Answer :
Let's analyze each statement about the function [tex]\( g \)[/tex] in detail:
1. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
For [tex]\( x < 2 \)[/tex], the function [tex]\( g(x) = \left(\frac{1}{2}\right)^{x-2} \)[/tex]. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( x-2 \)[/tex] also approaches negative infinity. Since [tex]\( \left(\frac{1}{2}\right)^{x-2} \)[/tex] represents an exponential function with a base between 0 and 1, it tends to expand towards positive infinity as the exponent becomes more negative. Therefore, this statement is true.
2. Function [tex]\( g \)[/tex] is increasing over the entire domain.
- For [tex]\( x < 2 \)[/tex], [tex]\( g(x) = \left(\frac{1}{2}\right)^{x-2} \)[/tex] is a decreasing exponential function.
- For [tex]\( x \geq 2 \)[/tex], [tex]\( g(x) = x^3 - 9x^2 + 27x - 25 \)[/tex] is a cubic function. To determine if it's increasing over this interval, we need to analyze its derivative [tex]\( g'(x) = 3x^2 - 18x + 27 \)[/tex], which is a quadratic function that can have positive, negative, or zero values depending on [tex]\( x \)[/tex].
Given that the first piece is decreasing and the second piece can vary, the function [tex]\( g \)[/tex] is not monotonic (not consistently increasing or decreasing) over its entire domain. Therefore, this statement is false.
3. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
For [tex]\( x \geq 2 \)[/tex], [tex]\( g(x) = x^3 - 9x^2 + 27x - 25 \)[/tex]. As [tex]\( x \)[/tex] approaches positive infinity, the term [tex]\( x^3 \)[/tex] dominates, leading the whole function towards positive infinity. Therefore, this statement is true.
4. Function [tex]\( g \)[/tex] includes an exponential piece and a quadratic piece.
The function [tex]\( g(x) \)[/tex] for [tex]\( x < 2 \)[/tex] is an exponential function, and for [tex]\( x \geq 2 \)[/tex] is a cubic (not quadratic) polynomial. Hence, the statement refers mistakenly to a quadratic instead of a cubic function. Therefore, this statement is false.
5. Function [tex]\( g \)[/tex] is continuous.
To determine continuity at [tex]\( x = 2 \)[/tex], we need:
[tex]\[ \lim_{{x \to 2^-}} g(x) = \lim_{{x \to 2^+}} g(x) = g(2) \][/tex]
[tex]\[ \lim_{{x \to 2^-}} g(x) = \left(\frac{1}{2}\right)^{2-2} = \left(\frac{1}{2}\right)^0 = 1 \][/tex]
[tex]\[ \lim_{{x \to 2^+}} g(x) = 2^3 - 9 \cdot 2^2 + 27 \cdot 2 - 25 = 8 - 36 + 54 - 25 = 1 \][/tex]
Hence, [tex]\( \lim_{{x \to 2}} g(x) = g(2) = 1 \)[/tex], showing that the function is continuous at [tex]\( x = 2 \)[/tex]. Therefore, this statement is true.
In summary:
- The first statement is true.
- The second statement is false.
- The third statement is true.
- The fourth statement is false.
- The fifth statement is true.
1. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
For [tex]\( x < 2 \)[/tex], the function [tex]\( g(x) = \left(\frac{1}{2}\right)^{x-2} \)[/tex]. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( x-2 \)[/tex] also approaches negative infinity. Since [tex]\( \left(\frac{1}{2}\right)^{x-2} \)[/tex] represents an exponential function with a base between 0 and 1, it tends to expand towards positive infinity as the exponent becomes more negative. Therefore, this statement is true.
2. Function [tex]\( g \)[/tex] is increasing over the entire domain.
- For [tex]\( x < 2 \)[/tex], [tex]\( g(x) = \left(\frac{1}{2}\right)^{x-2} \)[/tex] is a decreasing exponential function.
- For [tex]\( x \geq 2 \)[/tex], [tex]\( g(x) = x^3 - 9x^2 + 27x - 25 \)[/tex] is a cubic function. To determine if it's increasing over this interval, we need to analyze its derivative [tex]\( g'(x) = 3x^2 - 18x + 27 \)[/tex], which is a quadratic function that can have positive, negative, or zero values depending on [tex]\( x \)[/tex].
Given that the first piece is decreasing and the second piece can vary, the function [tex]\( g \)[/tex] is not monotonic (not consistently increasing or decreasing) over its entire domain. Therefore, this statement is false.
3. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
For [tex]\( x \geq 2 \)[/tex], [tex]\( g(x) = x^3 - 9x^2 + 27x - 25 \)[/tex]. As [tex]\( x \)[/tex] approaches positive infinity, the term [tex]\( x^3 \)[/tex] dominates, leading the whole function towards positive infinity. Therefore, this statement is true.
4. Function [tex]\( g \)[/tex] includes an exponential piece and a quadratic piece.
The function [tex]\( g(x) \)[/tex] for [tex]\( x < 2 \)[/tex] is an exponential function, and for [tex]\( x \geq 2 \)[/tex] is a cubic (not quadratic) polynomial. Hence, the statement refers mistakenly to a quadratic instead of a cubic function. Therefore, this statement is false.
5. Function [tex]\( g \)[/tex] is continuous.
To determine continuity at [tex]\( x = 2 \)[/tex], we need:
[tex]\[ \lim_{{x \to 2^-}} g(x) = \lim_{{x \to 2^+}} g(x) = g(2) \][/tex]
[tex]\[ \lim_{{x \to 2^-}} g(x) = \left(\frac{1}{2}\right)^{2-2} = \left(\frac{1}{2}\right)^0 = 1 \][/tex]
[tex]\[ \lim_{{x \to 2^+}} g(x) = 2^3 - 9 \cdot 2^2 + 27 \cdot 2 - 25 = 8 - 36 + 54 - 25 = 1 \][/tex]
Hence, [tex]\( \lim_{{x \to 2}} g(x) = g(2) = 1 \)[/tex], showing that the function is continuous at [tex]\( x = 2 \)[/tex]. Therefore, this statement is true.
In summary:
- The first statement is true.
- The second statement is false.
- The third statement is true.
- The fourth statement is false.
- The fifth statement is true.
