Join the IDNLearn.com community and get your questions answered by experts. Whether it's a simple query or a complex problem, our experts have the answers you need.
Sagot :
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]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.