At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get quick and reliable solutions to your questions from a community of experienced experts on our 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 appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
