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Consider the below function

$f(x)=\frac{x}{2x+1}\{x \neq- \frac{1}{2}\}$

Onto functions are those $\forall y \exists x(f(x)=y)$, means for all elements in co-domain we have a pre-image in the domain.

I particularly get stuck how to determine when a function is onto especially when the function is given as a mathematical expression.

Please guide.

user3767495
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2 Answers2

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"I particularly get stuck how to determine when a function is onto especially when the function is given as a mathematical expression."

As well you should as the codomain is not stated. It's impossible to state if the function is onto if the codomain is not stated.

The domain is $\mathbb R\setminus \{-\frac 12\}$ so if $f:\mathbb R\setminus \{-\frac 12\}\to \mathbb R$ this may or may not be onto. If $f:\mathbb R\setminus \{-\frac 12\}\to \mathbb C$ it most certain is not onto. ($f(x) = i$ has no solution) and if $f:\mathbb R\setminus \{-\frac 12\}\to f(\mathbb R\setminus \{-\frac 12\}) = \{f(x)|X\in \mathbb R\setminus \{-\frac 12\}\}$ must certainly is.

The unstated assumption is:

$f:\mathbb R\setminus \{-\frac 12\}\to \mathbb R$

and we need to prove/disprove for any $y \in \mathbb R$ that there exists one or more $x$ so that $f(x) = y$.

So if $\frac x{2x + 1} = y$ then $x = y*(2x+1)$ and...

$x = y*2x + y$

$x - 2yx = y$

$x(1-2y) = y$. If $1 - 2y\ne 0$ we have

$x = \frac {y}{1-2y}$ and so $f( \frac {y}{1-2y}) = y$ is possible so long as $\frac y{1-2y} \ne -\frac 12$.

i.e. if $2y = 2y - 1$ which would mean $0 = -1$ which is impossible.

So as long as $1-2y\ne 0$ then $x =\frac y{1-2y}$ is a solution to $f(x) = y$.

But what if $1-2y=0$ or $y=\frac 12$. Is it possible for

$\frac x{2x + 1} = \frac 12$? That would mean $2x = 2x + 1$ and that would mean $0 = 1$ which is impossible. So $f(x) = \frac 12$ has no solution. It is not onto if the codomain is $\mathbb R$.

However if $f:\mathbb R\setminus \{-\frac 12\}\to \mathbb R\setminus\{\frac 12\}=f(\mathbb R\setminus \{-\frac 12\})=\{f(x)|x \in \mathbb R\setminus \{-\frac 12\}\}$ then it is onto as for all $y\in \mathbb R; y \ne \frac 12$ then $\frac y{1- 2y}\in R\setminus \{-\frac 12\}$ and $f(\frac y{1-2y}) = y$

But it is conventional to assume the co-domain is, if not specified otherwise, is $\mathbb R$.

FWIS: All functions can be made onto by simply restricting them to the image of the function. That is to say $f: D \to f(D) = \{f(x)|x \in D\}$ is, by definition, onto. But it is conventional to assume the co-domains is $\mathbb R$ and not usually $f(D)$.

fleablood
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  • Nice....:) Can you help me with the function $f(x)=x^2+2x$. How do I determine whether it's onto? f is defined over $R \rightarrow R$ – user3767495 Nov 07 '18 at 12:43
  • Same way. For an arbitrary $y$ does $x^2 + 2x = y$ always have a real solution. $x^2 + 2x-y=0$ means $x=\frac {-2\pm\sqrt{4+4y^2}}2=-1\pm\sqrt{y+1}$. That has solutions only if $y+1\ge 0$ so if $y < -1$ there is no solution so no it is not onto. If you graph this, it is a parabola and in has it's minimum point at $(-1,-1)$ so it never gets below $-1$ so it... never gets below $-1$. It's image is $[-1,\infty)\subsetneq\mathbb R$. I wonder why you are having a mental block against this. – fleablood Nov 07 '18 at 15:08
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As noticed a function is completely defined when we specify what domain and codomain are that is

$$f:A\to B$$

Here we are assuming $A=\mathbb{R}$ but we need also to specify $B$.

Note that any function is by definition onto if and only if $B\subseteq$ range. Therefore in that case the key point is to determine what the range of $f$ is and then compare that with the codomain we are assuming for $f$.

As an alternative given the codomain if we can find a value $y$ such that $\not \exists x$ such that $y=f(x)$ it suffices to prove that $f$ is not onto.

user
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  • Good answer! I think the OP could use some help is determining what the range is, which is very similar to how you'd find a $y$ where $\not\exists x$ so $y=f(x)$. If you solve $\frac x{2x+1}=y$ for $x$ you find $x=\frac y{1-2y}$ which tells you both it is not defined at $y=-\frac 12$ and that the range is all reals except $\frac 12$. – fleablood Nov 07 '18 at 15:16
  • @fleablood Yes that's a good way to proceed! – user Nov 07 '18 at 18:18