- #1

DumpmeAdrenaline

- 71

- 1

Let f(x)=2-x if 0 ≤x ≤2 and 0 otherwise.

I want to describe the following functions 1) f(-x) 2) -f(x) 3) f(x+3) 4) f(x)+3 5) f(2x) 6) 2f(x) 7) f(2x+3)

The rule generates an output by multiplying the input by -1 and adding the result to 2.

1) f operates on the closed interval [0,2] as its domain. With f(-x) the domain then changes to [-2,0] so that upon multiplying each real number in this interval by -1 we obtain the same domain [0,2] and the same image [0,2] as f(x). f(-x)=2-(-x)=2+x.

2) -f(x) change the sign of the output keeping the input.

3) f(x+3) Each input in the closed interval [0,2] moves by 3 units, such that the domain changes to 3 ≤x+3 ≤5. However, under this domain the function generates an output of 0. As a result the domain of f(x+3) becomes [0,-1] so that upon adding 3 units to we wind up with the same domain and range as f(x).

4) Move every point up by 3 units

5) The domain of f(2x) shrinks to 0<x<1 so that when multiply each real number we obtain the same domain and range as f(x). f(2x)=2-2x

6) Multiply each output by 2 keeping the input.

7) Composed transformation of shrinking by 1/2 followed by a shift 3 units to the left resulting in a domain of [-3,-1].

In all of the above are we changing the rule f?