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I wanted to know why is that so. I am just lost and want a detailed explanation.

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**Short answer:** The second derivative at an inflection point may be zero but *it also may be undefined*.

**Longer answer:**

One definition of an inflection point is where the second derivative changes sign (from positive to negative or the reverse).

The Intermediate Value Theorem for derivatives says that if a derivative is defined on a closed interval then it assumes any intermediate value between the values at the endpoints of that interval. The second derivative is obviously a derivative, so if it is defined on any interval including the inflection point, if we look at a value on one side of the inflection point, which must be positive, and another value on the other side, which must be negative, there is some place where the second derivative is zero. The only place it can be zero is at the inflection point. Therefore, it is commonly said that the second derivative at the inflection point must be zero.

However, there is one more possibility. **The second derivative may not be defined at the inflection point**. This does not satisfy the assumptions of the Intermediate Value Theorem so there is no problem.

One example where the second derivative is undefined at an inflection point is $y=\sqrt[3] x$, where there is an inflection point at $x=0$ but the first and second derivatives are not defined there. You can see the curve is concave up on the left and concave down on the right.

By definition, inflection points are where a function *changes* concavity, or in other words where a function is neither concave up nor down but is (*often*) moving from one to the other. A positive second derivative corresponds to a function being concave up, and a negative corresponds to concave down, so it makes sense that it is when the second derivative is 0 that our function is changing concavity, and hence corresponds to an inflection point.

Perhaps this image can give you a visual idea of when you can see concavity changing, and a corresponding inflection point. It is also important to note (*as several answers have done already*) that an inflection point can also have an undefined second derivative.

Think about what the second derivative means. A positive second derivative means concave up, negative means concave down. Well, an inflection point is when the concavity switches. So naturally the second derivative has to equal zero at some point if our second derivative is going to switch signs.

It’s very analogous to a critical value.

An inflection point is the point where the concavity changes.

At the inflection point, the curve might go from concave down (with a decreasing slope) to concave up(with an increasing slope). A decreasing slope will give you a decreasing derivative and a negative second derivative. An increasing slope will give you an increasing derivative and a positive second derivative. At the inflection point, the second derivative changes from negative to positive, and must be zero.

If at the inflection point, the curve goes from concave up to concave down, then by same argument, the second derivative will change from positive to negative and must be zero.

Why is the first derivative of an extremum zero?

The reason is same for the second derivative at inflection point as well… because … slope is extremized at PI.

The first derivative at an inflection point locally is either a maximum or a minimum.

The second derivative at an inflection point vanishes.

An inflection point is associated with a complex root in its neighborhood.

An inflection point occurs on half profile of M type or W type, two inflection points occur on full profiles of M type or W type.

The last two are the ways I look at an inflection point, mentioned even if not directly in answer to the OP, for a broader perspective. Please comment

Graph of $ \sin (p \,x) / p + \sin (q\, x) /q = 0 ,( q = 2.112, p = 0.687 )$ with first and second derivatives of the function.

We require $ f \in C^2 $

By taylors theorem:

$f(x)=f(a)+f'(a)(x-a)+\frac{f”(a)}{2}(x-a)^2 +R$

Let $ f'(a)=0$

Now if $f”(a)>0$ then by continuity there is a small ball around $a$ such that $f”(a)$ is positive so $f(x+\epsilon) > f(x) $ and $f(x-\epsilon) > f(x) $.We do the same for $f”(a)<0$. So now $f”(a)=0$ and inflection points occur when $f”(a)=0$

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