I was given a simple function, very simple. f(x) = bla bla. With a constant and an X. The point was to find for which values of a(the constant) the function is not one-to-one. The domain of the function is limited from zero and upwards. Here is what I put down on the paper:

As long as a is negative, it will not be one-to-one. If a = -2. Then f(0) = 4 and f(4)=4. Thereby concluding that it's not one-to-one in it's domain as long as a < 0. And stating the point of one-to-one is that we shouldn't be able to reach the same output with different input.

I didn't get full score on this, what am I missing?

edit: f(x) = (x+a)^2

edit: I've added it! I not able to comment on my own post :(

  • $\begingroup$ You didn't mention what function you had. Without that crucial piece of information, we can only speculate why your teacher marked you the way s/he did. $\endgroup$ – J. M. is a poor mathematician Nov 18 '10 at 11:44
  • $\begingroup$ Can you show us the function? Maybe you're incorrect about what $a$ can and can't be? I'm just guessing, but in the absence of more specific information, I can't do any better. $\endgroup$ – Alex Basson Nov 18 '10 at 11:47
  • $\begingroup$ It's never one to one. Try drawing a bunch of horizontal lines on any graph of that function. $\endgroup$ – J. M. is a poor mathematician Nov 18 '10 at 11:49
  • $\begingroup$ @J.M. Note that the domain is taken to be $[0,\infty)$ in all cases. $\endgroup$ – Arturo Magidin Nov 18 '10 at 15:46
  • $\begingroup$ @Arturo: I know that ;) and I was merely giving @Algific a nudge on why a parabola cannot represent a one-to-one function. Of course, you're supposed to be drawing horizontal lines above the vertex, but whatever. ;) $\endgroup$ – J. M. is a poor mathematician Nov 18 '10 at 15:54

There are two things you are missing: you are missing the case in which $a\geq 0$; and you did not really justify your assertion when $a\lt 0$.

Technically, the only case you actually established was the case $a=-2$. That leaves quite a few values of $a$ untouched!

When one asserts that something happens "for all numbers less than $0$", it is almost never good enough to check what happens in one number less than $0$. That's called "argument by example", and examples almost never establish general cases. It would be like trying to argue that everyone in your class is going to fail by pointing to one student in the class who happens to be failing. It's certainly possible that all will fail, but just showing one person who will is not going to establish this.

So: first, you need to finish off the case $a\lt 0$: if $a$ is any number less than $0$, why is $f(x)=(x+a)^2$ not one-to-one on $[0,\infty)$? Because, no matter what $a$ is, you can always find at least two distinct points in $[0,\infty)$ on which the function takes the same value: $x=0$, and $x=-2a$ (which is positive because $a\lt 0$, and so both in the domain and different from $0$). Indeed, $f(0)=(0+a)^2 = a^2$, and $f(-2a) = (-2a+a)^2 = (-a)^2 = a^2=f(0)$. You'll note that this takes care of all the values of $a$ that are negative. (Note: These are not the only points you can use; for instance, $\frac{-a}{2} and $\frac{-3a}{2}$ will also have equal images. But to show that something is not one-to-one it does suffice to give a single example of two distinct points that map to the same thing, because the assertion "is not one-to-one" means "there is at least one pair $a,b$, with $a\neq b$ and $f(a)=f(b)$).

Now, we also need to deal with the case $a\geq 0$. What happens then? If $a\geq 0$, then $x+a\geq 0$ (since $x\geq 0$). If $f(x)=f(y)$, then we have: \begin{align*} f(x) &= f(y)\\ (x+a)^2 &= (y+a)^2\\ \sqrt{(x+a)^2} &= \sqrt{(y+a)^2}\\ |x+a|&=|y+a| \end{align*} (remember that $\sqrt{r^2}=|r|$). Now, because $x$, $y$, and $a$ are all nonnegative, so are $x+a$ and $y+a$, so they equal their absolute value: $|x+a|=x+a$ and $|y+a|=y+a$. So we get $x+a=y+a$, and so we deduce $x=y$. That is, if $a\geq 0$, then for $f(x)$ to equal $f(y)$ we must have $x=y$. This means that in this case, $f$ is one-to-one.

In summary: you forgot to check $a\geq 0$. And you did not fully establish $a\lt 0$. It's not enough to just "it is not one-to-one when $a\lt 0$", because this is only says what happens if $a\lt 0$; your answer is not presumed to be exhaustive (that is, we don't assume that you are telling us all the values in which it is not one-to-one) because you are not saying it is exhaustive (to do so, you would have to say "it is not one-to-one if and only if $a\lt 0$", and even then you would have to discuss what happens when $a\geq 0$).


The key here is that your domain is $[0,\infty)$. In your solution you gave a very loose reason why it is not one-to-one for $a < 0$, and did not say anything about the case where $a \geq 0$. For $a < 0$, consider $f(0)$ and $f(-2a)$. Both are well-defined, as they are contained in the domain. Then $$f(0) = (0 + a)^2 = a^2 = (-2a + a)^2 = f(-2a)$$ Since $0 \neq -2a$ by assumption, $f$ is not one-to-one for $a < 0$. To finish, you need to show that $f$ is one-to-one for $a \geq 0$. Hint: Derivatives should be useful here...


Try to draw its graphic for $a=0$: maybe you'll recognise something familiar, won't you? Then, try to figure out what happens when you put some $a\neq 0$ (and you'll see that this doesn't change anything at all about injectivity).

Then you could try to think about this equation:

$$ (x_1 +a )^2 = (x_2 + a)^2 \ . $$

Again: start with the case $a=0$, then see what happens for $a \neq 0$ (nothing new).

  • 1
    $\begingroup$ Indeed, it looks like x^2. When a=0. With the domain (0-infinity) we exclude negative numbers. With that in mind, couldn't it be one-to-one for some values of a? $\endgroup$ – Algific Nov 18 '10 at 12:14
  • $\begingroup$ It doesn't matter if $a>0$ or $a<0$. You're just moving the parabola to the right or to the left. $\endgroup$ – Agustí Roig Nov 18 '10 at 18:33

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