increasing in this interval.
Therefore, the function is increasing on the intervals (0, π/2) and (3π/2, 2π), and it is decreasing on the interval (π/2, 3π/2).
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To determine where the function is concave up and concave down, we need to find the inflection points by analyzing the second derivative.
1. Finding the inflection points:
Inflection points occur when the second derivative changes sign or is undefined. In this case, we need to solve for the values of x where f''(x) = 0.
f''(x) = -sin(x) + 2cos(x) = 0
Similar to the previous step, we can rearrange the equation to:
sin(x) = 2cos(x)
Using the identity sin^2(x) + cos^2(x) = 1, we can write:
sin^2(x) + (2sin^2(x)) = 1
3sin^2(x) = 1
sin^2(x) = 1/3
Solving for sin(x), we find two inflection points:
sin(x) = ±√(1/3)
x = arcsin(±√(1/3))
2. Analyzing the concavity using the second derivative:
To determine the concavity of the function, we need to analyze the sign of the second derivative in different intervals.
Interval 1: (0, arcsin(√(1/3)))
Pick a value x = 0 within this interval and substitute it into the second derivative:
f''(0) = -sin(0) + 2cos(0) = 2
Since f''(0) > 0, the function is concave up in this interval.
Interval 2: (arcsin(√(1/3)), arcsin(-√(1/3)))
Pick a value x = arcsin(√(1/3)) within this interval and substitute it into the second derivative:
f''(arcsin(√(1/3))) = -√(1/3) + 2√(2/3)
Since f''(arcsin(√(1/3))) < 0, the function is concave down in this interval.
Interval 3: (arcsin(-√(1/3)), 2π)
Pick a value x = 2π within this interval and substitute it into the second derivative:
f''(2π) = -sin(2π) + 2cos(2π) = 2
Since f''(2π) > 0, the function is concave up in this interval.
Therefore, the function is concave up on the intervals (0, arcsin(√(1/3))) and (arcsin(-√(1/3)), 2π), and it is concave down on the interval (arcsin(√(1/3)), arcsin(-√(1/3))).