1
$\begingroup$

I was looking through my calculus notes on my book regarding the properties of functions. I came upon two questions which appear to be unclear, so if anybody could explain to me what they mean and how to get to the answer it would be really nice.

1) A differentiable function which is not 1:1 has either a maximum or a minimum . Does this mean that whichever function which is NOT 1:1 (whatever that means) may have a min or a max?

2) A strictly increasing function has to be 1:1. Here again , i am not able to grasp the function of a 1:1 question.

Any explanation would be really usefull :)

$\endgroup$
4
  • $\begingroup$ An injective function (also called a one-to-one function) is a function $f~:~X\to Y$ such that for every $x_1,x_2$ if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$. In other words, if you take two different inputs, they must map to two different outputs. A function which is not injective must have at least one example of a pair of inputs $x_1,x_2$ such that despite $x_1\neq x_2$ you still have $f(x_1)=f(x_2)$. $\endgroup$
    – JMoravitz
    Jun 18, 2018 at 18:52
  • $\begingroup$ The first statement seems to be incorrect unless you change the phrase to be about local maximums or minimums. $f(x)=x^3-x=x(x+1)(x-1)$ is a differentiable function which is not one-to-one (e.g. $-1,0$ and $1$ all map to the same output of zero) but it is unbounded in both directions and as such has no max or min. The second statement is true since being strictly increasing means if $x_1<x_2$ then $f(x_1)<f(x_2)$ (for whatever < makes sense in the contexts of the domain and codomain) which directly implies that if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$ $\endgroup$
    – JMoravitz
    Jun 18, 2018 at 18:55
  • $\begingroup$ I grasped the concept of 1:1 with you answer well, but im still unsure about the first question ... i saw the question as stated above. However, wouldn't the question imply that a function which is not 1:1 , so many functions that actually exist, have either a maximum or minimum'? $\endgroup$ Jun 18, 2018 at 19:01
  • $\begingroup$ The question seems kind of unclear. Might i read it as if a function which is not 1:1 NEEDS to have a max or a min ? because in that case, i think the statement would indeed be false $\endgroup$ Jun 18, 2018 at 19:03

2 Answers 2

1
$\begingroup$

A 1-1 function means if $f(a) = f(b)$ then $a = b$. That for any value $c$ there are not two different $a,b$ so that $f(a) = c; f(b)=c$.

This makes 2) trivial. If $f$ is not 1-1 then there are some $a,b,$ $a\ne b$ and $f(a) = f(b)$ then either $a < b$ and $f(a) \not < f(b)$ so not increasing, or $b < a$ and $f(b) \not < f(a)$ so not increasing. So if $f$ is increasing it is impossible for $f$ to be not 1-1.

1) is refering to a local maximum/minimum. $f(x) = x^3 - x$ is not 1-1 because $f(0) = f(1) = f(-1) = 0$. But it has no global maximum/minimum. But is has a local maximum at $x=-\frac 1{\sqrt 3}$ and a local minimum $x = \frac 1{\sqrt 3}$ (It's a local minimum because all the $x$ near $\frac 1{\sqrt 3}$ yeild a larger $f(x)$. BUt not all the $x$ in the real numbers do. [Obviously $x = -10000$ is a much smaller result.]).

If $f$ is not 1-1 it might have local max/min but doesn't have to. But if it doesn't it can't be differentiable.

$f(x) = \begin{cases} x& x\in \mathbb Q\text{ and the denominator is odd}\\-x& x\in \mathbb Q\text{ and the denominator is even} \\ 0& x\not \in \mathbb Q\end{cases}$

Is not 1-1 as $f(x) = f(y) = 0$ if $x, y \not \in \mathbb Q$. But it has no max or min as every neighborhood of $x$ will have rationals with even and odd denominators so every neighborhood of $x$ will have $f$ values greater or less than $f(x)$.

$\endgroup$
6
  • $\begingroup$ Love the expalantion , thank you . just to make sure i grasped the concept, that would imply that the first statement is false and the second one true? $\endgroup$ Jun 18, 2018 at 19:53
  • $\begingroup$ Is it correct to say that a differentiable function which is not 1-1 might as well NOT have a local min /max? or would that be wrong? asuming what you just wrote, that if a 1-1 function doens't have a local min/max it can't be differentiable $\endgroup$ Jun 18, 2018 at 20:02
  • $\begingroup$ Finally implying that a DIFFERENTIABLE function (not 1-1) NEEDS to have a local min /max, because otherwise it wouldn't be differentiable? $\endgroup$ Jun 18, 2018 at 20:03
  • $\begingroup$ No, a differentiable function that is not 1-1 will have an $a < b$ so that $f(a) = f(b)$ and there must be a local max/min between a and b. That's the extreme value theorem. It's not so much that not having a max min means it wouldn't be differentiable so much as a differentable non 1-1 must have an extreme value. $\endgroup$
    – fleablood
    Jun 18, 2018 at 20:15
  • $\begingroup$ Okay , but is it always the case? meaning that a differentiable non 1-1 function must ALWAYS have a min /max according to the extreme value theorem? $\endgroup$ Jun 18, 2018 at 21:10
0
$\begingroup$

1) Any function, be it 1:1 or not , MAY have a minimum or a maximum.

2) 1:1 means that inputting different values yields different values as output. So if a function $f$ is (strictly) increasing,and $x, x'$ are in the domain of $f$, $x\ne x'$, say $x<x'$, then $f(x) <f(x')$, so they can't be equal.

$\endgroup$
7
  • $\begingroup$ Does that imply that the second question is indeed true? $\endgroup$ Jun 18, 2018 at 19:04
  • $\begingroup$ That every striclty increasing function is indeed one to one? $\endgroup$ Jun 18, 2018 at 19:11
  • $\begingroup$ Yes, absolutely. The same is true if, more generally, the function is strictly decreasing. $\endgroup$
    – Bernard
    Jun 18, 2018 at 19:12
  • $\begingroup$ Is it right to say , taking into consideration the first question , that a differentiable function NEEDS to have a local min /max, because otherwise it would not be differntiable? $\endgroup$ Jun 18, 2018 at 20:04
  • $\begingroup$ Why should it be so? Tthe function $f(x)=2x$ has neither a maximum nor a minimum on $\mathbf R$. $\endgroup$
    – Bernard
    Jun 18, 2018 at 20:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .