Piecewise function
x if −2≤x≤0;
x^2 if −2<x<0;
8-x. If if −2≤x≤3;
(d) Write down an analytic formula for f^{-1}, and state its domain and range.
Your pieces overlap. That is not the way to define a piecewise function.
Looks like your inverse function will also be piecewise, no? Just take the inverse on each piece.
There is a typo
8-x if 2≤x≤3
To find the inverse of a function, we need to switch the roles of the independent variable (x) and dependent variable (f(x)) and then solve for x. In this case, we have a piecewise function:
f(x) =
x if -2 ≤ x ≤ 0
x^2 if -2 < x < 0
8 - x if -2 ≤ x ≤ 3
To find the inverse, we need to find f^{-1}(y) such that y = f(x). We can split the function into three parts based on the given intervals:
For the first part (if -2 ≤ x ≤ 0), we have x = y.
For the second part (if -2 < x < 0), we have x^2 = y. To get x, we take the square root of both sides (keeping in mind that we need to consider both the positive and negative roots):
x = ±√(y).
For the third part (if -2 ≤ x ≤ 3), we have 8 - x = y. Solving for x, we subtract y from both sides:
x = 8 - y.
Putting these three parts together, we can write the inverse function as follows:
f^{-1}(y) =
y for -2 ≤ y ≤ 0
±√(y) for 0 < y < 4
8 - y for 4 ≤ y ≤ 9
The domain of f^{-1} is the range of the original function f(x), which is -2 ≤ x ≤ 3. Therefore, the domain of f^{-1} is -2 ≤ y ≤ 3.
The range of f^{-1} is the domain of the original function f(x), which consists of three separate intervals:
1. -2 ≤ y ≤ 0 (for x = y)
2. 0 < y < 4 (for x = ±√(y))
3. 4 ≤ y ≤ 9 (for x = 8 - y)
Therefore, the range of f^{-1} is -2 ≤ y ≤ 9.