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 :)
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 $f(x)$.