Discover a wealth of information and get your questions answered on IDNLearn.com. Join our community to receive timely and reliable responses to your questions from knowledgeable professionals.
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.
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
