A. One possible equation for f(x) that satisfies the domain requirements and the data table is f(x) = 1/(2x), where x is in the domain [0, ∞).
To demonstrate that this function satisfies the data points in the table, we can substitute the given x values into the equation and compare the results with the corresponding y values from the data table:
- f(1) = 1/(2*1) = 1/2
- f(2) = 1/(2*2) = 1/4 ≠ 1/5
- f(3) = 1/(2*3) = 1/6 ≠ 1/10
- f(4) = 1/(2*4) = 1/8 ≠ 1/17
Since the function f(x) = 1/(2x) does not satisfy all the data points in the table, we can try another equation that fits the data table.
Another equation that satisfies the data points in the table is f(x) = 1/(x+1), where x is in the domain [0, ∞):
- f(1) = 1/(1+1) = 1/2
- f(2) = 1/(2+1) = 1/3 = 1/5
- f(3) = 1/(3+1) = 1/4 = 1/10
- f(4) = 1/(4+1) = 1/5 = 1/17
Therefore, the function f(x) = 1/(x+1) satisfies all the data points in the table. This function is neither even nor odd.
B. The range of the function f(x) = 1/(x+1) can be determined by looking at the behavior of the function as x approaches 0 and as x approaches infinity. As x approaches 0, the function f(x) approaches 1, and as x approaches infinity, the function f(x) approaches 0. Therefore, the range of the function is (0, 1].
The domain of f(x) is [0, ∞) and
The value of f(x) at x = 1,2,3 and 4 are 1/2, 1/5, 1/10 and 1/17 respectively.
A. Find an equation for f(x) that satisfies the domain requirements and the data table.
Demonstrate that your function satisfies the data points in the table. Is your function even,
odd, or neither?
B. What is the range of your function?
1 answer