Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To determine which of these functions are even, let's recall the definition of an even function. A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex] for every [tex]\( x \)[/tex] in the domain of [tex]\( f \)[/tex].
Let’s examine each function step by step to see if it satisfies the condition of being even.
### A. [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex]
To determine if [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even, we need to check [tex]\( f(-\alpha) \)[/tex]:
[tex]\[ f(-\alpha) = 1 + \sec(-\alpha) \][/tex]
Since [tex]\(\sec(-\alpha) = \sec(\alpha)\)[/tex] (because the secant function is even), we have:
[tex]\[ f(-\alpha) = 1 + \sec(\alpha) = f(\alpha) \][/tex]
Thus, [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even.
### B. [tex]\( f(x) = x \cos(x) \)[/tex]
To determine if [tex]\( f(x) = x \cos(x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x) \cos(-x) \][/tex]
Since [tex]\(\cos(-x) = \cos(x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = (-x) \cos(x) = -x \cos(x) \][/tex]
Since [tex]\( f(-x) = -x \cos(x) \neq f(x) = x \cos(x) \)[/tex], the function [tex]\( f(x) = x \cos(x) \)[/tex] is not even.
### C. [tex]\( f(x) = \csc(x^2) \)[/tex]
To determine if [tex]\( f(x) = \csc(x^2) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \csc((-x)^2) = \csc(x^2) \][/tex]
Since [tex]\( f(-x) = \csc(x^2) = f(x) \)[/tex], the function [tex]\( f(x) = \csc(x^2) \)[/tex] is even.
### D. [tex]\( f(x) = \sin(2x) \)[/tex]
To determine if [tex]\( f(x) = \sin(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \sin(2(-x)) = \sin(-2x) \][/tex]
Since [tex]\(\sin(-2x) = -\sin(2x)\)[/tex] (because the sine function is odd), we have:
[tex]\[ f(-x) = -\sin(2x) \neq f(x) = \sin(2x) \][/tex]
Thus, the function [tex]\( f(x) = \sin(2x) \)[/tex] is not even.
### E. [tex]\( f(x) = \cos(2x) \)[/tex]
To determine if [tex]\( f(x) = \cos(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \cos(2(-x)) = \cos(-2x) \][/tex]
Since [tex]\(\cos(-2x) = \cos(2x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = \cos(2x) = f(x) \][/tex]
Thus, the function [tex]\( f(x) = \cos(2x) \)[/tex] is even.
### F. [tex]\( f(t) = \sec^2(t) - 1 \)[/tex]
To determine if [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even, we need to check [tex]\( f(-t) \)[/tex]:
[tex]\[ f(-t) = \sec^2(-t) - 1 \][/tex]
Since [tex]\(\sec(-t) = \sec(t)\)[/tex] (because the secant function is even), we have:
[tex]\[ \sec^2(-t) = \sec^2(t) \][/tex]
[tex]\[ f(-t) = \sec^2(t) - 1 = f(t) \][/tex]
Thus, the function [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even.
### Conclusion:
The even functions are:
- [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] (A)
- [tex]\( f(x) = \csc(x^2) \)[/tex] (C)
- [tex]\( f(x) = \cos(2x) \)[/tex] (E)
- [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] (F)
Therefore, the even functions are [tex]\(\boxed{A, C, E, F}\)[/tex].
Let’s examine each function step by step to see if it satisfies the condition of being even.
### A. [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex]
To determine if [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even, we need to check [tex]\( f(-\alpha) \)[/tex]:
[tex]\[ f(-\alpha) = 1 + \sec(-\alpha) \][/tex]
Since [tex]\(\sec(-\alpha) = \sec(\alpha)\)[/tex] (because the secant function is even), we have:
[tex]\[ f(-\alpha) = 1 + \sec(\alpha) = f(\alpha) \][/tex]
Thus, [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] is even.
### B. [tex]\( f(x) = x \cos(x) \)[/tex]
To determine if [tex]\( f(x) = x \cos(x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x) \cos(-x) \][/tex]
Since [tex]\(\cos(-x) = \cos(x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = (-x) \cos(x) = -x \cos(x) \][/tex]
Since [tex]\( f(-x) = -x \cos(x) \neq f(x) = x \cos(x) \)[/tex], the function [tex]\( f(x) = x \cos(x) \)[/tex] is not even.
### C. [tex]\( f(x) = \csc(x^2) \)[/tex]
To determine if [tex]\( f(x) = \csc(x^2) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \csc((-x)^2) = \csc(x^2) \][/tex]
Since [tex]\( f(-x) = \csc(x^2) = f(x) \)[/tex], the function [tex]\( f(x) = \csc(x^2) \)[/tex] is even.
### D. [tex]\( f(x) = \sin(2x) \)[/tex]
To determine if [tex]\( f(x) = \sin(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \sin(2(-x)) = \sin(-2x) \][/tex]
Since [tex]\(\sin(-2x) = -\sin(2x)\)[/tex] (because the sine function is odd), we have:
[tex]\[ f(-x) = -\sin(2x) \neq f(x) = \sin(2x) \][/tex]
Thus, the function [tex]\( f(x) = \sin(2x) \)[/tex] is not even.
### E. [tex]\( f(x) = \cos(2x) \)[/tex]
To determine if [tex]\( f(x) = \cos(2x) \)[/tex] is even, we need to check [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \cos(2(-x)) = \cos(-2x) \][/tex]
Since [tex]\(\cos(-2x) = \cos(2x)\)[/tex] (because the cosine function is even), we have:
[tex]\[ f(-x) = \cos(2x) = f(x) \][/tex]
Thus, the function [tex]\( f(x) = \cos(2x) \)[/tex] is even.
### F. [tex]\( f(t) = \sec^2(t) - 1 \)[/tex]
To determine if [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even, we need to check [tex]\( f(-t) \)[/tex]:
[tex]\[ f(-t) = \sec^2(-t) - 1 \][/tex]
Since [tex]\(\sec(-t) = \sec(t)\)[/tex] (because the secant function is even), we have:
[tex]\[ \sec^2(-t) = \sec^2(t) \][/tex]
[tex]\[ f(-t) = \sec^2(t) - 1 = f(t) \][/tex]
Thus, the function [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] is even.
### Conclusion:
The even functions are:
- [tex]\( f(\alpha) = 1 + \sec(\alpha) \)[/tex] (A)
- [tex]\( f(x) = \csc(x^2) \)[/tex] (C)
- [tex]\( f(x) = \cos(2x) \)[/tex] (E)
- [tex]\( f(t) = \sec^2(t) - 1 \)[/tex] (F)
Therefore, the even functions are [tex]\(\boxed{A, C, E, F}\)[/tex].
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
