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Magazine Chapter 8: Problem 2
True or False The graph of \(y=\sec x\) is one-to-one on the \(\operatorname{set}\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right]\)
Short Answer
Expert verified
False
Step by step solution
01
Understand the function and interval
The function given is the secant function: \(y = \sec x\). The interval specified is \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\). The goal is to determine if the secant function is one-to-one on this interval.
02
Analyze the behavior of the secant function
The secant function is the reciprocal of the cosine function: \(\sec x = \frac{1}{\cos x}\). It inherits the periodic properties and undefined points from the cosine function. Specifically, \(\sec x\) is undefined where \(\cos x = 0\), which occurs at \(x = \frac{\pi}{2}\).
03
Consider the interval \([0, \frac{\pi}{2})\)
On the interval \([0, \frac{\pi}{2})\), the cosine function \(\cos x\) decreases from 1 to 0, meaning the secant function \(\sec x\) ranges from 1 to +鈭�. Since \(y = \sec x\) is strictly increasing in this interval, it is one-to-one here.
04
Consider the interval \((\frac{\pi}{2}, \pi]\)
On the interval \((\frac{\pi}{2}, \pi]\), \(\cos x\) starts from 0 (just after \(\frac{\pi}{2}\)) and goes to -1 when approaching \(\pi\). Consequently, \(\sec x\) ranges from -鈭� to -1, and it is also strictly decreasing in this interval. Thus, it is one-to-one here as well.
05
Combine intervals \([0, \frac{\pi}{2})\) and \((\frac{\pi}{2}, \pi]\)
Both intervals, \([0, \frac{\pi}{2})\) and \((\frac{\pi}{2}, \pi]\), do not overlap and are individually one-to-one. However, the whole set \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\) is not one-to-one because there is a discontinuity at \(x = \frac{\pi}{2}\). The secant function jumps from \(+鈭瀄) to \(-鈭瀄) across \(\frac{\pi}{2}\).
06
Conclusion
Given the discontinuity at \(x = \frac{\pi}{2}\) and the fact that the secant function is not continuously one-to-one over the interval \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\), the graph of \(y = \sec x\) is not one-to-one over the entire interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
One-to-One Functions
A function is considered one-to-one if every output value (or y-value) is associated with exactly one input value (or x-value).
This means that no two different x-values can produce the same y-value.
A quick way to check if a function is one-to-one is to use the Horizontal Line Test.
If any horizontal line crosses the graph of the function more than once, the function is NOT one-to-one.
For example, the secant function, denoted as \(\sec x\), can be assessed for one-to-one behavior over specific intervals.
In the step-by-step solution, \[0, \frac{abla}{2} \) and \(\frac{abla}{2}, abla \] are examined in terms of the function's strict increase or decrease.
If it either strictly increases or strictly decreases within an interval, it ensures one-to-one behavior there.
This means that no two different x-values can produce the same y-value.
A quick way to check if a function is one-to-one is to use the Horizontal Line Test.
If any horizontal line crosses the graph of the function more than once, the function is NOT one-to-one.
For example, the secant function, denoted as \(\sec x\), can be assessed for one-to-one behavior over specific intervals.
In the step-by-step solution, \[0, \frac{abla}{2} \) and \(\frac{abla}{2}, abla \] are examined in terms of the function's strict increase or decrease.
If it either strictly increases or strictly decreases within an interval, it ensures one-to-one behavior there.
Interval Analysis
Interval analysis involves breaking down the domain into specific segments and examining the behavior of the function within these segments.
For the secant function, \(y = \sec x\), the given intervals are \[0, \frac{abla}{2} \) and \(\frac{abla}{2}, abla \].
Analyzing these intervals helps us understand how the function behaves around critical points and discontinuities.
In \[0, \frac{abla}{2} \), the cosine function decreases from 1 to 0, making the secant function range from 1 to +鈭�.
Here, \(\sec x\) is strictly increasing, thus one-to-one.
In \(\frac{abla}{2}, abla \], \(\sec x\) starts just after \(\frac{abla}{2}\) and decreases from -鈭� to -1.
As it strictly decreases, it remains one-to-one within this interval too.
For the secant function, \(y = \sec x\), the given intervals are \[0, \frac{abla}{2} \) and \(\frac{abla}{2}, abla \].
Analyzing these intervals helps us understand how the function behaves around critical points and discontinuities.
In \[0, \frac{abla}{2} \), the cosine function decreases from 1 to 0, making the secant function range from 1 to +鈭�.
Here, \(\sec x\) is strictly increasing, thus one-to-one.
In \(\frac{abla}{2}, abla \], \(\sec x\) starts just after \(\frac{abla}{2}\) and decreases from -鈭� to -1.
As it strictly decreases, it remains one-to-one within this interval too.
Discontinuities in Functions
Discontinuities are points where a function is not defined.
For the secant function, \(y = \sec x\), such discontinuities occur where the cosine function is zero because \(\sec x = \frac{1}{\cos x}\), and anything over zero is undefined.
These points are multiples of \(abla/2\).
For the specified set \[0, \frac{abla}{2} \) and \(\frac{abla}{2}, abla \], there is a discontinuity at \(\frac{abla}{2}\).
This causes the secant function to jump from +鈭� to -鈭�, making even a combined continuous interval not one-to-one.
For a function to be one-to-one over the entire interval, there can be no such jumps.
For the secant function, \(y = \sec x\), such discontinuities occur where the cosine function is zero because \(\sec x = \frac{1}{\cos x}\), and anything over zero is undefined.
These points are multiples of \(abla/2\).
For the specified set \[0, \frac{abla}{2} \) and \(\frac{abla}{2}, abla \], there is a discontinuity at \(\frac{abla}{2}\).
This causes the secant function to jump from +鈭� to -鈭�, making even a combined continuous interval not one-to-one.
For a function to be one-to-one over the entire interval, there can be no such jumps.
