How to determine function is onto

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.

• I deleted my answer, since for some reason I thought the formula was $2/(2x+1)$ instead of $x/(2x+1).$ – coffeemath Nov 7 '18 at 7:08

"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 $$f(D)$$.

• 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 7 '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 7 '18 at 15:08

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.

• 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 7 '18 at 15:16
• @fleablood Yes that's a good way to proceed! – gimusi Nov 7 '18 at 18:18