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Sagot :
Let's analyze the function [tex]\( f(x) \)[/tex] given by:
[tex]\[ f(x)=\left\{\begin{array}{ll} \left(\frac{1}{3}\right)^x, & x \leq 1 \\ -x^2 + 2x - 1, & x > 1 \end{array}\right. \][/tex]
### Statement: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
To see what happens as [tex]\( x \)[/tex] approaches positive infinity, we examine the function [tex]\( -x^2 + 2x - 1 \)[/tex] for large values of [tex]\( x \)[/tex] (since this is the part of the piecewise function used when [tex]\( x > 1 \)[/tex]).
[tex]\[ \lim_{x \to \infty} (-x^2 + 2x - 1) = -\infty \][/tex]
This shows that as [tex]\( x \)[/tex] approaches positive infinity, the function [tex]\( f(x) \)[/tex] approaches negative infinity, not positive infinity. Therefore, this statement is false.
### Statement: The domain of function [tex]\( f \)[/tex] is all real numbers.
The given function is defined for all real numbers [tex]\( x \)[/tex]. There are no restrictions on the values that [tex]\( x \)[/tex] can take. Therefore, this statement is true.
### Statement: Function [tex]\( f \)[/tex] is decreasing over the entire domain.
To determine if the function is decreasing over the entire domain, we look at both parts of the piecewise function:
1. For [tex]\( x \leq 1 \)[/tex], [tex]\( \left(\frac{1}{3}\right)^x \)[/tex] is a decreasing exponential function because [tex]\( \frac{1}{3} < 1 \)[/tex].
2. For [tex]\( x > 1 \)[/tex], [tex]\( -x^2 + 2x - 1 \)[/tex] is a quadratic function that opens downwards (concave down). We can determine the vertex of this parabola:
[tex]\[ x = \frac{-b}{2a} = \frac{-2}{2(-1)} = 1 \][/tex]
Since the vertex occurs at [tex]\( x = 1 \)[/tex] and for [tex]\( x > 1 \)[/tex], the parabola is decreasing. This means [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex] is decreasing for [tex]\( x > 1 \)[/tex].
Therefore, because both parts of the function are decreasing in their respective domains, the function [tex]\( f \)[/tex] is decreasing over the entire domain. This statement is true.
### Statement: Function [tex]\( f \)[/tex] is continuous.
To check the continuity at [tex]\( x = 1 \)[/tex], we need to ensure the left-hand limit and right-hand limit at [tex]\( x = 1 \)[/tex] are equal, and that they equal [tex]\( f(1) \)[/tex]:
- Left-hand limit as [tex]\( x \)[/tex] approaches 1 from the left:
[tex]\[ \lim_{x \to 1^-} \left(\frac{1}{3}\right)^x = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
- Right-hand limit as [tex]\( x \)[/tex] approaches 1 from the right:
[tex]\[ \lim_{x \to 1^+} (-x^2 + 2x - 1) = -(1)^2 + 2(1) - 1 = 0 \][/tex]
For the function to be continuous at [tex]\( x = 1 \)[/tex], the left-hand limit must equal the right-hand limit. Here, they are [tex]\( \frac{1}{3} \)[/tex] and 0, respectively, which are not equal. Thus, [tex]\( f \)[/tex] is not continuous at [tex]\( x = 1 \)[/tex]. This statement is false.
### Statement: The value of [tex]\( f(1) \)[/tex] is 2.
From the definition of the function:
[tex]\[ f(1) = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
The value of [tex]\( f(1) \)[/tex] is [tex]\( \frac{1}{3} \)[/tex], not 2. Therefore, this statement is false.
### Summary
The correct statements are:
- The domain of function [tex]\( f \)[/tex] is all real numbers.
- Function [tex]\( f \)[/tex] is decreasing over the entire domain.
[tex]\[ f(x)=\left\{\begin{array}{ll} \left(\frac{1}{3}\right)^x, & x \leq 1 \\ -x^2 + 2x - 1, & x > 1 \end{array}\right. \][/tex]
### Statement: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
To see what happens as [tex]\( x \)[/tex] approaches positive infinity, we examine the function [tex]\( -x^2 + 2x - 1 \)[/tex] for large values of [tex]\( x \)[/tex] (since this is the part of the piecewise function used when [tex]\( x > 1 \)[/tex]).
[tex]\[ \lim_{x \to \infty} (-x^2 + 2x - 1) = -\infty \][/tex]
This shows that as [tex]\( x \)[/tex] approaches positive infinity, the function [tex]\( f(x) \)[/tex] approaches negative infinity, not positive infinity. Therefore, this statement is false.
### Statement: The domain of function [tex]\( f \)[/tex] is all real numbers.
The given function is defined for all real numbers [tex]\( x \)[/tex]. There are no restrictions on the values that [tex]\( x \)[/tex] can take. Therefore, this statement is true.
### Statement: Function [tex]\( f \)[/tex] is decreasing over the entire domain.
To determine if the function is decreasing over the entire domain, we look at both parts of the piecewise function:
1. For [tex]\( x \leq 1 \)[/tex], [tex]\( \left(\frac{1}{3}\right)^x \)[/tex] is a decreasing exponential function because [tex]\( \frac{1}{3} < 1 \)[/tex].
2. For [tex]\( x > 1 \)[/tex], [tex]\( -x^2 + 2x - 1 \)[/tex] is a quadratic function that opens downwards (concave down). We can determine the vertex of this parabola:
[tex]\[ x = \frac{-b}{2a} = \frac{-2}{2(-1)} = 1 \][/tex]
Since the vertex occurs at [tex]\( x = 1 \)[/tex] and for [tex]\( x > 1 \)[/tex], the parabola is decreasing. This means [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex] is decreasing for [tex]\( x > 1 \)[/tex].
Therefore, because both parts of the function are decreasing in their respective domains, the function [tex]\( f \)[/tex] is decreasing over the entire domain. This statement is true.
### Statement: Function [tex]\( f \)[/tex] is continuous.
To check the continuity at [tex]\( x = 1 \)[/tex], we need to ensure the left-hand limit and right-hand limit at [tex]\( x = 1 \)[/tex] are equal, and that they equal [tex]\( f(1) \)[/tex]:
- Left-hand limit as [tex]\( x \)[/tex] approaches 1 from the left:
[tex]\[ \lim_{x \to 1^-} \left(\frac{1}{3}\right)^x = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
- Right-hand limit as [tex]\( x \)[/tex] approaches 1 from the right:
[tex]\[ \lim_{x \to 1^+} (-x^2 + 2x - 1) = -(1)^2 + 2(1) - 1 = 0 \][/tex]
For the function to be continuous at [tex]\( x = 1 \)[/tex], the left-hand limit must equal the right-hand limit. Here, they are [tex]\( \frac{1}{3} \)[/tex] and 0, respectively, which are not equal. Thus, [tex]\( f \)[/tex] is not continuous at [tex]\( x = 1 \)[/tex]. This statement is false.
### Statement: The value of [tex]\( f(1) \)[/tex] is 2.
From the definition of the function:
[tex]\[ f(1) = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
The value of [tex]\( f(1) \)[/tex] is [tex]\( \frac{1}{3} \)[/tex], not 2. Therefore, this statement is false.
### Summary
The correct statements are:
- The domain of function [tex]\( f \)[/tex] is all real numbers.
- Function [tex]\( f \)[/tex] is decreasing over the entire domain.
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