select the correct answer. consider function f. f(x) = { 2^x, x < 0; -x^2 - 4x + 1, 0 < x < 2; 1/2x + 3, x >…

Answer:

B. The domain is all real numbers.

Explanation:

Step1: Analyze domain of each part

The first - part $y = 2^{x}$ is defined for $x<0$, the second - part $y=-x^{2}-4x + 1$ is defined for $0 < x<2$, and the third - part $y=\frac{1}{2}x + 3$ is defined for $x>2$. Combining these intervals, the domain of $f(x)$ is $(-\infty,0)\cup(0,2)\cup(2,\infty)\cup{0}\cup{2}$, which is all real numbers $\mathbb{R}$.

Step2: Analyze option A

For $y = 2^{x}$ when $x<0$, it is increasing. For $y=-x^{2}-4x + 1=-(x + 2)^{2}+5$ when $0 < x<2$, its derivative $y^\prime=-2x - 4<0$ for $0 < x<2$, so it is decreasing. So the function is not increasing over its entire domain.

Step3: Analyze option C

As $x\rightarrow+\infty$, for $y=\frac{1}{2}x + 3$, $y\rightarrow+\infty$, but considering the part $y=-x^{2}-4x + 1$ for $0 < x<2$, the function does not have a simple increasing - to - infinity behavior over the whole domain.

Step4: Analyze option D

Check the continuity at $x = 0$: $\lim_{x\rightarrow0^{-}}2^{x}=1$, $\lim_{x\rightarrow0^{+}}-x^{2}-4x + 1 = 1$, but the function is not defined at $x = 0$. Also, check at $x = 2$: $\lim_{x\rightarrow2^{-}}-x^{2}-4x + 1=-4-8 + 1=-11$, $\lim_{x\rightarrow2^{+}}\frac{1}{2}x+3=1 + 3 = 4$. The function is not continuous.