Answer with explanation:
Given the function f from R to ![(0,\infty)](https://tex.z-dn.net/?f=%280%2C%5Cinfty%29)
f: ![R\rightarrow(0,\infty)](https://tex.z-dn.net/?f=R%5Crightarrow%280%2C%5Cinfty%29)
![-f(x)=x^2](https://tex.z-dn.net/?f=-f%28x%29%3Dx%5E2)
To prove that the function is objective from R to ![(0,\infty)](https://tex.z-dn.net/?f=%280%2C%5Cinfty%29)
Proof:
![f:(0,\infty )\rightarrow(0,\infty)](https://tex.z-dn.net/?f=f%3A%280%2C%5Cinfty%20%29%5Crightarrow%280%2C%5Cinfty%29)
When we prove the function is bijective then we proves that function is one-one and onto.
First we prove that function is one-one
Let ![f(x_1)=f(x_2)](https://tex.z-dn.net/?f=f%28x_1%29%3Df%28x_2%29)
![(x_1)^2=(x_2)^2](https://tex.z-dn.net/?f=%28x_1%29%5E2%3D%28x_2%29%5E2)
Cancel power on both side then we get
![x_1=x_2](https://tex.z-dn.net/?f=x_1%3Dx_2)
Hence, the function is one-one on domain [tex[(0,\infty)[/tex].
Now , we prove that function is onto function.
Let - f(x)=y
Then we get ![y=x^2](https://tex.z-dn.net/?f=y%3Dx%5E2)
![x=\sqrt y](https://tex.z-dn.net/?f=x%3D%5Csqrt%20y)
The value of y is taken from ![(0,\infty)](https://tex.z-dn.net/?f=%280%2C%5Cinfty%29)
Therefore, we can find pre image for every value of y.
Hence, the function is onto function on domain ![(0,\infty)](https://tex.z-dn.net/?f=%280%2C%5Cinfty%29)
Therefore, the given
is bijective function on
not on whole domain R .
Hence, proved.