Number of inflection points of $\log f(x)$ vs. $f(x)$ I am wondering, is there some (general) theorem or lemma (that I am not aware of) which would state something about number of inflection points of a logarithm of a positive function if we know the number of inflection points of the function (or the other way)?
In other words, if we know that a positive function $f(x)$ has $n$ inflection points, can we say something about number of inflection points of $\log f(x)$? Also the other way. If we know that $\log f(x)$ has $m$ inflection points, can we say something about number of inflection points of $f(x)$? Thanks.
 A: In general, there is no relationship.
For example, take $f(x) = e^{x+(\sin x)/10}$. One can check that $f$ has no inflection points but $\log f$ has infinitely many inflection points.
On the other hand, let $h$ be a solution to the differential equation $h'(x) = (-1 + (\sin x)/2) h(x)^2$ (with say $h(0)=1$, on say $x\in[0,\infty)$). Let $g(x)$ be an antiderivative of $h(x)$, and let $f(x) = e^{g(x)}$. The inflection points of $f$ occur precisely where $g''(x)+g'(x)^2=0$ and thus where $h'(x) = -h(x)^2$, which are the infinitely many multiples of $\pi$. However, one can show that $h$ is positive and decreasing on $[0,\infty)$; therefore $g''(x)=h'(x)$ is never zero, hence $g=\log f$ has no inflection points.
A: Horizontal inflexions of the original curve are in one-to-one correspondence with horizontal inflexions of the log curve. Apart from that, there is no connexion.
Consider the regular curve parametrised by $\gamma(t) = (t,x(t))$. The curve has an inflexion  if and only if the first two derivatives are linearly dependent, i.e. $\dot{\gamma}\parallel \ddot{\gamma}$. In otherwords: $\ddot{x}(t)=0$.
Next, consider the curve $\alpha(t)=(t,\ln x(t))$. We have $\dot{\alpha}(t) = (1,\dot{x}/x)$ and 
$$\ddot{\alpha}(t) = \left(0,\frac{x\ddot{x}-\dot{x}^2}{x^2}\right)$$. 
Assuming that $x>0$ then $\dot{\alpha}\parallel \ddot{\alpha}$ if, and only if, $x\ddot{x}-\dot{x}^2=0$. If $\gamma(t_0)$ is an inflexion then we also need $\dot{x}(t_0)=0$ for $\alpha(t_0)$ to be an inflexion. If $\alpha(t_0)$ is an inflexion then we need $\dot{x}(t_0)=0$ for $\gamma(t_0)$ to be an inflexion. 
Hence, for $x>0$, if $\dot{x}=0$ then $\gamma(t_0)$ is an inflexion if, and only if, $\alpha(t_0)$ is an inflexion.
The log curve $\alpha$ is free to have inflexions when the original $\gamma$ does not: we simply need $\dot{x}\neq 0$ at such points. Conversely, the original curve $\gamma$ is free to have inflexions when the log curve $\alpha$ does not: we just need $\dot{x} \neq 0$ at that point.
A: Write out the two successive derivatives (assuming $f\in\mathcal C^2$): $$\frac{\mathrm d^2}{\mathrm dx^2}(\log\circ f)=\frac{\mathrm d}{\mathrm dx}\frac{f'}{f}=\frac{f''f-(f')^2}{f^2}$$
We want this to be zero at some $x\in\mathbb R$: $$f''(x)f(x)-(f'(x))^2=0$$
So we can see that all the original (stationary) inflection points of $f$ restricted to $(0,\infty)$ are still inflection points, but there's no clear way of knowing whether there's more, without additional information about $f$. Notice that we may have lost any inflection points that were in the negative domain of $f$. We cannot conclude anything if $f'(x)\not = 0$.
A: See here----
Actually the points of inflection of a function $y=f(x)$ means,where the function $y=f(x)$ has its 2nd order derivative equals to zero.
or we can say mathematically where or for what values of  $x$ the function has its 2nd order derivative = 0.   OR  $f '' ( x ) = 0 $.
the values of $x$ for which the equation  $f'' (x) = 0$ is satisfied we can say those points in the domain of the function,it has the points of inflections or the $f '' (x)$ is $0$.
See where the function has its 2nd order derivative $f '' (x) = 0$ , at that values of $x$ the function $y = f(x)$ is changing its behavior from concave down to concave up or concave up to concave down.
Mathematically we can say if a function $y = f (x)$ has its $f '' (x)$ (2nd order derivative) less than $0$ or $f " (x) < 0$ or its $f '' (x)$ value is negative  in a certain scale of $x$  then you can say that  the function is just concave down or if we draw the curve of $y = f(x)$ in a co-ordinate system we will have a curve just having a concave down nature.
And similarly if the function has its $f'' (x ) > 0$ in certain scale of $x$ then we say that the function has a concave up characteristics and if we draw the graph of $y= f(x)$ we will have just a concave up picture of the function $y = f(x)$.
please go to the link : coz I dont have the software to upload answer with pic.
The link is the graph of a simple function to implement the concepts which I said earlier above.
The link is a picture of a graph of $y = x^3$.
You can go through the link or you can have a google search for it for better understanding along with my concepts....
graph
See here  at the point $x=0$ the $f ''( x )$ of the function $f(x) = x^3$
that is $f'' (x) = 3(x^2)$ is equal to zero. See $f '' (x)$ for $x=0$ is  $f'' (0) = 3(0^2) = 0$.
So you see there is the point of inflection at $x=0$ in the curve of the function $y = f(x)$.
and see if  $x>0$ the $f'' (x)= 3x^2$ is also greater than $0$. So it is concave up for $x>0$
and it is also true for $x<0$ and it is concave down when $x<0$.
But see here the log $f(x)$ will be:  let us consider  $g(x)  =  \log f(x) = \log {x^3} =  3\log x$.
Now if you have the second order derivative of $\log {x^3}$ or   $3\log x$ , you will have  $g'' (x) =  \frac{- 3 }{ x^2}$.
see the function $g(x) = \log f(x)$ has no such points where the second order derivative is 0. 
So, 
There is no points of inflections for the function $g(x) = \log f(x)$.
I was trying to make your concept on points of inflections and see there is no such direct dependency between the number of inflection points for the function $f(x)$
 and  $\log f(x)$.
the function $\log f(x)$ will have the point of inflection at that values of $x$ for which the equation:      $ \frac{( f(x) f " (x)  -   (f ' (x))^2 )}{  ((f(x))^2) }  = 0$ has a solution for $x$.
Or simply you can say if the equation :
        $$                                                                f(x) f ''(x)  -  (f ' (x) )^2 = 0$$
has any  solution for $x$ or not.
If this equation has any solution then the function $g(x) = \log (f(x))$ will have the points of inflection. And you can also draw the curve of  $y = f(x)$ or  $g(x) = \log f(x)$ by applying the 2nd order derivative criteria.
Hope the answer will help you a bit.Best of luck.
