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Find the following for the function:
[tex]\[ y = -5 \sec \left(x + \frac{\pi}{4}\right) \][/tex]

(a) Period:
The period is [tex]\( 2\pi \)[/tex]. (Type an exact answer in terms of [tex]\(\pi\)[/tex].)

(b) Phase Shift:
Select the correct choice below and fill in the answer box within your choice.

A. The phase shift is [tex]\(\square\)[/tex] units. (Type an exact answer in terms of [tex]\(\pi\)[/tex].)
B. The given function has no phase shift.

(c) Range:
The range of the function is [tex]\(\square\)[/tex].


Sagot :

Let's tackle each part of the question in detail:

### Part (a): Period of the Function
The function given is [tex]\( y = -5 \sec \left(x + \frac{\pi}{4}\right) \)[/tex].

The period of the secant function, [tex]\( \sec(x) \)[/tex], is [tex]\( 2\pi \)[/tex]. This property is retained even with modifications inside the argument of the secant function, such as horizontal shifts and reflections.

Therefore, the period of [tex]\( y = -5 \sec \left(x + \frac{\pi}{4}\right) \)[/tex] is:

[tex]\[ \boxed{2\pi} \][/tex]

### Part (b): Phase Shift
To find the phase shift, we look at the term inside the secant function. Specifically, we consider the function in the form [tex]\( \sec(x + C) \)[/tex], where [tex]\( C \)[/tex] is a constant that indicates the phase shift.

In our function, [tex]\( y = -5 \sec \left(x + \frac{\pi}{4}\right) \)[/tex], the term inside the secant is [tex]\((x + \frac{\pi}{4})\)[/tex].

Setting [tex]\((x + \frac{\pi}{4}) = 0\)[/tex] to determine the phase shift:

[tex]\[ x + \frac{\pi}{4} = 0 \][/tex]

[tex]\[ x = -\frac{\pi}{4} \][/tex]

This means the phase shift is:
[tex]\[ \boxed{-\frac{\pi}{4}} \][/tex]

### Part (c): Range of the Function
The range of the secant function, [tex]\( \sec(x) \)[/tex], is [tex]\((-\infty, -1] \cup [1, \infty)\)[/tex].

When multiplying [tex]\(\sec(x + \frac{\pi}{4})\)[/tex] by -5, the range notation changes. We need to examine the transformation of each segment of the range.

1. For the segment [tex]\((- \infty, -1]\)[/tex]:
- Multiplying by -5, which inverts the sign and stretches by a factor of 5:
[tex]\[ -5 \times \sec(x + \frac{\pi}{4}) \Rightarrow (-5, - \infty) \text{ (Switching endpoints directionally gives } -\infty \text{ to } -5\text{)} \][/tex]
This segment becomes:
[tex]\[ (- \infty, -5] \][/tex]

2. For the segment [tex]\([1, \infty)\)[/tex]:
- Multiplying by -5 similarly inverts and stretches:
[tex]\[ -5 \times \sec(x + \frac{\pi}{4}) \Rightarrow [-5, \infty] \text{ (Switching endpoints directionally gives } -5 \text{ to } \infty\text{)} \][/tex]
This segment becomes:
[tex]\[ [5, \infty) \][/tex]

Thus, the total range of the function [tex]\( y = -5 \sec \left(x + \frac{\pi}{4}\right) \)[/tex] is the union of these intervals:

[tex]\[ \boxed{(-\infty, -5] \cup [5, \infty)} \][/tex]

To summarize, the solutions to each part are:

(a) The period is [tex]\( 2\pi \)[/tex].

(b) The phase shift is [tex]\( -\frac{\pi}{4} \)[/tex] units.

(c) The range is [tex]\( (-\infty, -5] \cup [5, \infty) \)[/tex].
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