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Alenkinab [10]
3 years ago
6

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?
Mathematics
2 answers:
olasank [31]3 years ago
5 0

\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 )

Alex777 [14]3 years ago
4 0
<h2>Answer:</h2>
  • The inverse function is:

               f^{-1}(x)=\sqrt{x+4}+1

  • The domain of the inverse is:

                 x\geq -4

<h2>Step-by-step explanation:</h2>

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

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