Answer with explanation:
Given the function f from R to [tex](0,\infty)[/tex]
f: [tex]R\rightarrow(0,\infty)[/tex]
[tex]-f(x)=x^2[/tex]
To prove that the function is objective from R to [tex](0,\infty)[/tex]
Proof:
[tex]f:(0,\infty )\rightarrow(0,\infty)[/tex]
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 [tex]f(x_1)=f(x_2)[/tex]
[tex](x_1)^2=(x_2)^2[/tex]
Cancel power on both side then we get
[tex]x_1=x_2[/tex]
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 [tex]y=x^2[/tex]
[tex]x=\sqrt y[/tex]
The value of y is taken from [tex](0,\infty)[/tex]
Therefore, we can find pre image for every value of y.
Hence, the function is onto function on domain [tex](0,\infty)[/tex]
Therefore, the given [tex]f:R\rightarrow(0.\infty)[/tex] is bijective function on [tex](0,\infty)[/tex] not on whole domain R .
Hence, proved.