Find the domain and range of the following piecewise function. f(x)=x+4 if -4 < x < 3 2x-1 if 3 < x < 6
The Correct Answer and Explanation is:
To find the domain and range of the piecewise function f(x)={x+4if −4<x<32x−1if 3<x<6f(x) = \begin{cases} x + 4 & \text{if } -4 < x < 3 \\ 2x – 1 & \text{if } 3 < x < 6 \end{cases}f(x)={x+42x−1if −4<x<3if 3<x<6
Domain:
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For this function, we have two pieces:
- The first piece is x+4x + 4x+4, defined for −4<x<3-4 < x < 3−4<x<3.
- The second piece is 2x−12x – 12x−1, defined for 3<x<63 < x < 63<x<6.
The function is not defined at x=3x = 3x=3, since the interval −4<x<3-4 < x < 3−4<x<3 doesn’t include 333, and the second interval 3<x<63 < x < 63<x<6 starts just after 333. So, the domain is the union of these two intervals:
Domain: (−4,3)∪(3,6)(-4, 3) \cup (3, 6)(−4,3)∪(3,6)
Range:
The range of a function is the set of all possible output values (y-values). We will determine the range of each piece separately and then combine them.
- For the first piece, f(x)=x+4f(x) = x + 4f(x)=x+4, where −4<x<3-4 < x < 3−4<x<3:
- The smallest value of xxx is just greater than −4-4−4, so f(x)f(x)f(x) is just slightly greater than −4+4=0-4 + 4 = 0−4+4=0.
- The largest value of xxx is just less than 333, so f(x)f(x)f(x) is just slightly less than 3+4=73 + 4 = 73+4=7.
- For the second piece, f(x)=2x−1f(x) = 2x – 1f(x)=2x−1, where 3<x<63 < x < 63<x<6:
- The smallest value of xxx is just greater than 333, so f(x)f(x)f(x) is just slightly greater than 2(3)−1=52(3) – 1 = 52(3)−1=5.
- The largest value of xxx is just less than 666, so f(x)f(x)f(x) is just slightly less than 2(6)−1=112(6) – 1 = 112(6)−1=11.
Range:
The overall range is the union of the two ranges: (0,7)∪(5,11)(0, 7) \cup (5, 11)(0,7)∪(5,11)
Final Answer:
- Domain: (−4,3)∪(3,6)(-4, 3) \cup (3, 6)(−4,3)∪(3,6)
- Range: (0,7)∪(5,11)(0, 7) \cup (5, 11)(0,7)∪(5,11)
