in exercises 53 - 58, evaluate each piece - wise function at the given values of the independent…

Explanation:

Step1: Evaluate (f(-2)) for (f(x)=\begin{cases}3x + 5& \text{if }x<0\4x + 7& \text{if }x\geq0\end{cases})

Since (-2<0), use (f(x)=3x + 5). Substitute (x=-2) into (3x + 5): (3\times(-2)+5=-6 + 5=-1).

Step2: Evaluate (f(0)) for (f(x)=\begin{cases}3x + 5& \text{if }x<0\4x + 7& \text{if }x\geq0\end{cases})

Since (0\geq0), use (f(x)=4x + 7). Substitute (x = 0) into (4x+7): (4\times0+7=7).

Step3: Evaluate (f(3)) for (f(x)=\begin{cases}3x + 5& \text{if }x<0\4x + 7& \text{if }x\geq0\end{cases})

Since (3\geq0), use (f(x)=4x + 7). Substitute (x = 3) into (4x+7): (4\times3+7=12 + 7=19).

Answer:

a. (f(-2)=-1) b. (f(0)=7) c. (f(3)=19)

(We can follow the same - step - by - step process for the remaining piece - wise functions in the exercise set. For example, for (f(x)=\begin{cases}6x-1& \text{if }x<0\7x + 3& \text{if }x\geq0\end{cases}):

Step1: Evaluate (f(-3))

Since (-3<0), use (f(x)=6x-1). Substitute (x=-3) into (6x - 1): (6\times(-3)-1=-18-1=-19).

Step2: Evaluate (f(0))

Since (0\geq0), use (f(x)=7x + 3). Substitute (x = 0) into (7x+3): (7\times0+3=3).

Step3: Evaluate (f(4))

Since (4\geq0), use (f(x)=7x + 3). Substitute (x = 4) into (7x+3): (7\times4+3=28 + 3=31).

Answer for 54: a. (f(-3)=-19) b. (f(0)=3) c. (f(4)=31)

For (g(x)=\begin{cases}x + 3& \text{if }x\geq - 3\-(x + 3)& \text{if }x<-3\end{cases}):

Step1: Evaluate (g(0))

Since (0\geq - 3), use (g(x)=x + 3). Substitute (x = 0) into (x + 3): (0+3=3).

Step2: Evaluate (g(-6))

Since (-6<-3), use (g(x)=-(x + 3)). Substitute (x=-6) into (-(x + 3)): (-(-6 + 3)=-(-3)=3).

Step3: Evaluate (g(-3))

Since (-3\geq - 3), use (g(x)=x + 3). Substitute (x=-3) into (x + 3): (-3+3=0).

Answer for 55: a. (g(0)=3) b. (g(-6)=3) c. (g(-3)=0)

For (g(x)=\begin{cases}x + 5& \text{if }x\geq - 5\-(x + 5)& \text{if }x<-5\end{cases}):

Step1: Evaluate (g(0))

Since (0\geq - 5), use (g(x)=x + 5). Substitute (x = 0) into (x + 5): (0+5=5).

Step2: Evaluate (g(-6))

Since (-6<-5), use (g(x)=-(x + 5)). Substitute (x=-6) into (-(x + 5)): (-(-6 + 5)=-(-1)=1).

Step3: Evaluate (g(-5))

Since (-5\geq - 5), use (g(x)=x + 5). Substitute (x=-5) into (x + 5): (-5+5=0).

Answer for 56: a. (g(0)=5) b. (g(-6)=1) c. (g(-5)=0)

For (h(x)=\begin{cases}\frac{x^{2}-9}{x - 3}& \text{if }x\neq3\6& \text{if }x = 3\end{cases}):

Step1: Evaluate (h(5))

Since (5\neq3), use (h(x)=\frac{x^{2}-9}{x - 3}=\frac{(x + 3)(x - 3)}{x - 3}=x + 3) (for (x\neq3)). Substitute (x = 5) into (x + 3): (5+3=8).

Step2: Evaluate (h(0))

Since (0\neq3), use (h(x)=\frac{x^{2}-9}{x - 3}=x + 3). Substitute (x = 0) into (x + 3): (0+3=3).

Step3: Evaluate (h(3))

Since (x = 3), use (h(x)=6).

Answer for 57: a. (h(5)=8) b. (h(0)=3) c. (h(3)=6)

For (h(x)=\begin{cases}\frac{x^{2}-25}{x - 5}& \text{if }x\neq5\10& \text{if }x = 5\end{cases}):

Step1: Evaluate (h(7))

Since (7\neq5), use (h(x)=\frac{x^{2}-25}{x - 5}=\frac{(x + 5)(x - 5)}{x - 5}=x + 5) (for (x\neq5)). Substitute (x = 7) into (x + 5): (7+5=12).

Step2: Evaluate (h(0))

Since (0\neq5), use (h(x)=\frac{x^{2}-25}{x - 5}=x + 5). Substitute (x = 0) into (x + 5): (0+5=5).

Step3: Evaluate (h(5))

Since (x = 5), use (h(x)=10).

Answer for 58: a. (h(7)=12) b. (h(0)=5) c. (h(5)=10))