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Chapter 4: Problem 44

Give a graph of the function and identify the locations of all relative extrema and inflection points. Check your work with a graphing utility. $$ \sqrt{\tan x}, \quad 0 \leq x<\pi / 2 $$

Short Answer

Expert verified
The function has a relative minimum at \( x = 0 \), and there are no inflection points in the given interval.

Step by step solution

01

Understanding the Function

The function given is \( f(x) = \sqrt{\tan x} \) with the domain \( 0 \leq x < \pi/2 \). This function is only defined where \( \tan x \geq 0 \), which is satisfied on this interval.
02

Finding Critical Points

To find critical points, we first need the derivative. Using the chain rule and knowing \( f(x) = (\tan x)^{1/2} \), the derivative is \( f'(x) = \frac{1}{2}(\tan x)^{-1/2} \cdot \sec^2 x \). Set \( f'(x) = 0 \) to find critical points: this occurs when \( \tan x = 0 \), or \( x = 0 \).
03

Analyzing the Critical Point

Check the intervals around \( x = 0 \) to determine if it is a relative minimum or maximum. Since \( \tan x \) is increasing for \( 0 < x < \pi/2 \), \( f(x) \) has a local minimum at \( x = 0 \).
04

Finding Inflection Points

To find inflection points, consider the second derivative \( f''(x) \). First, simplify \( f'(x) \) to \( \frac{\sec^2 x}{2\sqrt{\tan x}} \) and then find \( f''(x) \). Set \( f''(x) = 0 \) and solve for \( x \). This step usually requires more computational work or graphing utility assistance.
05

Checking with Graphing Utility

Use a graphing utility to plot \( f(x) = \sqrt{\tan x} \) over \( 0 \leq x < \pi/2 \). Verify the critical points and any inflection points found. Observe the minimum at \( x = 0 \).
06

Interpreting the Graph

The graph confirms the results: \( x = 0 \) is a relative minimum. There are no inflection points on \( 0 < x < \pi/2 \) based on the graph, as the concavity does not change.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphing Functions
When studying calculus, graphing functions is a fundamental skill needed to visually understand functions' behaviors. By graphing, we can see how a function increases, decreases, and changes shape over its domain. For the function \( f(x) = \sqrt{\tan x} \) in the prescribed interval \( 0 \leq x < \pi/2 \), this graph helps reveal important characteristics.
Understanding graph interpretation is essential.
  • Plot the Function: Begin by plotting \( f(x) \) across its domain using a graphing utility, showcasing the curve's shape.
  • Domains and Ranges: Identify valid \( x \) values to confirm the function remains valid across the intended range.
  • Behavior Analysis: Investigate how \( f(x) \) reacts as it approaches boundaries (e.g., near \( x = \pi/2 \), where \( \tan x \) tends toward infinity).
A clear plot of \( f(x) = \sqrt{\tan x} \) will display steady growth from its minimum point at \( x = 0 \), trending upward.
Relative Extrema
Relative extrema refer to points on the graph of a function where the function reaches local maximum or minimum values within a specific interval. For \( f(x) = \sqrt{\tan x} \), identifying these can help understand where the function turns in its upward or downward path.
To find relative extrema:
  • Derivative Approach: Calculate the derivative \( f'(x) \). For our function, this derivative is \( f'(x) = \frac{1}{2}(\tan x)^{-1/2} \cdot \sec^2 x \).
  • Critical Points: Determine where \( f'(x) = 0 \). Critical points are candidates for relative extrema; in this case, \( x=0 \) is a critical point.
  • Test Intervals: Examine the function's behavior around \( x = 0 \) to conclude it's a relative minimum.
Since \( \tan x \) increases in the interval \( (0, \pi/2) \), \( f(x) \) shows a change from decreasing to increasing at \( x = 0 \), showing a local minimum there.
Inflection Points
Inflection points are locations on a curve where concavity changes between upward and downward, providing insights into how a function's rate of change is altering. However, not every function will have an inflection point, as we find in the case of \( f(x) = \sqrt{\tan x} \) across its defined interval.
For identifying inflection points:
  • Second Derivative: First, simplify the first derivative, then find the second derivative \( f''(x) \). This analysis helps determine concave behaviors.
  • Concavity Check: Look for \( f''(x) = 0 \) to find potential inflection points.
  • Graph Confirmation: Use a graphing utility to verify if the function changes concavity.
Upon graphing, \( f(x) = \sqrt{\tan x} \) visually reveals no concavity change within \( 0 < x < \pi/2 \); thus, no inflection points are present in this domain.

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