Select all the correct answers.

Consider the function [tex]g[/tex].
[tex]
g(x)=\left\{\begin{array}{ll}
\left(\frac{1}{2}\right)^x + 3, & x \ \textless \ 0 \\
-x^2 + 2, & x \geq 0
\end{array}\right.
[/tex]

Which statements are true about the function [tex]g[/tex]?

A. The function is continuous.
B. The [tex]y[/tex]-intercept is 2.
C. As [tex]x[/tex] approaches positive infinity, [tex]g(x)[/tex] approaches positive infinity.
D. The function is decreasing over its domain except for when [tex]x = 0[/tex].
E. The domain is all real numbers.



Answer :

Let's analyze the given function [tex]\( g \)[/tex]:

[tex]\[ g(x) = \left\{ \begin{array}{ll} \left(\frac{1}{2}\right)^x + 3, & x < 0 \\ -x^2 + 2, & x \geq 0 \end{array} \right. \][/tex]

1. Continuity of the function:
To check the continuity of [tex]\( g \)[/tex] at [tex]\( x = 0 \)[/tex], we need to verify if the left-hand limit and the right-hand limit at [tex]\( x = 0 \)[/tex] are equal, and if they are equal to [tex]\( g(0) \)[/tex].
- For [tex]\( x < 0 \)[/tex]: The left-hand limit is [tex]\( \left(\frac{1}{2}\right)^0 + 3 = 1 + 3 = 4 \)[/tex].
- For [tex]\( x \geq 0 \)[/tex]: The right-hand limit is [tex]\( -0^2 + 2 = 0 + 2 = 2 \)[/tex].

Since the left-hand limit (4) is not equal to the right-hand limit (2), the function is not continuous at [tex]\( x = 0 \)[/tex].
Therefore, the statement "The function is continuous" is false.

2. [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept is the value of [tex]\( g(x) \)[/tex] when [tex]\( x = 0 \)[/tex]. From the function definition for [tex]\( x \geq 0 \)[/tex],
[tex]\[ g(0) = -0^2 + 2 = 2. \][/tex]
Therefore, the statement "The [tex]\( y \)[/tex]-intercept is 2" is true.

3. Behavior as [tex]\( x \)[/tex] approaches positive infinity:
For [tex]\( x \geq 0 \)[/tex], [tex]\( g(x) = -x^2 + 2 \)[/tex].
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( -x^2 \)[/tex] will dominate, causing [tex]\( g(x) \)[/tex] to approach negative infinity.

Therefore, the statement "As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity" is false.

4. Decreasing nature of the function:
- For [tex]\( x < 0 \)[/tex], [tex]\( \left(\frac{1}{2}\right)^x \)[/tex] is a decreasing exponential function, so [tex]\( \left(\frac{1}{2}\right)^x + 3 \)[/tex] also decreases with [tex]\( x \)[/tex].
- For [tex]\( x \geq 0 \)[/tex], [tex]\( -x^2 + 2 \)[/tex] is a downward parabola, which is decreasing for all [tex]\( x > 0 \)[/tex]. At [tex]\( x = 0 \)[/tex], the function reaches its maximum value of 2.

Therefore, the statement "The function is decreasing over its domain except for when [tex]\( x = 0 \)[/tex]" is true.

5. Domain of the function:
The function is defined for all real numbers [tex]\( x \)[/tex] in both cases, [tex]\( x < 0 \)[/tex] and [tex]\( x \geq 0 \)[/tex].

Therefore, the statement "The domain is all real numbers" is true.

In summary, the true statements about the function [tex]\( g \)[/tex] are:
- The [tex]\( y \)[/tex]-intercept is 2.
- The function is decreasing over its domain except for when [tex]\( x = 0 \)[/tex].
- The domain is all real numbers.