How to verify whether function is surjective or injective I'm trying to learn how to verify whether a certain function is surjective / injective.
$g:\mathbb{R}\rightarrow \mathbb{R}$
$g(x)= 2x^5 +9$
How would I do so?
Thanks
 A: Your function is given by
$g:\mathbb{R}\to\mathbb{R}$, $g(x)=2x^5+9$.
We want to check if $g$ is injective/surjective. You have several ways to do that.
An analytic (more advanced) approach on surjectivity would to observe that:
$\lim_{x\to\infty} g(x)=\infty$ and
$\lim_{x\to -\infty} g(x)=-\infty$
By the mean-value theorem $g$ is surjective.
An analytic approach for injectivity would be to calculate
$g'(x)=10x^4$. We have that $g'(x)>0$ for every $x\in\mathbb{R}$. So $g$ is always increasing and thus injective.
As I said these methods are more advanced (but still pretty basic and are thought in a beginners analysis course).
I assume that you are starting out.
So an other approach is straight by definition.

For injectivity we have to show that if $g(a)=g(b)\Rightarrow a=b$

We have $g(a)=g(b)\Leftrightarrow 2a^5+9=2b^5+9\Leftrightarrow a^5-b^5=0\Leftrightarrow (a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)=0$
So either a-b=0 which implies a=b (what we had to show)
Or $a^4+a^3b+a^2b^2+ab^3+b^4=0$ but this is only $0$ iff $a=b=0$.
If $a=b=0$ there is nothing to show.
So let $a,b\neq 0$. Then we can seperate 3 cases. $a,b>0$, $a,b<0$ and (without loss of generality) $a>0$ and $b<0$

Without loss of generality means that we can assume $a>0$ and $b<0$ and do not need to show the other case $a<0$ and $b>0$. Because there is symmetrie in the expression and if the inequalities are swapped, we could just swap a and b too.

If $a,b>0$ then $a^4+a^3b+a^2b^2+ab^3+b^4>0$.
If $a,b<0$ then $a^4+a^3b+a^2b^2+ab^3+b^4>0$. Since every summand is positive. You might want to check that for yourself, if it is not clear.
If $a>0$ and $b<0$, then $\underbrace{a^4}_{>0}+\underbrace{a^3b}_{<0}+\underbrace{a^2b^2}_{>0}+\underbrace{ab^3}_{<0}+\underbrace{b^4}_{>0}$
Why is this now positive?
Well we have either $|a|\geq |b|$ or $|a|\leq |b|$.
If $|a|\geq |b|$. Then we have that $a^4+a^3b>0$. Since $a^3(a+b)>0$ since every factor is positiv. $a^3$ is positiv for sure, since $a>0$.
Why is $a+b>0$? Since $|a|\geq |b|$. Think about that.

Keep in mind that I am avoiding the use of the fifth root $\sqrt[5]{\square}$, because you might not know that concept yet in your class.

Else we could just use that here and deduce $a=b$ that way.
The approach above is the most 'basic'.
After all we showed that $g(a)=g(b)$ implies $a=b$. So $g$ is injective.

$g$ is surjective is showed like this:
By definition we have to show that for every $y\in\mathbb{R}$ (value range) there is some $x\in\mathbb{R}$ (definition range) such that $g(x)=y$.

So we have to solve the equation $2x^5+9=y$ for $x$.
This is done easily with the use of the fifth root (which we avoided earlier, but unfortunatly can not(!) avoid here).
We get: $x=\sqrt[5]{\frac{y-9}{2}}$.
Indeed:
$g(\sqrt[5]{\frac{y-9}{2}})=2\left(\sqrt[5]{\frac{y-9}{2}}\right)^5+9=2\frac{y-9}{2}+9=y$.

Notice that we have shown more here. We jus calculated the inverse function of $g$. Which is given by $g^{-1}:\mathbb{R}\to\mathbb{R}$, $g^{-1}(x)=\sqrt[5]{\frac{x-9}{2}}$

We can check this by showing:
$g(g^{-1}(y))=y$ and $g^{-1}(g(x))=x$
Then $g^{-1}$ is indeed the inverse to $g$ and this shows immediatly that $g$ is not just surjective, but also injective!
A: Since $f'(x)=10x^4$ maintains a positive sign (except at finitely many points) for all real $x,$ it follows that the function $f(x)$ is monotonic on the real axis, and therefore injective there.
For surjectivity, note that as $x\to\pm\infty,$ we have that $f(x)\to\pm\infty$ too respectively. Since the function is continuous, it follows by IVT that it is surjective.
A: Since in $\Bbb R$, the following functions are bijective, so is their composition:
$$x\mapsto x^5\\ x\mapsto 2x\\ x\mapsto x+9$$
A: hint
You can put
$$f(x)=x^5$$
and
$$h(x)=2x+9$$
with $ g = h \circ f $.
it is easy to prove that $f$ and $ h$ are bijectives.
