Respuesta :
[tex]\text{We have been given the function }\\ f(x)=(x-1)^2-4\\ \text{Let f(x)=y}\\ y=(x-1)^2-4\\ \text{Swap x and y, we get}\\ \\ x=(y-1)^2-4\\ \text{Now we solve for y}\\ (y-1)^2=x+4\\ \\ y-1=\pm \sqrt{x+4}\\ \\ y=1\pm \sqrt{x+4}\\ \\ f^{-1}(x)=1\pm \sqrt{x+4}\\[/tex]
[tex]x+4\geq 0\\ x\geq -4\\ \text{Thus, the domain is given by}\\ x\in [-4,\infty )[/tex]
Answer:
- The inverse function is:
[tex]f^{-1}(x)=\sqrt{x+4}+1[/tex]
- The domain of the inverse is:
[tex]x\geq -4[/tex]
Step-by-step explanation:
The function f(x) is given by:
[tex]f(x)=(x-1)^2-4[/tex]
The domain is restricted to x≤1
so that the function is one-one.
Also, the range of function is:
[tex](x-1)^2\geq 0\\\\\\(x-1)^2-4\geq -4[/tex]
Now, as the function is one-one and it is onto as well.
Hence, the inverse of the function exist and it is a function as well.
and we find the inverse function as follows:
Keep f(x)=y
[tex](x-1)^2-4=y[/tex]
interchange x and y
[tex](y-1)^2-4=x[/tex]
Now we solve for y
[tex](y-1)^2=x+4\\\\y-1=\sqrt{x+4}\\\\y=\sqrt{x+4}+1[/tex]
Hence, the inverse function is:
[tex]f^{-1}(x)=\sqrt{x+4}+1[/tex]
and the domain of inverse function is the range of the function.
Hence the domain is:
[tex]x\geq -4[/tex]
