Consider the function: f(x) = StartLayout enlarged left-brace 1st Row 1st column seven-halves + 2x, 2nd column x less-than-or-equal-to negative 1 2nd row 1st column negative 5 + StartFraction 3 x Over 2 EndFraction, 2nd column Negative 1 less-than x less-than 3 Third row 1st column one-fourth x, 2nd column x greater-than-or-equal-to 3 EndLayout A number line goes from negative 5 to 5.What are these values?

Question

f(−3) =___
f(−1) =___
f(3) =___

all the answers are fractions by the way the blanks

Answer 1

To evaluate the function \( f(x) \) for the given values, we need to determine which piece of the piecewise function to use based on the value of \( x \).

The function \( f(x) \) is defined as follows:

\[ f(x) = \begin{cases} \frac{7}{2} + 2x & \text{for } x \leq -1 \ -5 + \frac{3x}{2} & \text{for } -1 < x < 3 \ \frac{1}{4}x & \text{for } x \geq 3 \end{cases} \]

Now let's evaluate \( f(-3) \), \( f(-1) \), and \( f(3) \).

1. Finding \( f(-3) \):

Since \( -3 \leq -1 \), we use the first case: \[ f(-3) = \frac{7}{2} + 2(-3) = \frac{7}{2} - 6 = \frac{7}{2} - \frac{12}{2} = \frac{-5}{2} \]

2. Finding \( f(-1) \):

Since \( -1 \) is at the boundary and \( x \leq -1 \), we use the first case again: \[ f(-1) = \frac{7}{2} + 2(-1) = \frac{7}{2} - 2 = \frac{7}{2} - \frac{4}{2} = \frac{3}{2} \]

3. Finding \( f(3) \):

Since \( 3 \geq 3 \), we use the third case: \[ f(3) = \frac{1}{4}(3) = \frac{3}{4} \]

Putting it all together, we have:

• \( f(-3) = \frac{-5}{2} \)
• \( f(-1) = \frac{3}{2} \)
• \( f(3) = \frac{3}{4} \)

So the final answers are:

• \( f(-3) = \frac{-5}{2} \)
• \( f(-1) = \frac{3}{2} \)
• \( f(3) = \frac{3}{4} \)

Answer 2

Which graph represents the piecewise-defined function f(x) = StartLayout enlarged left-brace 1st Row 1st column negative 1.5 x + 3.5, 2nd column x less-than 2 2nd row 1st column 4 + x, 2nd column x greater-than-or-equal-to 2 EndLayout? 1. On a coordinate plane, a piecewise function has 2 lines. The first line has an open circle at (2, 0) and then goes up through (1, 2) with an arrow instead of an endpoint. The second line has a closed circle at (2, 6) and goes up through (3, 7) with an arrow instead of an endpoint. 2. On a coordinate plane, a piecewise function has 2 lines. The first line has a closed circle at (2, 0) and then goes up through (1, 2) with an arrow instead of an endpoint. The second line has an open circle at (2, 6) and then goes up through (3, 6) with an arrow instead of an endpoint. 3. On a coordinate plane, a piecewise function has 2 lines. The first line has a closed circle at (2, 6) then goes down through (1, 5) with an arrow instead of an endpoint. The second line has an open circle at (2, 0) and goes up through (5, 4) with an arrow instead of an endpoint. 4. On a coordinate plane, a piecewise function has 2 lines. The first line has an open circle at (2, 6) and then goes down through (1, 5) with an arrow instead of an endpoint. The second line has a closed circle at (2, 0) and then goes up through (5, 4) with an arrow instead of an endpoint.

Answer 3

To determine which graph represents the piecewise-defined function

\[ f(x) = \begin{cases} -1.5x + 3.5 & \text{for } x < 2 \ 4 + x & \text{for } x \geq 2 \end{cases} \]

we will analyze the two parts of the function.

1. For \( x < 2 \):

The equation is \( f(x) = -1.5x + 3.5 \).

• The y-intercept (when \( x = 0 \)) is \( f(0) = 3.5 \).
• When \( x = 2 \), \( f(2) = -1.5(2) + 3.5 = -3 + 3.5 = 0.5 \).
• However, at \( x = 2 \) the function is not defined, so we will represent this point with an open circle at \( (2, 0.5) \).

2. For \( x \geq 2 \):

The equation is \( f(x) = 4 + x \).

• When \( x = 2 \), \( f(2) = 4 + 2 = 6 \).
• Therefore, this point should be represented with a closed circle at \( (2, 6) \).
• As \( x \) increases, the line will continue upwards since the slope is positive.

Now, we can summarize the key points:

• At \( x = 2 \), there is a closed circle at \( (2, 6) \) for the second piece.
• For \( x < 2 \), the line approaches \( (2, 0.5) \) but does not touch it (open circle).

Now let's check the options:

1. Option 1: Open circle at \( (2, 0) \); closed circle at \( (2, 6) \) - This is not correct.

2. Option 2: Closed circle at \( (2, 0) \) and open circle at \( (2, 6) \) - This is not correct.

3. Option 3: Closed circle at \( (2, 6) \) and open circle at \( (2, 0) \) - This is not correct.

4. Option 4: Open circle at \( (2, 6) \) and closed circle at \( (2, 0) \) - This is not correct.

It seems like none of the options are correct based on our analysis.

The correct representation based on the definitions of the pieces would have a closed circle at \( (2, 6) \) for the second piece and an open circle at \( (2, 0.5) \) for the limit of the first piece.

If you meant the correct function values through the analysis, please consider that there may be some inconsistencies in the options as described. The open and closed circles are mixed up based on the function definitions. If you have further context or a need for a follow-up, please provide the details!

Answer 4

its option 1

Answer 5

Thank you for your patience. Let’s clarify the function \( f(x) \) and how it relates to Option 1.

The piecewise function we are considering is:

\[ f(x) = \begin{cases} -1.5x + 3.5 & \text{for } x < 2 \ 4 + x & \text{for } x \geq 2 \end{cases} \]

1. For \( x < 2 \) (use \( f(x) = -1.5x + 3.5 \)):

   • This is a line with a slope of -1.5 and a y-intercept of 3.5.
   • As \( x \) approaches 2 from the left, we can find the value: \[ f(2) = -1.5(2) + 3.5 = -3 + 3.5 = 0.5. \]
   • Since \( f(x) \) is defined for \( x < 2 \), the point \( (2, 0.5) \) should be represented with an open circle.

2. For \( x \geq 2 \) (use \( f(x) = 4 + x \)):

   • This line starts at \( x = 2 \): \[ f(2) = 4 + 2 = 6. \]
   • Since \( f(x) \) is defined for \( x \geq 2 \), this point should be represented with a closed circle at \( (2, 6) \).

Given this analysis:

• Closed circle at \( (2, 6) \) for the line \( f(x) = 4 + x \) (for \( x \geq 2 \)).
• Open circle at \( (2, 0.5) \) for the line \( f(x) = -1.5x + 3.5 \) (for \( x < 2 \)).

Option 1 states:

• Open circle at (2, 0) and goes through (1, 2) (which actually does not match our calculation for the first piece).
• Closed circle at (2, 6) and continues upwards through (3, 7).

It appears that perhaps the y-coordinate at \( (2, 0) \) in Option 1 is incorrectly labeled when it should align with our calculated limit of \( 0.5 \).

Although we both recognize that Option 1 signifies the general style of piecewise representation for the second part is right with the closed circle at \( (2, 6) \), the exact values and proper identification of the points could be misleading.

Conclusion:

Given the choices, if Option 1 was presented with an open circle at a point which should represent a non-valid transition (like \( (2, 0) \)), it might still convey the sense of the defined sections visually with the line types correctly.

If that was the intended visual or standard for your exam or text, it seems there may have been a misinterpretation around what constituted the function values accurately at those critical points. Thank you for pointing that out, and I'm glad you confirmed Option 1 aligns in some visual aspect even if the values deviate slightly on workings!