Question

complete the following for ( f(x) = \begin{cases} 5x & \text{if } 0 leq x < 3 \\ x + 2 & \text{if } 3 leq x leq 5 end{cases} )
(a) determine the domain of ( f ).
(b) evaluate ( f(-1) ), ( f(1) ), and ( f(6) ).
(c) graph ( f ).
(d) is ( f ) continuous on its domain?

(a) choose the correct domain of ( f ) below.
○ a. ( square leq x leq square )
● b. all real numbers

Explanation:

Part (a)

Step1: Analyze the piecewise function's intervals

The function \( f(x) \) is defined in two pieces: one for \( 0 \leq x < 3 \) and another for \( 3 \leq x \leq 5 \). To find the domain, we combine these intervals.
The first interval starts at \( x = 0 \) (inclusive) and goes up to but not including \( x = 3 \). The second interval starts at \( x = 3 \) (inclusive) and goes up to \( x = 5 \) (inclusive). Combining these, the domain is from \( x = 0 \) to \( x = 5 \), inclusive. So the domain is \( 0 \leq x \leq 5 \), which means option A is correct with the first box as \( 0 \) and the second box as \( 5 \).

Part (b)

Step1: Evaluate \( f(-1) \)

The function \( f(x) \) is defined for \( 0 \leq x < 3 \) and \( 3 \leq x \leq 5 \). Since \( -1 \) is not in either of these intervals, \( f(-1) \) is undefined.

Step2: Evaluate \( f(1) \)

Since \( 1 \) is in the interval \( 0 \leq x < 3 \), we use the first piece \( f(x)=5x \). Substitute \( x = 1 \): \( f(1)=5(1)=5 \).

Step3: Evaluate \( f(6) \)

Since \( 6 \) is not in the interval \( 3 \leq x \leq 5 \) (it's greater than \( 5 \)), \( f(6) \) is undefined.

Part (c)

Step1: Graph the first piece (\( 0 \leq x < 3 \), \( f(x)=5x \))

This is a linear function with slope \( 5 \). At \( x = 0 \), \( f(0)=5(0)=0 \) (closed dot). At \( x = 3 \), \( f(3)=5(3)=15 \) (open dot, since \( x = 3 \) is not included in this piece). Draw a line from \( (0, 0) \) to \( (3, 15) \) with a closed dot at \( (0, 0) \) and open dot at \( (3, 15) \).

Step2: Graph the second piece (\( 3 \leq x \leq 5 \), \( f(x)=x + 2 \))

This is a linear function with slope \( 1 \). At \( x = 3 \), \( f(3)=3 + 2 = 5 \) (closed dot, since \( x = 3 \) is included here). At \( x = 5 \), \( f(5)=5 + 2 = 7 \) (closed dot). Draw a line from \( (3, 5) \) to \( (5, 7) \) with closed dots at both endpoints.

Part (d)

Answer:

Step1: Check continuity at \( x = 3 \)

For a function to be continuous at a point, the left - hand limit, right - hand limit, and the function value at that point must be equal.

• Left - hand limit as \( x \to 3^{-} \): Use the first piece \( f(x)=5x \). \( \lim_{x\to 3^{-}}f(x)=\lim_{x\to 3^{-}}5x = 5\times3=15 \).
• Right - hand limit as \( x \to 3^{+} \): Use the second piece \( f(x)=x + 2 \). \( \lim_{x\to 3^{+}}f(x)=\lim_{x\to 3^{+}}(x + 2)=3 + 2 = 5 \).
• Function value at \( x = 3 \): \( f(3)=3 + 2 = 5 \).

Since the left - hand limit (\( 15 \)) is not equal to the right - hand limit (\( 5 \)) (and also not equal to the function value at \( x = 3 \)), the function is not continuous at \( x = 3 \). Since \( x = 3 \) is in the domain of \( f \), the function \( f \) is not continuous on its domain.

Final Answers

(a) The domain is \( 0\leq x\leq5 \), so the first box is \( 0 \) and the second box is \( 5 \).

(b) \( f(-1) \) is undefined, \( f(1)=5 \), \( f(6) \) is undefined.

(c) Graph as described in the steps above.

(d) No, \( f \) is not continuous on its domain.