Wanting to know my reasons to the questions below are justified for it to be bijective/not
Please correct me if I am wrong as I am trying to learn bijectivty and solving problems
- $f(x)=x^4+2x^2+1$ with $f: [0,\infty) \to [0,\infty)$
We know that for a function to be bijective it must satisfy both injectivity (one to one) and subjectivity (onto)
For this 1) I believe that this is not bijective although the domain lets this graph satisfy injective properties since it is a one to one function, the image of this graph is not equal to the codomain it is not sujective.
2.
I believe that this function with the given domain and codomain is bijective since it satifies one to one and also surjectivity because the image of this graph corresponds to the codomain. Since whatever even value x is inputted the function is always push out odd integers.
- Is there a bijective function within this domain and codomain?
$f:[-1,1] \to [-10000,10000]$
if $f(x)=10000x$
I believe that this would be a bijective function because first of all it is a linear function and it satisfies one to one and within this domain the image of the function and the codomain is equal.
I am not entirely sure of my answers and would like some clarification if wrong :)
Thank you
(1) If $f:[0,\infty) \to [0,\infty)$, then $f(x)=x^4+2x^2+1 \ge 1$ hence it is not onto.
(2) Assuming $E$ and $T$ both contain negative numbers as well, let $f:E\to T$ be given by $f(x) = 3x/2+21$. It's easy to see that $f^{-1}:T \to E$ will be given by $f^{-1}(t) = 2(t-21)/3 = 2t/3-14$. Note that both $f$ (on $E$) and $f^{-1}$ (on $T$) are well-defined functions, which makes $f$ bijective.
If $E$ and $T$ don't have negative numbers, this is not true, can you find a counter-example?
(3) correct