Answer:
The domain of f(x) is x>-2, so the range of f-1(x) is y>-2
The range of f(x) is y>-1, so the domain of the f-1(x) is x>-1
Step-by-step explanation:
To solve this question, we need to know the definition of domain and range.
The domain of a function is defined as all the possible input x-values while the range is defined as all the possible output y-values.
For f(x), domain for x is; x ≥ 1 and range is; y ≥ 2
For f⁻¹(x); domain is; x ≥ 2 and range is; x ≥ 1
We are given the function;
f(x) = [tex](\sqrt{x - 1} ) + 2[/tex]
Now, when x = 1, we will have zero inside the square root sign and this means that any x-value less than 1 will make the value inside the square root symbol to be negative which will result in an imaginary root.
Thus, we can say that, the domain for x is; x ≥ 1
Putting x = 1, we can find the least range value;
f(x) = 0 + 2
f(x) = 2
Thus, the range will be; y ≥ 2
Now, we want to find the range of f⁻¹(x)
This means an inverse of the original equation.
Thus;
f⁻¹(x) = [tex](x - 2)^{2} + 1[/tex]
Let's put x = 2;
f⁻¹(x) = (2 -2)² + 1
f⁻¹(x) = 1
Since x=2 gives the bracket to be zero, it means the domain is; x ≥ 2 and range is; x ≥ 1
Read more at; brainly.com/question/24522291
