15 pts. Prove that the function f from R to (0, oo) is bijective if - f(x)=x2 if r- Hint: each piece of the function helps to "cover" information to break your proof(s) into cases. part of (0, oo).. you may want to use this

Respuesta :

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.

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