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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]
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