Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals 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 hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.