Question

Consider the function y = 9 - x2, where x ≥ 3. What is the inverse of the function? What is the domain of the inverse? Show all of your work for full credit.

Answer 1

Answer:

The inverse of the function is $y^{-1}=\sqrt{9-x}$.

The domain of the inverse function is $D:(-\infty,0],\{x|x\in \mathbb{R}\}$

Step-by-step explanation:

Given : Function $y=9-x^2$ where, $x\geq 3$

To find : What is the inverse of the function? What is the domain of the inverse?

Solution :

Function $y=9-x^2$

To find the inverse we interchange the value of x and y,

$x=9-y^2$

Now, we get the value of y

$y^2=9-x$

$y=\pm\sqrt{9-x}$

As $x\geq 3$ so x>0

$y=\sqrt{9-x}$

The inverse of the function is $y^{-1}=\sqrt{9-x}$.

The domain of the inverse is the range of the original function.

The range is defined as the set of all possible value of y.

As $x\geq 3$

Squaring both side,

$x^2\geq 9$

Subtract $x^2$ both side,

$9-x^2\leq 0$

$y\leq 0$

The range of the function is $R:(-\infty,0],\{y|y\in \mathbb{R}\}$

The domain of the inverse function is $D:(-\infty,0],\{x|x\in \mathbb{R}\}$

Answer 2

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