Problem 60 - Precalculus: A Concise Course

Question

Use the graph of the function to determine whether the function is even, odd, or neither. Verify your answer algebraically. $$ g(x)=\csc x $$

Step-by-Step Solution

Answer

The function \( g(x) = \csc x \) is odd.

Step 1: Interpret the Graph of the Function

Draw the graph of the function \( g(x) = \csc x \), which is the reciprocal of the sine function. It's an infinitely repeating pattern of U and inverted U shapes, with vertical asymptotes at the integer multiples of \( \pi \) and the maximum and minimum values of each U-shape touching the points where the sine function equals 1 and -1, respectively. Notice that the function is not symmetric with respect to the y-axis or the origin.

Step 2: Verify algebraically if the function is even

An even function is defined as having the property \( f(-x) = f(x) \) for all \( x \) in its domain. Substitute \( -x \) for \( x \) in \( g(x) \) to get \( g(-x) = \csc(-x) \). Because the cosecant is the reciprocal of the sine and the sine function is odd, the negative sign doesn't go away, i.e., \( \csc(-x) = -\csc(x) \). Hence, \( g(-x) \neq g(x) \), so \( g(x) \) is not an even function.

Step 3: Verify algebraically if the function is odd

An odd function is defined as having the property \( f(-x) = -f(x) \) for all \( x \) in its domain. We already found that \( g(-x) = -\csc(x) = -g(x) \), so \( g(x) \) is an odd function.

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc x \), is one of the six fundamental trigonometric functions. It is defined as the reciprocal of the sine function, which means \( \csc x = \frac{1}{\sin x} \). Because the sine function approaches zero at its zeros, the cosecant function has vertical asymptotes at those points, creating undefined values.
The cosecant function exhibits a unique shape on a graph. It shows a repeating pattern of U and inverted U shapes, interspersed with vertical asymptotes. These asymptotes occur at integer multiples of \( \pi \) (e.g., \( 0, \pi, 2\pi \), etc.), which are the zeros of the sine function. Additionally, the peaks and valleys of these U-shaped curves touch the points where the sine function is either 1 or -1.

Function Symmetry

Understanding the symmetry of a function involves looking at how the graph behaves across the y-axis and the origin. We call a function even if it reflects across the y-axis, meaning \( f(-x) = f(x) \) for its entire domain. An odd function exhibits point symmetry through the origin, meaning \( f(-x) = -f(x) \).
For the function \( g(x) = \csc x \), we need to investigate these properties. Substituting \( -x \) into \( \csc x \) yields \( \csc(-x) = \frac{1}{\sin(-x)} \), and by sine's odd nature, this becomes \( -\csc x \). Consequently, since \( \csc(-x) = -\csc x \), \( g(x) \) meets the criteria for being an odd function, not an even one.

Graph Interpretation

Interpreting the graph of \( g(x) = \csc x \) gives visual insight into its behavior and properties. The graph features repeating U and inverted U patterns, each associated with sine's waves' maxima and minima. The vertical asymptotes mark the places where the graph shoots up to infinity and down to negative infinity around loci like \( 0 \) and \( \pi \).

• These asymptotes correspond to where sine equals zero. As a result, the cosecant function does not have values, creating the asymptotic lines.
• In between these points of discontinuity, the U and inverted U shapes mirror each other, emphasizing the function's periodic and symmetric qualities.

Analyzing these properties reveals the hospitable intervals where the function is defined and helps in understanding its genuine odd symmetry as seen in its graph.

Algebraic Verification

Algebraically verifying whether a function is odd, even, or neither requires manipulating the function expression. For \( g(x) = \csc x \), this means evaluating \( g(-x) \) and comparing it to \( g(x) \).
The algebraic approach involves substituting \( -x \) into the original function to get \( \csc(-x) \). Using the fact that \( \sin(-x) = -\sin x \), we find that \( \csc(-x) = \frac{1}{-\sin x} = -\csc x \). This result confirms that \( g(-x) = -g(x) \), satisfying the condition of an odd function.
This process demonstrates why establishing a function's oddness or evenness isn't just about looking at graphs but also confirming it with solid algebraic reasoning. This makes analysis accurate and grounded in reliable mathematical principles.