Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Let's analyze each option to determine which one must be an even function, given that [tex]\( g(x) \)[/tex] is an odd function. Remember that a function [tex]\( g(x) \)[/tex] is odd if [tex]\( g(-x) = -g(x) \)[/tex], and a function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex].
### Option 1: [tex]\( f(x) = g(x) + 2 \)[/tex]
To check if [tex]\( f(x) = g(x) + 2 \)[/tex] is an even function, we need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = g(-x) + 2 \][/tex]
Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Thus,
[tex]\[ f(-x) = -g(x) + 2 \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = g(x) + 2 \quad \text{and} \quad f(-x) = -g(x) + 2 \][/tex]
Since [tex]\( f(-x) \ne f(x) \)[/tex], [tex]\( f(x) = g(x) + 2 \)[/tex] is not an even function.
### Option 2: [tex]\( f(x) = g(x) + g(x) \)[/tex]
To check if [tex]\( f(x) = g(x) + g(x) \)[/tex] is an even function:
[tex]\[ f(x) = 2g(x) \][/tex]
We need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 2g(-x) = 2(-g(x)) = -2g(x) \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 2g(x) \quad \text{and} \quad f(-x) = -2g(x) \][/tex]
Since [tex]\( f(-x) \ne f(x) \)[/tex], [tex]\( f(x) = g(x) + g(x) \)[/tex] is not an even function.
### Option 3: [tex]\( f(x) = g(x)^2 \)[/tex]
To check if [tex]\( f(x) = g(x)^2 \)[/tex] is an even function:
[tex]\[ f(x) = g(x)^2 \][/tex]
We need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (g(-x))^2 \][/tex]
Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Thus,
[tex]\[ f(-x) = (-g(x))^2 = g(x)^2 \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = g(x)^2 \quad \text{and} \quad f(-x) = g(x)^2 \][/tex]
Since [tex]\( f(-x) = f(x) \)[/tex], [tex]\( f(x) = g(x)^2 \)[/tex] is indeed an even function.
### Option 4: [tex]\( f(x) = -g(x) \)[/tex]
To check if [tex]\( f(x) = -g(x) \)[/tex] is an even function:
[tex]\[ f(x) = -g(x) \][/tex]
We need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = -g(-x) \][/tex]
Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Thus,
[tex]\[ f(-x) = -(-g(x)) = g(x) \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = -g(x) \quad \text{and} \quad f(-x) = g(x) \][/tex]
Since [tex]\( f(-x) \ne f(x) \)[/tex], [tex]\( f(x) = -g(x) \)[/tex] is not an even function.
### Conclusion
The function that must be an even function is [tex]\( f(x) = g(x)^2 \)[/tex]. Therefore, the correct option is:
[tex]\[ \boxed{3} \][/tex]
### Option 1: [tex]\( f(x) = g(x) + 2 \)[/tex]
To check if [tex]\( f(x) = g(x) + 2 \)[/tex] is an even function, we need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = g(-x) + 2 \][/tex]
Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Thus,
[tex]\[ f(-x) = -g(x) + 2 \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = g(x) + 2 \quad \text{and} \quad f(-x) = -g(x) + 2 \][/tex]
Since [tex]\( f(-x) \ne f(x) \)[/tex], [tex]\( f(x) = g(x) + 2 \)[/tex] is not an even function.
### Option 2: [tex]\( f(x) = g(x) + g(x) \)[/tex]
To check if [tex]\( f(x) = g(x) + g(x) \)[/tex] is an even function:
[tex]\[ f(x) = 2g(x) \][/tex]
We need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 2g(-x) = 2(-g(x)) = -2g(x) \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 2g(x) \quad \text{and} \quad f(-x) = -2g(x) \][/tex]
Since [tex]\( f(-x) \ne f(x) \)[/tex], [tex]\( f(x) = g(x) + g(x) \)[/tex] is not an even function.
### Option 3: [tex]\( f(x) = g(x)^2 \)[/tex]
To check if [tex]\( f(x) = g(x)^2 \)[/tex] is an even function:
[tex]\[ f(x) = g(x)^2 \][/tex]
We need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (g(-x))^2 \][/tex]
Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Thus,
[tex]\[ f(-x) = (-g(x))^2 = g(x)^2 \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = g(x)^2 \quad \text{and} \quad f(-x) = g(x)^2 \][/tex]
Since [tex]\( f(-x) = f(x) \)[/tex], [tex]\( f(x) = g(x)^2 \)[/tex] is indeed an even function.
### Option 4: [tex]\( f(x) = -g(x) \)[/tex]
To check if [tex]\( f(x) = -g(x) \)[/tex] is an even function:
[tex]\[ f(x) = -g(x) \][/tex]
We need to evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = -g(-x) \][/tex]
Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Thus,
[tex]\[ f(-x) = -(-g(x)) = g(x) \][/tex]
Comparing [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = -g(x) \quad \text{and} \quad f(-x) = g(x) \][/tex]
Since [tex]\( f(-x) \ne f(x) \)[/tex], [tex]\( f(x) = -g(x) \)[/tex] is not an even function.
### Conclusion
The function that must be an even function is [tex]\( f(x) = g(x)^2 \)[/tex]. Therefore, the correct option is:
[tex]\[ \boxed{3} \][/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.