Problem 62 - Precalculus

Question

In Exercises \(57-66,\) state the domain and range of the functions. $$y=1-2 \sec \left(\frac{1}{2} x+\pi\right)$$

Step-by-Step Solution

Answer

Domain: All real numbers except \(x = -\pi + 2k\pi\); Range: \((-\infty, -1] \cup [3, \infty)\).

Step 1: Understanding the Function

The given function is a secant function, which is based on the reciprocal of the cosine function. This function is transformed by a horizontal stretch, horizontal shift, and vertical shift. The general form of the secant function is given by \(y = a \sec(bx - c) + d\), where \(a\), \(b\), \(c\), and \(d\) are constants.

Step 2: Determining Intervals of the Cosine Function

Since \(\sec(x) = \frac{1}{\cos(x)}\), we need to identify where \(\cos\left(\frac{1}{2}x + \pi\right)\) is equal to zero because at these points \(\sec\left(\frac{1}{2}x + \pi\right)\) will be undefined. The cosine function is zero at odd multiples of \(\frac{\pi}{2}\). Set the equation inside the cosine equal to these odd multiples:\[ \frac{1}{2}x + \pi = \frac{\pi}{2} + k\pi, \text{ where } k \text{ is an integer.}\]

Step 3: Solve for x-values where the function is undefined

Solve the equation from Step 2 for \(x\):\[\frac{1}{2}x + \pi = \frac{\pi}{2} + k\pi\]\[\frac{1}{2}x = \frac{\pi}{2} - \pi + k\pi = -\frac{\pi}{2} + k\pi\]\[x = -\pi + 2k\pi.\] At these values, the function is undefined.

Step 4: Determine the Domain of the Secant Function

The domain of the secant function will exclude the \(x\)-values that make the cosine zero, which are given by \(x = -\pi + 2k\pi\). Therefore, the domain of \(y = 1 - 2\sec\left(\frac{1}{2}x + \pi\right)\) is all real numbers except \(x = -\pi + 2k\pi\).

Step 5: Determine the Range of the Function

The range of the basic secant function \(\sec(x)\) is \((\infty, -1] \cup [1, \infty)\). This range is transformed vertically by \(-2\) and shifted by \(1\) in our function. So, we calculate:For \(y = 1 - 2\sec(z)\), rewrite the ranges: \(1 - 2(-1) \rightarrow 1 + 2 = 3,\)\(1 - 2(1) \rightarrow 1 - 2 = -1.\)Thus, the range is \(( -\infty, -1 ] \cup [ 3, \infty)\).

Key Concepts

Domain of Trigonometric Functions

When we talk about the domain of trigonometric functions, we are mainly referring to the set of input values for which the function is defined. Secant (\( \sec(x) \)) is one such function. It is defined everywhere that the cosine function (\( \cos(x) \)) is not zero. This is because \( \sec(x) = \frac{1}{\cos(x)} \), and division by zero is not allowed.
To find where the cosine is zero, we identify the angles for which \( \cos(x) = 0 \). These occur at odd multiples of \( \frac{\pi}{2} \), which are:

• \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \)
• More generally, \( x = \frac{\pi}{2} + k\pi \) for integer \( k \).

However, in our specific function \( y = 1 - 2 \sec\left(\frac{1}{2}x + \pi\right) \), the argument of cosine is transformed to \( \frac{1}{2}x + \pi \). Setting it equal to the \( \frac{\pi}{2} + k\pi \) values, you solve for \( x \) to find:

• \( x = -\pi + 2k\pi \).

Thus, the domain is all real numbers except these specific \( x \)-values.

Range of Trigonometric Functions

The range of a trigonometric function is the set of all possible output values. For the basic secant function, \( \sec(x) \), the range is \((\infty, -1] \cup [1, \infty)\). This means secant can yield values either less than or equal to \(-1\) or greater than or equal to \(1\).
In our given function \( y = 1 - 2 \sec\left(\frac{1}{2}x + \pi\right) \), this range is modified due to the various transformations. We perform a vertical stretch by \(-2\) and a vertical shift by adding \(1\). This affects the range as follows:

• For values in \([-1, \infty)\), \(-2 \times (-1) + 1 = 3\), hence transformed to \([3, \infty)\)
• For values in \((\infty, -1]\), \(-2 \times 1 + 1 = -1\), hence transformed to \((\infty, -1]\)

Therefore, the transformed range of our function is \(( -\infty, -1 ] \cup [ 3, \infty)\).

Function Transformation

Function transformation involves altering a function in terms of shifts, stretches, or compressions. When you take the secant function, \( \sec(x) \), and apply transformations, you change how it appears on a graph and change the mathematical expression.
Let's break down the transformations in \( y = 1 - 2 \sec\left(\frac{1}{2}x + \pi\right) \):

• Horizontal Stretch and Shift: The \( \frac{1}{2}x \) inside the secant represents a horizontal stretch. It doubles the cycle length. The \(+\pi\) denotes a horizontal shift to the left by \(\pi\) units.
• Vertical Stretch: The \(-2\) before the secant implicates a vertical stretch. Negative sign inverts the function and multiplies the outputs by \(2\).
• Vertical Shift: Adding \(1\) shifts the entire function upward by \(1\) unit.

These transformations lead to a trigonometric function that is significantly altered. The function's cycle length, range, and characteristic points shift and stretch based on these modifications, making graph interpretation crucial for understanding these complex combinations.