Let's analyze the given function [tex]\( g(x) \)[/tex] which is defined as:
[tex]\[ g(x)=\left\{\begin{array}{ll}
\left(\frac{1}{2}\right)^{x-2}, & x<2 \\
x^3-9 x^2+27 x-25, & x \geq 2
\end{array}\right. \][/tex]
We need to determine which of the following statements about function [tex]\( g \)[/tex] are true.
1. Function [tex]\( g \)[/tex] includes an exponential piece and a quadratic piece.
The function [tex]\( g \)[/tex] is piecewise defined. For [tex]\( x < 2 \)[/tex], it is given by [tex]\( \left(\frac{1}{2}\right)^{x-2} \)[/tex], which is an exponential function. For [tex]\( x \geq 2 \)[/tex], it is given by [tex]\( x^3 - 9x^2 + 27x - 25 \)[/tex], which is a polynomial function (specifically cubic).
Therefore, the statement is True.
2. Function [tex]\( g \)[/tex] is increasing over the entire domain.
To determine if [tex]\( g \)[/tex] is increasing over the entire domain, we need to check the behavior of both pieces of the function:
- For [tex]\( x < 2 \)[/tex], [tex]\( g(x) = \left(\frac{1}{2}\right)^{x-2} \)[/tex] is an exponential function with a base less than 1, which means it is decreasing.
- For [tex]\( x \geq 2 \)[/tex], [tex]\( g(x) = x^3 - 9x^2 + 27x - 25 \)[/tex] is a cubic polynomial, and cubic polynomials can have regions where they increase and regions where they decrease.
Since [tex]\( g(x) \)[/tex] is not increasing in the region [tex]\( x < 2 \)[/tex] and the behavior in [tex]\( x \geq 2 \)[/tex] can also include non-increasing intervals, this statement is False.
3. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
Looking at the behavior of [tex]\( g \)[/tex] for [tex]\( x \geq 2 \)[/tex]:
- [tex]\( g(x) = x^3 - 9x^2 + 27x - 25 \)[/tex]
As [tex]\( x \to \infty \)[/tex], the term [tex]\( x^3 \)[/tex] dominates the polynomial, causing the function to approach [tex]\( \infty \)[/tex].
Therefore, this statement is True.
4. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
For [tex]\( x < 2 \)[/tex]:
- [tex]\( g(x) = \left(\frac{1}{2}\right)^{x-2} \)[/tex]
As [tex]\( x \to -\infty \)[/tex], the exponent [tex]\( x - 2 \)[/tex] becomes very large negatively, which means [tex]\( \left(\frac{1}{2}\right)^{x-2} \to \infty \)[/tex].
Therefore, this statement is False.
5. Function [tex]\( g \)[/tex] is continuous.
To check the continuity of [tex]\( g \)[/tex] at [tex]\( x = 2 \)[/tex], we need to ensure that the left-hand limit as [tex]\( x \)[/tex] approaches 2 from the left is equal to the right-hand limit as [tex]\( x \)[/tex] approaches 2 from the right and that these are equal to [tex]\( g(2) \)[/tex].
- Left-hand limit:
[tex]\[
\lim_{x \to 2^-} \left(\frac{1}{2}\right)^{x-2} = \left(\frac{1}{2}\right)^{2-2} = 1
\][/tex]
- Right-hand limit:
[tex]\[
\lim_{x \to 2^+} (x^3 - 9x^2 + 27x - 25) = 2^3 - 9 \cdot 2^2 + 27 \cdot 2 - 25 = 1
\][/tex]
- [tex]\( g(2) \)[/tex]:
[tex]\[
g(2) = 2^3 - 9 \cdot 2^2 + 27 \cdot 2 - 25 = 1
\][/tex]
Since the left-hand limit, right-hand limit, and [tex]\( g(2) \)[/tex] are all equal, [tex]\( g \)[/tex] is continuous at [tex]\( x = 2 \)[/tex] and thus continuous over its entire domain.
Therefore, this statement is True.
Finally, the following statements are true:
- Function [tex]\( g \)[/tex] includes an exponential piece and a quadratic piece. (True)
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity. (True)
- Function [tex]\( g \)[/tex] is continuous. (True)
