Problem 8 - Essential Calculus

Question

Identify the open intervals on which the function is increasing or decreasing. $$ h(x)=\cos \frac{x}{2}, 0

Step-by-Step Solution

Answer

The function is decreasing on the open interval (0, \(2 \pi\)).

Step 1: Find the derivative

The derivative of h(x) using the chain rule is \(h'(x) = -\frac{1}{2} \sin \frac{x}{2}\).

Step 2: Find critical points

To find the critical points, set \(h'(x)\) equal to zero, \(-\frac{1}{2} \sin \frac{x}{2} = 0\). Solving this gives \(x = 0\) and \(x = 2 \pi\). However those are the end points and not included in the interval.

Step 3: Test intervals

Plug a number between 0 and 2 \(\pi\) into \(h'(x)\) to see if you get a positive or negative number. It's simple to check this with adding the sine function, where we know that \(x = \pi\) is a decreasing function. Therefore, the entire interval from (0, \(2 \pi\)) is decreasing.

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It is one of the fundamental concepts in calculus, akin to the slope of a line for linear functions. To find the derivative of a function like \(h(x) = \cos \frac{x}{2}\), we essentially want to determine how much \(h(x)\) changes as \(x\) changes slightly.
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For our example function, the derivative \(h'(x)\) was found using the chain rule, resulting in \(-\frac{1}{2} \sin \frac{x}{2}\). The negative sign indicates that the function is decreasing on the interval that we're observing. Derivatives are crucial because they help us determine many features of a function, including critical points and intervals where the function increases or decreases.

Chain Rule

The chain rule is a rule in calculus for differentiating compositions of functions. It is vital when dealing with functions that are nested within one another, such as \(\cos \frac{x}{2}\).
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In our scenario, we differentiated \(\cos \frac{x}{2}\) by first considering the outer function (cosine) and the inner function (\(\frac{x}{2}\)).
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Here's how it works:

• Differentiate the outer function, \(\cos(u)\), to get \(-\sin(u)\).
• Differentiate the inner function \(\frac{x}{2}\), which gives us \(\frac{1}{2}\).
• Multiply these results: \(-\sin(u)\times \frac{1}{2} = -\frac{1}{2} \sin \frac{x}{2}\).

This method simplifies the differentiation process, especially when function compositions become complex.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are important because they often indicate where the function could switch from increasing to decreasing or vice versa.
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For \(h'(x) = -\frac{1}{2} \sin \frac{x}{2}\), setting the derivative equal to zero helps find critical points. Solving \(-\frac{1}{2} \sin \frac{x}{2} = 0\) gives critical points at \(x = 0\) and \(x = 2 \pi\).
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While these values are critical in principle, they lie outside our open interval (0, \(2 \pi\)), meaning they don't impact the behavior of \(h(x)\) within its defined range. Still, they are worth noting when sketching or deeply analyzing the function's behavior.

Increasing and Decreasing Functions

Determining where a function is increasing or decreasing is key in understanding its overall behavior. A function is increasing on intervals where its derivative is positive, and decreasing where its derivative is negative.
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For our function \(h(x) = \cos \frac{x}{2}\), with the derivative \(h'(x) = -\frac{1}{2} \sin \frac{x}{2}\), we observe its sign over the interval (0, \(2 \pi\)).
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Since \(-\frac{1}{2} \sin \frac{x}{2}\) remains negative throughout this interval (because \(\sin \frac{x}{2}\) is positive from \(0\) to \(\pi\), turning the entire expression negative), it confirms that \(h(x)\) is decreasing on the entire interval from \(0\) to \(2 \pi\). Understanding increase and decrease helps anticipate and describe graph shapes and function behaviors.