Given:
The function is
[tex]f(x)=-\dfrac{1}{2}\sqrt{x+3},x\geq -3[/tex]
To find:
The inverse of the given function.
Solution:
We have,
[tex]f(x)=-\dfrac{1}{2}\sqrt{x+3}[/tex]
Put f(x)=y.
[tex]y=-\dfrac{1}{2}\sqrt{x+3}[/tex]
Interchange x and y.
[tex]x=-\dfrac{1}{2}\sqrt{y+3}[/tex]
Isolate the variable y.
[tex]-2x=\sqrt{y+3}[/tex]
[tex](-2x)^2=y+3[/tex]
[tex]4x^2-3=y[/tex]
[tex]y=4x^2-3[/tex]
Putting [tex]y=f^{-1}(x)[/tex], we get
[tex]f^{-1}(x)=4x^2-3[/tex]
For [tex]x\geq -3[/tex],
[tex]x+3\geq 0[/tex]
[tex]\sqrt{x+3}\geq 0[/tex]
[tex]-\dfrac{1}{2}\sqrt{x+3}\leq 0[/tex]
[tex]f(x)\leq 0[/tex]
It means for function [tex]f(x)\leq 0[/tex] and for inverse function [tex]x\leq 0[/tex] because the range of the function is the domain of inverse function.
Therefore, [tex]f^{-1}(x)=4x^2-3[/tex] for [tex]x\leq 0[/tex].
