Answer:
The range of [tex]f(x)[/tex] is [tex]f(x)\in[0,3][/tex]
[tex]f^{-1}(x)=-\sqrt{-x^2+9}[/tex] with a domain [tex]x\in[0,3][/tex]
Step-by-step explanation:
Domain: set of input values (x-values)
Range: set of output values (y-values)
Given function:
[tex]f(x)=\sqrt{9-x^2}\quad(x\in[-3,0])[/tex]
Therefore, as the original function has a restricted domain, its range is also restricted:
[tex]f(-3)=\sqrt{9-(-3)^2}=0[/tex]
[tex]f(0)=\sqrt{9-0}=3[/tex]
[tex]\therefore\textsf{Range}\:f(x)\in[0,3][/tex]
To determine the inverse of a function
Change [tex]x[/tex] to [tex]y[/tex]:
[tex]\implies x=\sqrt{9-y^2}[/tex]
Square both sides:
[tex]\implies x^2=9-y^2[/tex]
Switch sides:
[tex]\implies 9-y^2=x^2[/tex]
Subtract 9 from both sides:
[tex]\implies -y^2=x^2-9[/tex]
Divide both sides by -1:
[tex]\implies y^2=-x^2+9[/tex]
Therefore:
[tex]\implies y=\sqrt{-x^2+9}\textsf{ and }y=-\sqrt{-x^2+9}[/tex]
As the range of the inverse function is the same as the domain of the original function:
[tex]\implies f^{-1}(x)=-\sqrt{-x^2+9}[/tex] only as the range is [-3, 0]
The domain of the inverse function is the same as the range of the original function.
[tex]\therefore\textsf{Domain of}\:f^{-1}(x):x \in [0,3][/tex]
The inverse of a function is ordinarily the reflection of the original function in the line [tex]y=x[/tex].
**Please see attached graph**
