Question

The function f(x)=(x-1)^2-4 is not one-to-one. If you restrict the domain for f(x) to x(less than or equal to symbol)1, what is its inverse function and the domain for the inverse?

Answer 1

$\text{We have been given the function }\\ f(x)=(x-1)^2-4\\ \text{Let f(x)=y}\\ y=(x-1)^2-4\\ \text{Swap x and y, we get}\\ \\ x=(y-1)^2-4\\ \text{Now we solve for y}\\ (y-1)^2=x+4\\ \\ y-1=\pm \sqrt{x+4}\\ \\ y=1\pm \sqrt{x+4}\\ \\ f^{-1}(x)=1\pm \sqrt{x+4}$

$x+4\geq 0\\ x\geq -4\\ \text{Thus, the domain is given by}\\ x\in [-4,\infty )$

Answer 2

Answer:

• The inverse function is:

               $f^{-1}(x)=\sqrt{x+4}+1$

• The domain of the inverse is:

                 $x\geq -4$

Step-by-step explanation:

The function f(x) is given by:

                 $f(x)=(x-1)^2-4$

The domain is restricted to x≤1

so that the function is one-one.

Also, the range of  function is:

$(x-1)^2\geq 0\\\\\\(x-1)^2-4\geq -4$

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

$(x-1)^2-4=y$

interchange x and y

$(y-1)^2-4=x$

Now we solve for y

$(y-1)^2=x+4\\\\y-1=\sqrt{x+4}\\\\y=\sqrt{x+4}+1$

Hence, the inverse function is:

          $f^{-1}(x)=\sqrt{x+4}+1$

and the domain of inverse function is the range of the function.

Hence the domain is:

                $x\geq -4$