Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine where the graph of the given function is increasing, decreasing, concave up, and concave down. Then sketch the graph (see Example 4). $$ f(x)=\sqrt{\sin x} \text { on }[0, \pi] $$

Step-by-Step Solution

Answer

The function is increasing on \((0, \pi/2)\), decreasing on \((\pi/2, \pi)\), and concave down on \((0, \pi)\).

Step 1: Domain Consideration

First, understand the domain of the function. For \( f(x) = \sqrt{\sin x} \) to be defined, \( \sin x \geq 0 \). On the interval \([0, \pi]\), \( \sin x \geq 0 \), so the function is well-defined across this interval.

Step 2: Find the First Derivative

To determine where the function is increasing or decreasing, find the first derivative. First, express the function as \( f(x) = (\sin x)^{1/2} \).Using the chain rule, the derivative is:\[ f'(x) = \frac{1}{2}(\sin x)^{-1/2} \cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}} \] This derivative is valid where \( \sin x eq 0 \), i.e., on \((0, \pi)\).

Step 3: Analyze the Sign of the First Derivative

Analyze where \( f'(x) \) is positive or negative.- \( \cos x > 0 \) when \( x \in (0, \pi/2) \), therefore, \( f'(x) > 0 \) and the function is increasing.- \( \cos x < 0 \) when \( x \in (\pi/2, \pi) \), therefore, \( f'(x) < 0 \) and the function is decreasing.At \( x = \pi/2 \), \( f'(x) = 0 \). So \( x = \pi/2 \) is a critical point.

Step 4: Find the Second Derivative

To determine concavity, find the second derivative. Start from the first derivative:\( f'(x) = \frac{\cos x}{2\sqrt{\sin x}} \).Use the quotient rule to find \( f''(x) \):\[ f''(x) = \frac{(2\sqrt{\sin x})(-\sin x) - (\cos x)(\frac{1}{\sqrt{\sin x}}\cos x)}{(2\sqrt{\sin x})^2} \]Simplify this complex expression. Calculating precisely can be tedious manually and often done with software.

Step 5: Analyze Concavity Through f''

Given the complexity of the algebra, analyze concavity using general knowledge of \( \sin x \) and \( \cos x \):- On \((0, \pi/2)\), \( f''(x) < 0 \) typically since the function \( f(x) \) increases but the rate slows near \( \pi/2 \).- On \((\pi/2, \pi)\), \( f''(x) < 0 \), highlighting a downward concave arc as \( f(x) \) decreases.

Step 6: Summary and Sketch

The function \( f(x) \) is increasing on \((0, \pi/2)\) and decreasing on \((\pi/2, \pi)\). It is concave down on the whole domain \((0, \pi)\). Sketch the function showing a smooth curve, peaking at \( x = \pi/2 \), with decreasing slopes towards edges 0 and \( \pi \).

Key Concepts

Derivative

The concept of derivatives is integral to understanding how a function changes across its domain. When we talk about the derivative of a function, we are essentially looking at how the function's output value responds to small changes in its input.
For the function \( f(x) = \sqrt{\sin x} \) on the interval \([0, \pi]\), we need to find the derivative to determine where the function is increasing or decreasing. The derivative \( f'(x) = \frac{\cos x}{2\sqrt{\sin x}} \) tells us about the rate of change of the function.

• When \( f'(x) > 0 \), the function is increasing.
• When \( f'(x) < 0 \), the function is decreasing.
• Critical points, where \( f'(x) = 0 \), indicate potential local maxima or minima.

Understanding these changes allows us to create a sketch of the graph that accurately reflects the behavior of the function over its domain.

Concavity

Concavity tells us about the curvature of the function's graph – whether it bends upwards or downwards. This is determined using the second derivative \( f''(x) \).
In the context of our function \( f(x) = \sqrt{\sin x} \), computing the second derivative is more complex, but insightful:

• When \( f''(x) > 0 \), the graph is concave up, resembling a cup.
• When \( f''(x) < 0 \), the graph is concave down, like an arch.

For \( f(x) = \sqrt{\sin x} \), it's known that the function is concave down on the entire interval \((0, \pi)\). This means the function's graph forms a downward-facing arch throughout this range. Understanding concavity helps to refine our graph sketch, showing where the function speeds up or slows down in its increase or decrease.

Function Analysis

Analyzing a function involves studying its various properties like domain, derivatives, and critical points. For \( f(x) = \sqrt{\sin x} \), understanding the domain \([0, \pi]\) is our first step. Here, the function is well-defined because \( \sin x \geq 0 \).
Next, we look at the derivative to find where the function is increasing or decreasing. We noted earlier that \( f'(x) > 0 \) for \( x \in (0, \pi/2) \), indicating an increase, and \( f'(x) < 0 \) for \( x \in (\pi/2, \pi) \), indicating a decrease.
Critical points, like \( x = \pi/2 \) where \( f'(x) = 0 \), help to mark transitions from increasing to decreasing behaviors. By analyzing both \( f'(x) \) and \( f''(x) \), we get a comprehensive view of how the function behaves – which is crucial for accurate graph sketching.

Graph Sketching

Graph sketching is a visual representation of everything you've calculated and understood about the function. For \( f(x) = \sqrt{\sin x} \), we begin by noting key features:

• Interval of increase: from \( 0 \) to \( \pi/2 \).
• Interval of decrease: from \( \pi/2 \) to \( \pi \).
• Overall concavity: concave down on \((0, \pi)\).
• Critical point at \( x = \pi/2 \): the peak of the graph.

Using these insights, draw the graph with a smooth, downward-arching curve that rises then falls within the domain. Highlight the peak at \( x = \pi/2 \) and ensure the curve dimensions echo the concave nature observed. Graph sketching brings the abstract function into a concrete form, helping visualize both the behavior and the changes across its domain.