Answer:
f(x)
Domain: all real numbers
Range:![y\ge -2](https://tex.z-dn.net/?f=y%5Cge%20-2)
![f^{-1}(x)](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29)
Domain: ![y\ge -2](https://tex.z-dn.net/?f=y%5Cge%20-2)
Range:all real numbers
Step-by-step explanation:
Given ![f(x)=x^2-2](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2-2)
The domain of this function, refers to all values of x for which f(x) is defined.
The given function is a quadratic polynomial.
Polynomial functions are defined everywhere, therefore the domain is all real numbers.
The range refers to the values of y, for which x is defined
Let ![y=x^2-2](https://tex.z-dn.net/?f=y%3Dx%5E2-2)
Solve for x;
![x=\pm \sqrt{y+2}](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Csqrt%7By%2B2%7D)
This function is defined for ![y+2\ge0](https://tex.z-dn.net/?f=y%2B2%5Cge0)
This implies that; ![y\ge-2](https://tex.z-dn.net/?f=y%5Cge-2)
Inverse Function
The domain of f(x) becomes the range of
and the range of f(x) becomes the domain of
.
Domain: ![y\ge -2](https://tex.z-dn.net/?f=y%5Cge%20-2)
Range:all real numbers
Or see it in details
Let ![y=x^2-2](https://tex.z-dn.net/?f=y%3Dx%5E2-2)
interchange x and y;
![x=y^2-2](https://tex.z-dn.net/?f=x%3Dy%5E2-2)
![x+2=y^2](https://tex.z-dn.net/?f=x%2B2%3Dy%5E2)
![\pm \sqrt{x+2}=y](https://tex.z-dn.net/?f=%5Cpm%20%5Csqrt%7Bx%2B2%7D%3Dy)
![f^{-1}(x)=\pm \sqrt{x+2}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%3D%5Cpm%20%5Csqrt%7Bx%2B2%7D)
Domain: ![x\ge-2](https://tex.z-dn.net/?f=x%5Cge-2)
Range:
Let
![y=\pm \sqrt{x+2}](https://tex.z-dn.net/?f=y%3D%5Cpm%20%5Csqrt%7Bx%2B2%7D)
![\implies y^2-2=x](https://tex.z-dn.net/?f=%5Cimplies%20y%5E2-2%3Dx)
x is defined for all y-values.
523.6 m3 i hope this helps mark me as brainliest