Let $f$ be an odd function such that:
$f(0)=0$
$f'(x)>0$ if $x>0$
$f(x) < 0$ if $x > 0$
$f$ has an inflection point in $0$.
Question : how can it be that
$f'$ is not defined in 0
$f"$ is not defined in 0
first derivative: shouldn't it be infinity at $0$ ? second derivative: shouldn't it be $0$ at $0$ ?
I guess the function goes straight up, for a while, at $0$ ? But that isn't really possible, because one $x$ would have more than a single $y$ value, thus wouldn't be a function anymore. So how would such a function possibly look? I goes straight up only in 1 point, x?
Thanks!
"f' is not defined [at] 0"
The prompt tells us that $f$ is negative and increasing for $x>0$. Since $f$ is an odd function, $f$ is positive and increasing (see below for why this is true) for $x<0$. We thus see that $f$ cannot be continuous at x = 0. Therefore, $f^\prime$ is not defined at $x=0$.
"f'' is not defined [at] 0"
Because $f^\prime$ isn't defined at $x=0$.
Why $f^\prime$ is increasing for $x<0$; i.e., why the derivative of an odd function is even
By definition, $f$ being odd means that $f(-x)=-f(x)$. Differentiating both sides gives us $-f^\prime(-x)=-f^\prime(x) \iff f^\prime(-x)=f^\prime(x)$. So the derivative of an odd function is an even function. $\ \square$
Let me know if you need further explanation.
Added later
Here's an example of such a function: $f(x)=\frac{-1}{x}$
Edit of the above
I forgot that $f(0)=0$, so consider this instead: $f(x)=\begin{cases} 0 & x=0 \\ \frac{-1}{x} & x\not=0 \end{cases} $