What Are the Inflection Points of a Second Derivative Function?

[greg_rack]

Homework Statement

$$y=\sqrt[3]{e^x-1}$$

Relevant Equations

Theorems for second derivatives

Since the index of the root is odd, the domain is going to be $R$, and I can calculate the second derivative to be:
$$y''=\frac{1}{3}\times \frac{e^x(e^x-3)}{3(e^x-1)^{\frac{5}{3}}}$$
Studying the sign of this function, it results positive for $x<0 \vee x>ln(3)$, so the main function will be convex in these intervals.

My doubt comes with the inflection points: $ln(3)$ will surely be one since it's a root of the nominator and in its neighborhood the derivative switches sign, but what about $0$, for which the second derivative is not zero, but instead undefined(denominator=0)?
How do I treat such critical points?

 

[Mark44]

My doubt comes with the inflection points: $ln(3)$ will surely be one since it's a root of the nominator and in its neighborhood the derivative switches sign, but what about $0$, for which the second derivative is not zero, but instead undefined(denominator=0)?

The first derivative is also undefined at x = 0. An inflection point is a point for which the slope changes sign; i.e., changes from pos. to neg. or from neg. to pos.

 

[SammyS]

Homework Statement:: $$y=\sqrt[3]{e^x-1}$$
Relevant Equations:: Theorems for second derivatives

Since the index of the root is odd, the domain is going to be $R$, and I can calculate the second derivative to be:
$$y''=\frac{1}{3}\times \frac{e^x(e^x-3)}{3(e^x-1)^{\frac{5}{3}}}$$
Studying the sign of this function, it results positive for $x<0 \vee x>ln(3)$, so the main function will be convex in these intervals.

My doubt comes with the inflection points: $ln(3)$ will surely be one since it's a root of the nominator and in its neighborhood the derivative switches sign, but what about $0$, for which the second derivative is not zero, but instead undefined(denominator=0)?
How do I treat such critical points?

At an inflection point, the concavity of the function changes between opening up and opening down. Thus, the sign of the second derivative is different on each side of the inflection point. Of course the function must be continuous there, but the derivatives may be undefined at an inflection point.

 

[greg_rack]

Got it! @Mark44 @SammyS
This thing of undefined derivatives is always confusing me up... I often can't really get the geometrical/mathematical meaning of an undefined derivative.

Anyway, in conclusion, the research for inflection points should only be based on analyzing the neighborhoods of critical points(being either roots or points excluded from the domain of the derivative), and looking for changes in concavity?

 

[Delta2]

I often can't really get the geometrical/mathematical meaning of an undefined derivative.

Here are a few tips that might help you with this:

A function f whose derivative at a point $x_0$ doesn't exist usually implies at least one of the following things:

1. The function f is not defined at the point $x_0$ i.e $f(x_0)$ doesn't exist. In this case the graph of f(x) will have a gap at the point $x=x_0$.
2. The function f is defined at the point $x_0$ but it is not continuous at the point $x_0$. In this case the graph of f(x) will have a jump at the point $x=x_0$
3. The function f is defined at the point $x_0$ and is continuous at the point $x_0$ but the derivative does not exist at this point. In this case the graph of f(x) will have a sharp corner at the point $x=x_0$. In order for the derivative to exist all changes in the slope of the graph must be done in a smooth curved way, not with sharp corners. Typical example of such function is $f(x)=|x|$ at the point $x_0=0$.

It also worth to take a look at the following wikipedia article
Weierstrass function - Wikipedia
which is a function that is continuous everywhere but nowhere differentiable.

 

[greg_rack]

Here are a few tips that might help you with this:

A function f whose derivative at a point $x_0$ doesn't exist usually implies at least one of the following things:

1. The function f is not defined at the point $x_0$ i.e $f(x_0)$ doesn't exist. In this case the graph of f(x) will have a gap at the point $x=x_0$.
2. The function f is defined at the point $x_0$ but it is not continuous at the point $x_0$. In this case the graph of f(x) will have a jump at the point $x=x_0$
3. The function f is defined at the point $x_0$ and is continuous at the point $x_0$ but the derivative does not exist at this point. In this case the graph of f(x) will have a sharp corner at the point $x=x_0$. In order for the derivative to exist all changes in the slope of the graph must be done in a smooth curved way, not with sharp corners. Typical example of such function is $f(x)=|x|$ at the point $x_0=0$.

It also worth to take a look at the following wikipedia article
Weierstrass function - Wikipedia
which is a function that is continuous everywhere but nowhere differentiable.

That was really exhaustive and helpful, thank you!

And could we write something similar for the second derivative, related to the original function? In which cases may that one be undefined, and why?

 

[Delta2]

The same three hold for $f''(x_0)$ existence but in addition two more things

1. f' is defined at $x_0$ but f' is not continuous at $x_0$. Unfortunately I don't know how to reduce this condition in terms of the original function f.
2. f' is defined at $x_0$ and f' is continuous at $x_0$ but f' is not differentiable at $x_0$ because the graph of f' has sharp corner at $x_0$. Again I don't know how to reduce this condition in terms of the original function f.

 

[SammyS]

3. The function f is defined at the point $x_0$ and is continuous at the point $x_0$ but the derivative does not exist at this point. In this case the graph of f(x) will have a sharp corner at the point $x=x_0$. In order for the derivative to exist all changes in the slope of the graph must be done in a smooth curved way, not with sharp corners. Typical example of such function is $f(x)=|x|$ at the point $x_0=0$.

This situation ( function f is defined at the point $x_0$ and is continuous at the point $x_0$ but the derivative does not exist at this point) can also occur if the tangent line at $x_0$ is vertical. That is the case for the function in this thread.
@greg_rack: Have you graphed the function?

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