# $f$ differentiable $5$ times around $x=a,\ f'(a)=f''(a)=f'''(a)=0,\ f^{(4)}(x) <0 \Rightarrow x=a$ is either a local minimum or a local maximum point

So I've been trying to prove the following statement:

Let $$f$$ be a function such that it is differentiable $$5$$ times around $$x=a$$. Prove or disprove, that if $$f'(a)=f''(a)=f'''(a)=0$$ and $$f^{(4)}(x)<0$$ then $$x=a$$ is either a local maximum or a local minimum point.

Since I couldn't find a counterexample, I believe this statement is false, and $$x=a$$ must be a maximum point, but I'm not certainly sure since I have no idea how to prove this. I thought using Taylor but it doesn't seem to help that much.

Thank you very much!.

• There's a theorem that says if the lowest-order nonzero derivative is of odd order, then you have a point of inflection. It's a critical point, so it's either a local extremum or a point of inflection. Jan 14, 2019 at 16:24
• But isn't the lowest-order nonzero derivative here is of even order? Jan 14, 2019 at 16:42
• Some heuristics; I’m not entirely sure it is correct but it feels like it is in the right direction... $f^{(4)}(x)<0$ implies $f^{(2)}(x)$ is convex down. In particular, I think the condition that $f’(a)=0$ means $f’’(a)$ is a maximum, which means $f’(x)$ changes sign in a neighborhood of $a$, and ultimately means $a$ is a local maximum. Again, take this with a grain of salt, I’m just out walking my dog :) Jan 14, 2019 at 17:30
• @AmitZach: Right, I was thinking we could use a contrapositive. But that would have to go the other way. Clayton's reasoning is the informal reasoning I was thinking of, for sure, but I'm not sure how to make it rigorous. Jan 14, 2019 at 17:47

The following theorem was proved by Colin Maclaurin in 1742.

Let $$f$$ be a real-valued function defined on an open interval $$J$$ which is $$(n-1)$$-times continuously differentiable in a neighborhood of a point $$a \in J$$ and for which moreover $$f^{(n)}(a)$$ exists.

Assume $$f'(a) = f''(a) = \dots f^{(n-1)}(a) = 0$$ and $$f^{(n)}(a) \ne 0$$. Then:

1) If $$n$$ is even, then $$f$$ has a local maximum [resp. local minimum] at $$a$$ if $$f^{(n)}(a) < 0$$ [resp. $$f^{(n)}(a) > 0$$].

2) If $$n$$ is odd, then $$f$$ has an inflection point at $$a$$.

So you see that your $$f$$ has a local maximum at $$a$$.

Edited:

The proof is based on Taylor's theorem. Under the above assumptions it is a bit tedious, so let us restrict to the special case that $$f$$ is $$n$$-times continuously differentiable in a neighborhood of $$a$$. This covers your question since you assume that $$f$$ is $$5$$-times differentiable.

We can write $$f(a + h) = f(a) + \dfrac{h^n}{n!}f^{(n)}(a + \theta h)$$, $$0 < \theta < 1$$. That means $$f(a + h) - f(a) = \dfrac{h^n}{n!}f^{(n)}(a + \theta h) .$$ Since $$f^{(n)}$$ is continuous in $$a$$ and $$f^{(n)}(a) \ne 0$$, we see for $$\lvert h \rvert < \epsilon$$ the sign of $$f^{(n)}(a + \theta h)$$ agrees with the sign of $$f^{(n)}(a)$$. This shows that for $$n$$ even $$f(a + h) - f(a)$$ has the same sign in a neigborhood of $$a$$. For $$n$$ odd the sign is different on both sides of $$a$$.

• Do you know if it was proven in the manner I described? Jan 14, 2019 at 18:30
• Yes, I edit my answer. Jan 14, 2019 at 18:31

Here is an alternate proof which doesn't require Taylor's Theorem (although admittedly it does use a lot of the same ideas, just in a much simpler form).

Since $$f^{(4)}(a) < 0$$, there exists some $$\delta > 0$$ such that whenever $$0 < |x - a| < \delta$$, we have $$\frac{f^{(3)}(x)}{x - a} = \frac{f^{(3)}(x) - f^{(3)}(a)}{x-a} < 0$$ (for example, using $$\epsilon := -\frac{1}{2} f^{(4)}(a)$$ in the definition of limit). This implies that $$f^{(3)}(x) < 0$$ for $$a < x < a + \delta$$, whereas $$f^{(3)}(x) > 0$$ for $$a - \delta < x < a$$. Now, for $$a < x < a + \delta$$, we have $$f''(x) = f''(x) - f''(a) = \int_a^x f^{(3)}(t)\,dt < 0$$; $$f'(x) = f'(x) - f'(a) = \int_a^x f''(t)\,dt < 0$$; and finally, $$f(x) = f(a) + \int_a^x f'(t)\,dt < f(a)$$. Similarly, for $$a - \delta < x < a$$, we have $$f''(x) = f''(x) - f''(a) = \int_a^x f^{(3)}(t)\,dt = -\int_x^a f^{(3)}(t)\,dt < 0$$; $$f'(x) = f'(x) - f'(a) = \int_a^x f''(t)\,dt = -\int_x^a f''(t)\,dt > 0$$; and finally, $$f(x) = f(a) + \int_a^x f'(t)\,dt = f(a) - \int_x^a f'(t)\,dt < f(a)$$.

Putting together these facts, we conclude that $$f(a)$$ is the maximum value of $$f$$ for $$x \in (a - \delta, a + \delta)$$, i.e. $$f$$ has a local maximum at $$x=a$$.