Get comprehensive solutions to your problems with IDNLearn.com. Our experts are ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
To determine the range of the piecewise function [tex]\( f(x) \)[/tex] given as:
[tex]\[ f(x) = \begin{cases} 3 & \text{if } x < 0 \\ x^2 + 2 & \text{if } 0 \leq x < 2 \\ \frac{1}{2}x + 5 & \text{if } x \geq 2 \end{cases} \][/tex]
Let's analyze each piece of the piecewise function to find the possible values the function can take.
1. For [tex]\( x < 0 \)[/tex]:
- If [tex]\( x \)[/tex] is less than 0, the function [tex]\( f(x) \)[/tex] is always 3.
- Hence, the value 3 is included in the range.
2. For [tex]\( 0 \leq x < 2 \)[/tex]:
- The function in this interval is [tex]\( f(x) = x^2 + 2 \)[/tex].
- Evaluate the function at the endpoints:
- At [tex]\( x = 0 \)[/tex], [tex]\( f(0) = 0^2 + 2 = 2 \)[/tex].
- As [tex]\( x \)[/tex] approaches but does not reach 2, [tex]\( f(x) \approx 2^2 + 2 = 6 \)[/tex] (since [tex]\( f(x) \)[/tex] is continuous and smoothing approaching this value from the left).
- Therefore, the values taken by [tex]\( f(x) \)[/tex] range from 2 to just under 6 (exclusive).
3. For [tex]\( x \geq 2 \)[/tex]:
- The function in this interval is [tex]\( f(x) = \frac{1}{2}x + 5 \)[/tex].
- Evaluate the function at [tex]\( x = 2 \)[/tex]:
- At [tex]\( x = 2 \)[/tex], [tex]\( f(2) = \frac{1}{2}(2) + 5 = 1 + 5 = 6 \)[/tex].
- As [tex]\( x \)[/tex] continues to increase, [tex]\( \frac{1}{2}x + 5 \)[/tex] increases without bound towards positive infinity.
- Therefore, for [tex]\( x \geq 2 \)[/tex], [tex]\( f(x) \)[/tex] starts from 6 and goes to infinity.
Combining the ranges from all intervals:
- From [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 3 \)[/tex].
- From [tex]\( 0 \leq x < 2 \)[/tex], [tex]\( f(x) \)[/tex] spans the interval [tex]\([2, 6)\)[/tex] (includes 2 but values approach 6).
- For [tex]\( x \geq 2 \)[/tex], [tex]\( f(x) \)[/tex] spans from 6 to infinity, including 6.
Thus, the overall range of the function is:
[tex]\[ \{3\} \cup [2, 6) \cup [6, \infty) \][/tex]
Simplifying, we notice the overlapping point at 6. So, the final range combining all segments is:
[tex]\[ \{3\} \cup [2, \infty) \][/tex]
Therefore, the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{[2, \infty)} \text{ including the value 3.} \][/tex]
[tex]\[ f(x) = \begin{cases} 3 & \text{if } x < 0 \\ x^2 + 2 & \text{if } 0 \leq x < 2 \\ \frac{1}{2}x + 5 & \text{if } x \geq 2 \end{cases} \][/tex]
Let's analyze each piece of the piecewise function to find the possible values the function can take.
1. For [tex]\( x < 0 \)[/tex]:
- If [tex]\( x \)[/tex] is less than 0, the function [tex]\( f(x) \)[/tex] is always 3.
- Hence, the value 3 is included in the range.
2. For [tex]\( 0 \leq x < 2 \)[/tex]:
- The function in this interval is [tex]\( f(x) = x^2 + 2 \)[/tex].
- Evaluate the function at the endpoints:
- At [tex]\( x = 0 \)[/tex], [tex]\( f(0) = 0^2 + 2 = 2 \)[/tex].
- As [tex]\( x \)[/tex] approaches but does not reach 2, [tex]\( f(x) \approx 2^2 + 2 = 6 \)[/tex] (since [tex]\( f(x) \)[/tex] is continuous and smoothing approaching this value from the left).
- Therefore, the values taken by [tex]\( f(x) \)[/tex] range from 2 to just under 6 (exclusive).
3. For [tex]\( x \geq 2 \)[/tex]:
- The function in this interval is [tex]\( f(x) = \frac{1}{2}x + 5 \)[/tex].
- Evaluate the function at [tex]\( x = 2 \)[/tex]:
- At [tex]\( x = 2 \)[/tex], [tex]\( f(2) = \frac{1}{2}(2) + 5 = 1 + 5 = 6 \)[/tex].
- As [tex]\( x \)[/tex] continues to increase, [tex]\( \frac{1}{2}x + 5 \)[/tex] increases without bound towards positive infinity.
- Therefore, for [tex]\( x \geq 2 \)[/tex], [tex]\( f(x) \)[/tex] starts from 6 and goes to infinity.
Combining the ranges from all intervals:
- From [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 3 \)[/tex].
- From [tex]\( 0 \leq x < 2 \)[/tex], [tex]\( f(x) \)[/tex] spans the interval [tex]\([2, 6)\)[/tex] (includes 2 but values approach 6).
- For [tex]\( x \geq 2 \)[/tex], [tex]\( f(x) \)[/tex] spans from 6 to infinity, including 6.
Thus, the overall range of the function is:
[tex]\[ \{3\} \cup [2, 6) \cup [6, \infty) \][/tex]
Simplifying, we notice the overlapping point at 6. So, the final range combining all segments is:
[tex]\[ \{3\} \cup [2, \infty) \][/tex]
Therefore, the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{[2, \infty)} \text{ including the value 3.} \][/tex]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.