Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Let's analyze each statement step by step:
1. Function [tex]\( f \)[/tex] is continuous.
To determine if the function [tex]\( f \)[/tex] is continuous at [tex]\( x = 1 \)[/tex], we must check if the left-hand limit equals the right-hand limit at this point.
For [tex]\( x \leq 1 \)[/tex]:
[tex]\[ f(x) = \left(\frac{1}{3}\right)^x \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
For [tex]\( x > 1 \)[/tex]:
[tex]\[ f(x) = -x^2 + 2x - 1 \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -(1)^2 + 2(1) - 1 = -1 + 2 - 1 = 0 \][/tex]
Since [tex]\(\frac{1}{3} \neq 0\)[/tex], the left-hand limit does not equal the right-hand limit at [tex]\( x = 1 \)[/tex]. Therefore, the function [tex]\( f \)[/tex] is not continuous.
2. The domain of function [tex]\( f \)[/tex] is all real numbers.
The function is defined for all [tex]\( x \leq 1 \)[/tex] by [tex]\( \left( \frac{1}{3} \right)^x \)[/tex] and for all [tex]\( x > 1 \)[/tex] by [tex]\( -x^2 + 2x - 1 \)[/tex]. Both parts are defined for their respective domains without any restrictions. Therefore, the domain of function [tex]\( f \)[/tex] is all real numbers.
3. Function [tex]\( f \)[/tex] is decreasing over the entire domain.
- For [tex]\( x \leq 1 \)[/tex], [tex]\( f(x) = \left( \frac{1}{3} \right)^x \)[/tex] is decreasing because [tex]\( \frac{1}{3} < 1 \)[/tex].
- For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. The derivative is:
[tex]\[ f'(x) = -2x + 2 \][/tex]
At [tex]\( x = 1 \)[/tex], [tex]\( f'(1) = -2(1) + 2 = 0 \)[/tex]. For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) > 0 \)[/tex] indicating an increasing function, and for [tex]\( x > 1 \)[/tex], [tex]\( f'(x) < 0 \)[/tex], indicating a decreasing function. However, since it’s not consistently decreasing over [tex]\( x \leq 1 \)[/tex] and [tex]\( x > 1 \)[/tex] domains, [tex]\( f \)[/tex] is not decreasing over the entire domain.
4. The value of [tex]\( f(1) \)[/tex] is 0.
At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 0 \][/tex]
This is based on the right-hand expression, even though the left-hand expression yields [tex]\(\frac{1}{3}\)[/tex]. From either part, it has previously been confirmed that the value of [tex]\( f(1) \)[/tex] directly here is 0.
5. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. As [tex]\( x \to +\infty \)[/tex]:
[tex]\[ -x^2 \text{ term dominates, leading to } f(x) \to -\infty \][/tex]
Therefore, as [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \)[/tex] does not approach positive infinity, but rather negative infinity.
Based on this detailed analysis:
Correct selections:
- The domain of function [tex]\( f \)[/tex] is all real numbers.
- The value of [tex]\( f(1) \)[/tex] is 0.
1. Function [tex]\( f \)[/tex] is continuous.
To determine if the function [tex]\( f \)[/tex] is continuous at [tex]\( x = 1 \)[/tex], we must check if the left-hand limit equals the right-hand limit at this point.
For [tex]\( x \leq 1 \)[/tex]:
[tex]\[ f(x) = \left(\frac{1}{3}\right)^x \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
For [tex]\( x > 1 \)[/tex]:
[tex]\[ f(x) = -x^2 + 2x - 1 \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -(1)^2 + 2(1) - 1 = -1 + 2 - 1 = 0 \][/tex]
Since [tex]\(\frac{1}{3} \neq 0\)[/tex], the left-hand limit does not equal the right-hand limit at [tex]\( x = 1 \)[/tex]. Therefore, the function [tex]\( f \)[/tex] is not continuous.
2. The domain of function [tex]\( f \)[/tex] is all real numbers.
The function is defined for all [tex]\( x \leq 1 \)[/tex] by [tex]\( \left( \frac{1}{3} \right)^x \)[/tex] and for all [tex]\( x > 1 \)[/tex] by [tex]\( -x^2 + 2x - 1 \)[/tex]. Both parts are defined for their respective domains without any restrictions. Therefore, the domain of function [tex]\( f \)[/tex] is all real numbers.
3. Function [tex]\( f \)[/tex] is decreasing over the entire domain.
- For [tex]\( x \leq 1 \)[/tex], [tex]\( f(x) = \left( \frac{1}{3} \right)^x \)[/tex] is decreasing because [tex]\( \frac{1}{3} < 1 \)[/tex].
- For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. The derivative is:
[tex]\[ f'(x) = -2x + 2 \][/tex]
At [tex]\( x = 1 \)[/tex], [tex]\( f'(1) = -2(1) + 2 = 0 \)[/tex]. For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) > 0 \)[/tex] indicating an increasing function, and for [tex]\( x > 1 \)[/tex], [tex]\( f'(x) < 0 \)[/tex], indicating a decreasing function. However, since it’s not consistently decreasing over [tex]\( x \leq 1 \)[/tex] and [tex]\( x > 1 \)[/tex] domains, [tex]\( f \)[/tex] is not decreasing over the entire domain.
4. The value of [tex]\( f(1) \)[/tex] is 0.
At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 0 \][/tex]
This is based on the right-hand expression, even though the left-hand expression yields [tex]\(\frac{1}{3}\)[/tex]. From either part, it has previously been confirmed that the value of [tex]\( f(1) \)[/tex] directly here is 0.
5. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. As [tex]\( x \to +\infty \)[/tex]:
[tex]\[ -x^2 \text{ term dominates, leading to } f(x) \to -\infty \][/tex]
Therefore, as [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \)[/tex] does not approach positive infinity, but rather negative infinity.
Based on this detailed analysis:
Correct selections:
- The domain of function [tex]\( f \)[/tex] is all real numbers.
- The value of [tex]\( f(1) \)[/tex] is 0.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.