Answer :
To determine which of the given piecewise functions is continuous across its entire domain, we need to check the continuity at the points where the definition of the function changes, specifically at the points [tex]\( x = -2 \)[/tex] and [tex]\( x = 4 \)[/tex].
### Function A
[tex]\[ f_A(x)=\left\{ \begin{array}{ll} x-2, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 25-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_A(-2) = -2 - 2 = -4 \)[/tex]
- From the right: [tex]\( f_A(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\(-4 \neq 2\)[/tex], [tex]\( f_A(x) \)[/tex] is not continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_A(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_A(4) = 25 - 3(4) = 13 \)[/tex]
- Since [tex]\( 8 \neq 13\)[/tex], [tex]\( f_A(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex].
### Function B
[tex]\[ f_B(x)=\left\{ \begin{array}{ll} x+6, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 20-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_B(-2) = -2 + 6 = 4 \)[/tex]
- From the right: [tex]\( f_B(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\( 4 \neq 2\)[/tex], [tex]\( f_B(x) \)[/tex] is not continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_B(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_B(4) = 20 - 3(4) = 8 \)[/tex]
- Since [tex]\( 8 = 8 \)[/tex], [tex]\( f_B(x) \)[/tex] is continuous at [tex]\( x = 4 \)[/tex].
### Function C
[tex]\[ f_C(x)=\left\{ \begin{array}{ll} x+4, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 20-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_C(-2) = -2 + 4 = 2 \)[/tex]
- From the right: [tex]\( f_C(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\( 2 = 2 \)[/tex], [tex]\( f_C(x) \)[/tex] is continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_C(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_C(4) = 20 - 3(4) = 8 \)[/tex]
- Since [tex]\( 8 = 8 \)[/tex], [tex]\( f_C(x) \)[/tex] is continuous at [tex]\( x = 4 \)[/tex].
### Function D
[tex]\[ f_D(x)=\left\{ \begin{array}{ll} x+4, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 25-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_D(-2) = -2 + 4 = 2 \)[/tex]
- From the right: [tex]\( f_D(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\( 2 = 2 \)[/tex], [tex]\( f_D(x) \)[/tex] is continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_D(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_D(4) = 25 - 3(4) = 13 \)[/tex]
- Since [tex]\( 8 \neq 13\)[/tex], [tex]\( f_D(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex].
After checking continuity at the transition points for all piecewise functions, only function [tex]\( C \)[/tex] is continuous across its domain.
Therefore, the correct answer is:
c. [tex]\( f(x)=\left\{\begin{array}{ll}x+4, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 20-3 x, & 4 \leq x \leq 8\end{array}\right.\)[/tex]
### Function A
[tex]\[ f_A(x)=\left\{ \begin{array}{ll} x-2, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 25-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_A(-2) = -2 - 2 = -4 \)[/tex]
- From the right: [tex]\( f_A(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\(-4 \neq 2\)[/tex], [tex]\( f_A(x) \)[/tex] is not continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_A(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_A(4) = 25 - 3(4) = 13 \)[/tex]
- Since [tex]\( 8 \neq 13\)[/tex], [tex]\( f_A(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex].
### Function B
[tex]\[ f_B(x)=\left\{ \begin{array}{ll} x+6, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 20-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_B(-2) = -2 + 6 = 4 \)[/tex]
- From the right: [tex]\( f_B(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\( 4 \neq 2\)[/tex], [tex]\( f_B(x) \)[/tex] is not continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_B(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_B(4) = 20 - 3(4) = 8 \)[/tex]
- Since [tex]\( 8 = 8 \)[/tex], [tex]\( f_B(x) \)[/tex] is continuous at [tex]\( x = 4 \)[/tex].
### Function C
[tex]\[ f_C(x)=\left\{ \begin{array}{ll} x+4, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 20-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_C(-2) = -2 + 4 = 2 \)[/tex]
- From the right: [tex]\( f_C(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\( 2 = 2 \)[/tex], [tex]\( f_C(x) \)[/tex] is continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_C(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_C(4) = 20 - 3(4) = 8 \)[/tex]
- Since [tex]\( 8 = 8 \)[/tex], [tex]\( f_C(x) \)[/tex] is continuous at [tex]\( x = 4 \)[/tex].
### Function D
[tex]\[ f_D(x)=\left\{ \begin{array}{ll} x+4, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 25-3 x, & 4 \leq x \leq 8 \end{array} \right. \][/tex]
- At [tex]\( x = -2 \)[/tex]:
- From the left: [tex]\( f_D(-2) = -2 + 4 = 2 \)[/tex]
- From the right: [tex]\( f_D(-2) = 0.5(-2)^2 = 2 \)[/tex]
- Since [tex]\( 2 = 2 \)[/tex], [tex]\( f_D(x) \)[/tex] is continuous at [tex]\( x = -2 \)[/tex].
- At [tex]\( x = 4 \)[/tex]:
- From the left: [tex]\( f_D(4) = 0.5(4)^2 = 8 \)[/tex]
- From the right: [tex]\( f_D(4) = 25 - 3(4) = 13 \)[/tex]
- Since [tex]\( 8 \neq 13\)[/tex], [tex]\( f_D(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex].
After checking continuity at the transition points for all piecewise functions, only function [tex]\( C \)[/tex] is continuous across its domain.
Therefore, the correct answer is:
c. [tex]\( f(x)=\left\{\begin{array}{ll}x+4, & -4 \leq x<-2 \\ 0.5 x^2, & -2 \leq x<4 \\ 20-3 x, & 4 \leq x \leq 8\end{array}\right.\)[/tex]
