Question

for the function f(x)= (x^2) - 3, x >= 0, find the following : (1) the inverse on the given interval and write it in the form y=f^(-1)x and (2) verify the relationships: f(f^(-1)x))=x and f^(-1)(f(x))=x

Answer 1

since f(x)=x^2-3 is a parabola=> it is not one to one unless there is a domain restriction

so you have the domain restriction:x>=0

now we have a 1 to 1 function=>our function has inverse

we need to solve y=x^2-3 for x

add 3 on both sides

y+3=x^2

take square root of both sides

\[\pm \sqrt{y+3}=x\]

we need to decide which is the inverse the \[\sqrt{x+3}=f^{-1}(x) or -\sqrt{x+3}=f^{-1}(x)\]
recall the domain of our function f(x)=x^2-3 is [0,inf) and our range is [-3,inf)
so this means the inverse function has domain [-3,inf) and range [0,inf)

so we don't want no negative output for f inverse since we have range [0,inf) for f inverse
so thus the inverse function is \[f^{-1}(x)=\sqrt{x+3}\]