Problem 34
Question
Analyze the trigonometric function f over the specified interval, stating where f is increasing, decreasing, concave up, and concave down, and stating the x-coordinates of all inflection points. Confirm that your results are consistent with the graph of f generated with a graphing utility. \(f(x)=\sec x \tan x ;(-\pi / 2, \pi / 2)\)
Step-by-Step Solution
Verified Answer
f is concave down on (-π/2, 0) and concave up on (0, π/2). No inflection points in (-π/2, π/2).
1Step 1: Identify Derivative
First, identify the derivative of the function \( f(x) = \sec x \tan x \) to find intervals where it is increasing or decreasing. Using the product rule, we get \( f'(x) = \sec x \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(\sec x) \). This simplifies to \( f'(x) = \sec x \sec^2 x + \tan x \sec x \tan x \).
2Step 2: Simplify Derivative
Simplify the expression for \( f'(x) \). It becomes \( f'(x) = \sec^3 x + \sec x \tan^2 x \), which simplifies further to \( f'(x) = \sec x (\sec^2 x + \tan^2 x) \). Knowing \( \tan^2 x = \sec^2 x - 1 \), substitute this to get \( f'(x) = \sec^3 x + \sec x (\sec^2 x - 1) = 2\sec^3 x - \sec x \).
3Step 3: Solve for Critical Points
Set \( f'(x) = 0 \) to find critical points: \( 2\sec^3 x - \sec x = 0 \), which simplifies to \( \sec x (2\sec^2 x - 1) = 0 \). \( \sec x = 0 \) has no solution since \( \sec x \) is never zero. Thus, \( 2\sec^2 x - 1 = 0 \) leads to \( \sec^2 x = \frac{1}{2} \), giving no real solutions. Thus, no critical points on the interval (-\(\pi/2\), \(\pi/2\)).
4Step 4: Second Derivative for Concavity
Find the second derivative \( f''(x) \) to test for concavity and inflection points. We start from \( f'(x) = 2\sec^3 x - \sec x \). By differentiating, \( f''(x) = 6\sec^4 x \tan x - 2\sec^3 x \tan x \). Simplified, \( f''(x) = 4\sec^5 x \tan x \).
5Step 5: Analyze the Concavity
Since \( \sec x \) and \( \tan x \) are undefined at some points within \((-\pi/2, \pi/2)\), analyze intervals based on known behavior of \( \sec x \) and \( \tan x \). For \( x \in (0, \pi/2) \), \( f''(x) > 0 \), indicating \( f(x) \) is concave up. For \( x \in (-\pi/2, 0) \), \( f''(x) < 0 \), indicating \( f(x) \) is concave down.
6Step 6: Determine Inflection Points
Identify points where concavity changes from up to down or vice-versa. Since derivative solutions showed no real critical points in \((0, \pi/2)\) and \((-\pi/2, 0)\), and \( f''(x) \) does not equal zero, there are no inflection points within \((-\pi/2, \pi/2)\).
7Step 7: Final Step: Verify with Graph
Graph \( f(x) = \sec x \tan x \) using a graphing utility over the interval \((-\pi/2, \pi/2)\) to ensure findings. Look for visual confirmation of increasing, decreasing, concave up, concave down, and absence of inflection points.
Key Concepts
CalculusConcavityCritical PointsDerivativeInflection Points
Calculus
Calculus is a branch of mathematics that focuses on the study of change through derivatives and integrals. In this problem, the function we are examining is a trigonometric function, specifically, \( f(x) = \sec x \tan x \), over the interval \((-\pi/2, \pi/2)\). Calculus helps us assess how functions behave, whether they are increasing or decreasing, and their concavity.
Through derivatives, we learn about the slopes and curvature of the function. Derivatives allow us to determine the function's behavior by finding rates of change. In this exercise, we use derivatives to understand where the secant-tangent function increases or decreases. Integrals, on the other hand, measure accumulated quantities or areas under curves, but are not the focus of this particular problem.
Overall, calculus provides the tools to deeply understand and predict the behavior of complex functions and their graphs.
Through derivatives, we learn about the slopes and curvature of the function. Derivatives allow us to determine the function's behavior by finding rates of change. In this exercise, we use derivatives to understand where the secant-tangent function increases or decreases. Integrals, on the other hand, measure accumulated quantities or areas under curves, but are not the focus of this particular problem.
Overall, calculus provides the tools to deeply understand and predict the behavior of complex functions and their graphs.
Concavity
Concavity is a concept in calculus that describes the direction in which a curve bends. It tells us about the shape of the graph of a function in specific intervals.
A function is concave up on an interval if its graph bends upward, like a cup, indicating that the second derivative \( f''(x) \) is positive. Conversely, it is concave down if the graph bends downward, showing that \( f''(x) \) is negative.
For the function \( f(x) = \sec x \tan x \), the analysis of the second derivative reveals:
A function is concave up on an interval if its graph bends upward, like a cup, indicating that the second derivative \( f''(x) \) is positive. Conversely, it is concave down if the graph bends downward, showing that \( f''(x) \) is negative.
For the function \( f(x) = \sec x \tan x \), the analysis of the second derivative reveals:
- For \( x \in (0, \pi/2)\), \( f''(x) > 0 \), indicating concave up.
- For \( x \in (-\pi/2, 0)\), \( f''(x) < 0 \), indicating concave down.
Critical Points
Critical points of a function occur where its first derivative is zero or undefined. They are essential in identifying where a function changes from increasing to decreasing, or vice versa.
In the given function \( f(x) = \sec x \tan x \), we find that the critical points occur when the first derivative \( f'(x) = 2\sec^3 x - \sec x \) equals zero. However, in this particular interval, there are no real solutions to this equation.
This means there are no critical points within \((-\pi/2, \pi/2)\). Critical points can be vital for finding local maxima and minima, but in this case, the absence of critical points simplifies our analysis of the function's behavior.
In the given function \( f(x) = \sec x \tan x \), we find that the critical points occur when the first derivative \( f'(x) = 2\sec^3 x - \sec x \) equals zero. However, in this particular interval, there are no real solutions to this equation.
This means there are no critical points within \((-\pi/2, \pi/2)\). Critical points can be vital for finding local maxima and minima, but in this case, the absence of critical points simplifies our analysis of the function's behavior.
Derivative
The derivative of a function indicates the rate of change at any point on its graph. It tells us the slope of the tangent line to the function at any point.
For the function \( f(x) = \sec x \tan x \), the first derivative is crucial for understanding where the function increases or decreases.
Using the product rule, we derived the first derivative as \( f'(x) = 2\sec^3 x - \sec x \). From this, the solution explores the behavior of the function extensively across its domain, \((-\pi/2, \pi/2)\).
The derivative serves as a foundational tool in calculus, revealing much about the dynamics of a function. It aids immensely in analyzing and predicting function behavior, which is why the exploration of \( f'(x) \) is a significant part of the solution.
For the function \( f(x) = \sec x \tan x \), the first derivative is crucial for understanding where the function increases or decreases.
Using the product rule, we derived the first derivative as \( f'(x) = 2\sec^3 x - \sec x \). From this, the solution explores the behavior of the function extensively across its domain, \((-\pi/2, \pi/2)\).
The derivative serves as a foundational tool in calculus, revealing much about the dynamics of a function. It aids immensely in analyzing and predicting function behavior, which is why the exploration of \( f'(x) \) is a significant part of the solution.
Inflection Points
Inflection points occur where the graph of a function changes its concavity, i.e., from concave up to concave down or vice versa. These points are found where the second derivative \( f''(x) \) changes sign.
To locate inflection points for \( f(x) = \sec x \tan x \), we first find the second derivative. After simplifying, \( f''(x) = 4\sec^5 x \tan x \).
In our interval, \((-\pi/2, \pi/2)\), \( f''(x) \) switches from positive to negative when crossing zero but doesn’t actually equal zero at any point that realizes a change in concavity. Thus, the function has no inflection points in this range. Identifying inflection points is important in understanding the overall shape of the graph, even though, in this case, there are none to be found.
To locate inflection points for \( f(x) = \sec x \tan x \), we first find the second derivative. After simplifying, \( f''(x) = 4\sec^5 x \tan x \).
In our interval, \((-\pi/2, \pi/2)\), \( f''(x) \) switches from positive to negative when crossing zero but doesn’t actually equal zero at any point that realizes a change in concavity. Thus, the function has no inflection points in this range. Identifying inflection points is important in understanding the overall shape of the graph, even though, in this case, there are none to be found.
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