Problem 15
Question
Identify the open intervals on which the function is increasing or decreasing.
$$
y=x-2 \cos x, \quad 0
Step-by-Step Solution
Verified Answer
The function \(y=x-2 \cos x\) is increasing on the intervals \(0
1Step 1: Differentiate the function
Find the derivative of the function \( y=x-2 \cos x \) by applying the rules of differentiation. The derivative of \( x \) is 1, and the derivative of \( -2 \cos x \) is \(2 \sin x\). Therefore, the derivative is \( y'=1+2 \sin x \).
2Step 2: Find critical points
Set the derivative equal to zero and solve to find the critical points. So, solve the equation \( 1+2 \sin x = 0 \). This gives \( \sin x = -1/2 \). The solutions to this equation in the interval \( 0 < x < 2\pi \) are \( x=7\pi/6, 11\pi/6 \).
3Step 3: Determine intervals of increase and decrease
Test the intervals between the critical points (i.e., the intervals \(0
Key Concepts
Differential CalculusCritical PointsFunction Analysis
Differential Calculus
Differential calculus is a subfield of calculus concerned with the study of the rates at which quantities change. It is the mathematical way of finding the derivative of a function. In simple terms, when you have a certain function that represents a curve on a graph, differential calculus helps you find the slope of the line tangent to the curve at any point.
When assessing if a function like \(y = x - 2 \cos x\) is increasing or decreasing, we use differential calculus to find the derivative which represents the slope. For instance, the derivative of the aforementioned function is \(y' = 1 + 2 \sin x\). A positive derivative means that the function is climbing upwards as you move from left to right, indicating an increasing interval. Conversely, a negative derivative implies a decreasing interval where the function slopes downwards as you move along the x-axis.
When assessing if a function like \(y = x - 2 \cos x\) is increasing or decreasing, we use differential calculus to find the derivative which represents the slope. For instance, the derivative of the aforementioned function is \(y' = 1 + 2 \sin x\). A positive derivative means that the function is climbing upwards as you move from left to right, indicating an increasing interval. Conversely, a negative derivative implies a decreasing interval where the function slopes downwards as you move along the x-axis.
Critical Points
Critical points are essential in understanding the changes in direction of a graph of a function. They are the points where the function's derivative is zero or undefined. Identifying these points can reveal where the function might achieve local maximums and minimums, or change from increasing to decreasing or vice versa.
To find the critical points of \(y = x - 2 \cos x\), we set the derivative \(y' = 1 + 2 \sin x\) equal to zero. This leads us to \(\sin x = -1/2\), and the solutions within \(0 < x < 2\pi\) result in critical points at \(x=7\pi/6\) and \(x=11\pi/6\). These critical points are where the function's growth rate changes, serving as boundaries to distinguish between increasing and decreasing intervals.
To find the critical points of \(y = x - 2 \cos x\), we set the derivative \(y' = 1 + 2 \sin x\) equal to zero. This leads us to \(\sin x = -1/2\), and the solutions within \(0 < x < 2\pi\) result in critical points at \(x=7\pi/6\) and \(x=11\pi/6\). These critical points are where the function's growth rate changes, serving as boundaries to distinguish between increasing and decreasing intervals.
Function Analysis
Function analysis is a method of studying the properties of functions such as limits, continuity, and differentiability, in addition to their intervals of increase and decrease. During function analysis, we often use test points to determine the nature of the function around its critical points. Test points are selected values from the intervals formed by critical points to examine the behavior of a derivative.
For the function \(y = x - 2 \cos x\), after finding the critical points, we check the sign of the derivative \(y' = 1 + 2 \sin x\) at test points from intervals \(0
For the function \(y = x - 2 \cos x\), after finding the critical points, we check the sign of the derivative \(y' = 1 + 2 \sin x\) at test points from intervals \(0
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