select all the correct answers.\nwhich statements are true about function g?\n$$ g ( x ) = left{…

select all the correct answers.\nwhich statements are true about function g?\n$$ g ( x ) = left{ \begin{array} { l l } { ( \frac { 1 } { 2 } ) ^ { x - 2 } , } & { x < 2 } \\ { x ^ { 3 } - 9 x ^ { 2 } + 27 x - 25 , } & { x geq 2 } end{array} \right. $$\nfunction g is continuous.\nfunction g includes an exponential piece and a quadratic piece.\nas x approaches positive infinity, g ( x ) approaches positive infinity.\nas x approaches negative infinity, g ( x ) approaches positive infinity.\nfunction g is increasing over the entire domain.

select all the correct answers.\nwhich statements are true about function g?\n$$ g ( x ) = left{ \begin{array} { l l } { ( \frac { 1 } { 2 } ) ^ { x - 2 } , } & { x < 2 } \\ { x ^ { 3 } - 9 x ^ { 2 } + 27 x - 25 , } & { x geq 2 } end{array} \right. $$\nfunction g is continuous.\nfunction g includes an exponential piece and a quadratic piece.\nas x approaches positive infinity, g ( x ) approaches positive infinity.\nas x approaches negative infinity, g ( x ) approaches positive infinity.\nfunction g is increasing over the entire domain.

Answer

Explanation:

Step1: Check continuity at (x = 2)

  • Left - hand limit: (\lim_{x\rightarrow2^{-}}g(x)=\left(\frac{1}{2}\right)^{2 - 2}=1)
  • Right - hand limit: (\lim_{x\rightarrow2^{+}}g(x)=2^{3}-9\times2^{2}+27\times2 - 25=8-36 + 54-25 = 1)
  • (g(2)=2^{3}-9\times2^{2}+27\times2 - 25 = 1) Since (\lim_{x\rightarrow2^{-}}g(x)=\lim_{x\rightarrow2^{+}}g(x)=g(2)), the function is continuous.

Step2: Analyze the types of pieces

The first piece (y = (\frac{1}{2})^{x - 2}) is an exponential function. The second piece (y=x^{3}-9x^{2}+27x - 25) is a cubic function (not quadratic).

Step3: Analyze the end - behavior as (x\rightarrow+\infty)

For (x\geq2), (g(x)=x^{3}-9x^{2}+27x - 25). As (x\rightarrow+\infty), the leading term (x^{3}) dominates. Since the coefficient of (x^{3}) is (1>0), (\lim_{x\rightarrow+\infty}g(x)=+\infty)

Step4: Analyze the end - behavior as (x\rightarrow-\infty)

For (x < 2), (g(x)=(\frac{1}{2})^{x - 2}=2^{2 - x}). As (x\rightarrow-\infty), (2 - x\rightarrow+\infty), so (\lim_{x\rightarrow-\infty}g(x)=+\infty)

Step5: Analyze the increasing/decreasing nature

  • For (x < 2), (g(x)=(\frac{1}{2})^{x - 2}), and (g^{\prime}(x)=(\frac{1}{2})^{x - 2}\ln(\frac{1}{2})<0) (function is decreasing for (x < 2))
  • For (x\geq2), (g^{\prime}(x)=3x^{2}-18x + 27=3(x^{2}-6x + 9)=3(x - 3)^{2}\geq0) (function is non - decreasing for (x\geq2))

Answer:

Function (g) is continuous, As (x) approaches positive infinity, (g(x)) approaches positive infinity, As (x) approaches negative infinity, (g(x)) approaches positive infinity.