Derivation of a Higher Order Derivative Test

[bagasme]

TL;DR

How can we generalize first and second-order derivative tests to higher derivatives?

Hello,

In second-order derivative test, the test is inconclusive when $f''(c)=0$, so we had to generalize to higher-order derivative test.

I was wondering how such tests can be generalized and derived?

For example, how can I prove that $f(x)=x^4$ have minimum at 0?

Bagas

 

[Mark44]

Summary:: How can we generalize first and second-order derivative tests to higher derivatives?

Hello,

In second-order derivative test, the test is inconclusive when $f''(c)=0$, so we had to generalize to higher-order derivative test.

I was wondering how such tests can be generalized and derived?

For example, how can I prove that $f(x)=x^4$ have minimum at 0?

For this example, there's no need for a test involving higher derivatives. It's easy to see that this function has its minimum at 0 because, if x > 0, then f'(x) > 0, and if x < 0, then f'(x) < 0. In other words, f is decreasing on $(-\infty, 0)$ and is increasing on $(0, \infty)$. It's as simple as that.

 

[bagasme]

For this example, there's no need for a test involving higher derivatives. It's easy to see that this function has its minimum at 0 because, if x > 0, then f'(x) > 0, and if x < 0, then f'(x) < 0. In other words, f is decreasing on $(-\infty, 0)$ and is increasing on $(0, \infty)$. It's as simple as that.

Nope, I just want higher derivative test derivation.

 

[FactChecker]

Assume that the function is analytic. If the first non-zero derivative at the point $x_0$ is the n'th derivative, then the function behaves locally like $f(x_0)+(f^{(n)}(x_0)/n!) (x-x_0)^n$. So $x_0$ is a local minimum if $n$ is even and $f^{(n)}(x_0) \gt 0$.

 

[bagasme]

I saw the derivation on SE forum, is the explanation right?

 

[FactChecker]

I saw the derivation on SE forum, is the explanation right?

Yes. In fact, it points out a correction that I must make by adding $f(x_0)$ to my post. I will edit and fix my post.