Answer :
To find the values of [tex]\( f(x) \)[/tex] at specific points, we need to evaluate the given piecewise function at those points.
The function is defined as follows:
[tex]\[ f(x)= \begin{cases} \frac{7}{2} + 2x, & \text{if } x \leq -1 \\ -5 + \frac{3x}{2}, & \text{if } -1 < x < 3 \\ \frac{1}{4}x, & \text{if } x \geq 3 \end{cases} \][/tex]
Step-by-step, let's evaluate [tex]\( f(-3), f(-1), \)[/tex] and [tex]\( f(3) \)[/tex]:
1. Evaluating [tex]\( f(-3) \)[/tex]:
Since [tex]\( -3 \leq -1 \)[/tex], we use the first piece of the function:
[tex]\[ f(x) = \frac{7}{2} + 2x. \][/tex]
Plugging in [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = \frac{7}{2} + 2(-3) = \frac{7}{2} - 6. \][/tex]
To subtract, we need a common denominator:
[tex]\[ \frac{7}{2} = 3.5 \quad \text{and} \quad 6 = 6 \times \frac{2}{2} = \frac{12}{2}. \][/tex]
So,
[tex]\[ f(-3) = \frac{7}{2} - \frac{12}{2} = \frac{7 - 12}{2} = \frac{-5}{2} = -2.5. \][/tex]
Thus,
[tex]\[ f(-3) = -2.5. \][/tex]
2. Evaluating [tex]\( f(-1) \)[/tex]:
Since [tex]\( x = -1 \)[/tex] falls on the boundary [tex]\( x \leq -1 \)[/tex], we again use the first piece of the function:
[tex]\[ f(x) = \frac{7}{2} + 2x. \][/tex]
Plugging in [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{7}{2} + 2(-1) = \frac{7}{2} - 2. \][/tex]
Converting [tex]\( 2 \)[/tex] to a fraction:
[tex]\[ 2 = 2 \times \frac{2}{2} = \frac{4}{2}. \][/tex]
So,
[tex]\[ f(-1) = \frac{7}{2} - \frac{4}{2} = \frac{7 - 4}{2} = \frac{3}{2} = 1.5. \][/tex]
Thus,
[tex]\[ f(-1) = 1.5. \][/tex]
3. Evaluating [tex]\( f(3) \)[/tex]:
For [tex]\( x = 3 \)[/tex], we use the third piece of the function since [tex]\( x \geq 3 \)[/tex]:
[tex]\[ f(x) = \frac{1}{4}x. \][/tex]
Plugging in [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{4} \times 3 = \frac{3}{4}. \][/tex]
Converting to decimal form:
[tex]\[ \frac{3}{4} = 0.75. \][/tex]
Thus,
[tex]\[ f(3) = 0.75. \][/tex]
Summarizing the results:
[tex]\[ \begin{aligned} f(-3) &= -2.5, \\ f(-1) &= 1.5, \\ f(3) &= 0.75. \end{aligned} \][/tex]
The function is defined as follows:
[tex]\[ f(x)= \begin{cases} \frac{7}{2} + 2x, & \text{if } x \leq -1 \\ -5 + \frac{3x}{2}, & \text{if } -1 < x < 3 \\ \frac{1}{4}x, & \text{if } x \geq 3 \end{cases} \][/tex]
Step-by-step, let's evaluate [tex]\( f(-3), f(-1), \)[/tex] and [tex]\( f(3) \)[/tex]:
1. Evaluating [tex]\( f(-3) \)[/tex]:
Since [tex]\( -3 \leq -1 \)[/tex], we use the first piece of the function:
[tex]\[ f(x) = \frac{7}{2} + 2x. \][/tex]
Plugging in [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = \frac{7}{2} + 2(-3) = \frac{7}{2} - 6. \][/tex]
To subtract, we need a common denominator:
[tex]\[ \frac{7}{2} = 3.5 \quad \text{and} \quad 6 = 6 \times \frac{2}{2} = \frac{12}{2}. \][/tex]
So,
[tex]\[ f(-3) = \frac{7}{2} - \frac{12}{2} = \frac{7 - 12}{2} = \frac{-5}{2} = -2.5. \][/tex]
Thus,
[tex]\[ f(-3) = -2.5. \][/tex]
2. Evaluating [tex]\( f(-1) \)[/tex]:
Since [tex]\( x = -1 \)[/tex] falls on the boundary [tex]\( x \leq -1 \)[/tex], we again use the first piece of the function:
[tex]\[ f(x) = \frac{7}{2} + 2x. \][/tex]
Plugging in [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{7}{2} + 2(-1) = \frac{7}{2} - 2. \][/tex]
Converting [tex]\( 2 \)[/tex] to a fraction:
[tex]\[ 2 = 2 \times \frac{2}{2} = \frac{4}{2}. \][/tex]
So,
[tex]\[ f(-1) = \frac{7}{2} - \frac{4}{2} = \frac{7 - 4}{2} = \frac{3}{2} = 1.5. \][/tex]
Thus,
[tex]\[ f(-1) = 1.5. \][/tex]
3. Evaluating [tex]\( f(3) \)[/tex]:
For [tex]\( x = 3 \)[/tex], we use the third piece of the function since [tex]\( x \geq 3 \)[/tex]:
[tex]\[ f(x) = \frac{1}{4}x. \][/tex]
Plugging in [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{4} \times 3 = \frac{3}{4}. \][/tex]
Converting to decimal form:
[tex]\[ \frac{3}{4} = 0.75. \][/tex]
Thus,
[tex]\[ f(3) = 0.75. \][/tex]
Summarizing the results:
[tex]\[ \begin{aligned} f(-3) &= -2.5, \\ f(-1) &= 1.5, \\ f(3) &= 0.75. \end{aligned} \][/tex]
