Question

Determine f(x) is one to one, onto or both
Please, help

Answer 1

\[f(x)\begin{cases}x^2~~~if~~~x\geq 0\\-x^2 ~~~if~~~x<0\end{cases}\]
Domain f ={x\in R, x>0}
f: R^+ --> R

Answer 2

Algebra way, please.

Answer 3

The one-to-one part shouldn't be too hard.

To show a function is one-to-one, you assume \(f(a)=f(b)\) implies \(a=b\). (Sometimes it's easier to use the converse by assuming \(a\not=b\) implies \(f(a)\not=f(b)\).)

Suppose \(a,b>0\), then
\[\begin{align*}
f(a)&=f(b)\\
a^2&=b^2\\
a&=b
\end{align*}\]
Clearly if \(a>0\) and \(b<0\), then you can never have \(a=b\), so we ignore this case altogether. The case that both \(a,b<0\) is the same as above.

Answer 4

I got this part, thank you
How to prove onto part?

Answer 5

One more question: is it not that a^2 = b^2 --> a = \(\pm b\)??

Answer 6

And from the beginning of the proof, we assume f(a) = f(b) and it leads to a =b or a = -b. The last case is not good. :)

Answer 7

That would be true, but we're assuming that both \(a\) and \(b\) are non-negative from the start so that we can assign \(f(x)=x^2\). Since \(a,b>0\), we know that \(-a,-b<0\), but \(f(x)\not=x^2\) for negative numbers. Does that make sense?

Answer 8

Oh yea, I got you. Thanks for making it clear. :)

Answer 9

For showing the function is onto: Set \(y=f(x)\), and suppose \(x\ge0\) for the moment. Clearly, \(x\in\mathbb{R}_+\cup\{0\}\) (all positive reals plus zero). We want to establish that, at least for \(f(x)=x^2\), \(x\ge0\), we have \(f(x)\in\mathbb{R}_+\cup\{0\}\) as well.
\[\begin{align*}
y&=x^2\\
\sqrt y=x
\end{align*}\]
\(\sqrt y\) is defined only for non-negative real values of \(y\), so we're done.

Doing the same for the other piece of \(f(x)\), we assume \(y=f(x)\) and \(x<0\). We have to establish that for all \(x\in\mathbb{R}_-\) (all negative reals), we also have \(y\in\mathbb{R}_-\). The proof for this is the same as (or similar to) the above proof.

Answer 10

What if the domain is R? Is it the same with above?

Answer 11

Not sure what you mean by that... Do you mean if \(f(x)=x^2\) for \(x\in\mathbb{R}\) ?

Answer 12

yes

Answer 13

Well in that case, the function would not be onto.
\[y=x^2~~\implies~~\pm\sqrt y=x\]
\(y\) must be non-negative for the square root to be defined. The fact that \(x\) is also non-negative means \(y=x^2\) cannot take on negative values. But negative values are a subset of \(\mathbb{R}\), which means \(y\in\mathbb{R}_+\cup\{0\}\), but in general \(y\not\in\mathbb{R}\). Since the domain isn't the same as the range, \(y=x^2\) is not onto.

Answer 14

Thank you so much. I need you confirm my understanding about this :
I have to negate this " \(\forall x, \exists n ~~| ~~a\leq x\leq b\)"
when negating, how to deal with the last part? \(a\leq x\leq b\)??

Answer 15

I do a > x or b <x . Am I right?

Answer 16

\(a\le x\le b\) means \(x\) is bounded between \(a\) and \(b\). The negation would be that \(x\) is outside of the interval \([a,b]\), or \(x<a\) and \(x>b\). You're right :)

Answer 17

oh yeah!! thank you.

Answer 18

You're welcome!