The second and third derivitive tests give unexpected $0$

Question

The final question on a mock paper I recently did was the following:

The curve $y=f(x)$ has $\frac{\text{d}y}{\text{d}x}=kx(x-3)^3$, where k is a negative constant.

There is a stationary point at $x=3$

Classify this stationary point.

My thought process:
$$\begin{align} \text{Find the nature of the point}\\ f'(x) & = kx(x-3)^3 \\ & = kx(x^3-9x^2+27x-27) \\ & = kx^4-9kx^3+27kx^2-27kx \\ f''(x) & = 4kx^3-27kx^2+54kx-27k \\ f''(3) & = 4k\cdot(3)^3-27k\cdot(3)^2+54k\cdot(3)-27k \\ & = 0 \\ & \therefore x=3 \text{ is an inflection.} \\ \text{Find the direction of inflection} \\ f'''(x)& = 12kx^2-54kx+54k \\ f'''(3)& = 12k\cdot(3)^2-54k\cdot(3)+54k \\ & = 0 \end{align}$$

If $f'''(x)>0$ it is an increacing inflection, $f'''(x)<0$ it is decreacing inflection, but $f'''(3)=0$ gives me no information.

After the paper, I found that nobody else used the second derivitive method and all found the point was a maximum. Did I make an error in differentiation, or do the second and third derivitives sometimes not work?

Edit: the mark scheme mentions the fourth derivitive, but doesn't explain why it's used.

Answer 1

To test if a point is a point of inflection (using derivatives as opposed to comparing gradients) you need the lower-order non-zero derivative to be odd. If the lower-order non-zero derivative is even then you have an undulation point.

If you evaluate $f''''(3)$ you'll get $18k$ so the point is not a point of inflection but is instead an undulation point.

Further reading

Answer 2

Note that the second derivative test doesn't always work. If $f(x)=x^{20}$, $x=0$ is clearly the location of a local minimum. But $f^{(n)}(0)=0$ for all $n$ less than 20. If $f''(x)$ is zero at a stationary point, this can mean an inflection point, but it doesn't have to. This is one case where the second derivative test is inconclusive.

Answer 3

Actually, your fellow students were right: The function $f$ does not have a point of inflection there. As a matter of fact, the derivative that we see does have a point of inflection, but the question is about the function itself.

Recall that it is sufficient for a local maximum that $f'(x)=0$ and $f''(x)<0$. As well it is sufficient for an increasing point of inflection to have $f''(x)=0$ and $f'''(x)>0$. However, these conditions are not necessary. Here we have an example that shows that sometimes we have to dig deeper: The more general case is that we have $f'(x)=f''(x)=\ldots =f^{(n)}(x)=0$ and $f^{(n+1)}(x)\ne 0$.

• If in such a case, $n$ is odd, then we have a local extremum (and the sign of $f^{(n+1)}(x)$ decides between minimum and maximum: $f^{(n+1)}(x)<0$ means local maximum, $f^{(n+1)}(x)>0$ means local minimum).
• If on the other hand $n$ is even, then our stationary point is a point of inflection. (And again $f^{(n+1)}(x)$ determines the type of inflection).
• For completeness' sake, it is possible that a function is infinitely differentiable and that all derivatives are $=0$ at some point without $f$ being constant. The function $$f(x)=\begin{cases}ae^{-1/x}&x>0\\be^{1/x}&x<0\\0&x=0\end{cases}$$ has this property at $x=0$ and by choice of $a,b$ we can make that point a local minimum or maximum or a point of inflection.
• Not to mention that also functions that are not differentiable as often as the sufficient criteria need can have local extrema. Just look at $f(x9=|x|$, where $f'(0)$ does not exist.

So back to the problem at hand: Since we find $f'(3)=f''(3)=f'''(3)=0$, we should compute $f''''(3)$ and see what happens.

Alternatively, we forget about the standard methods to check for local extrema and verify directly that $f(x)<f(3)$ for all $x$ with $x>0$ and $x\ne 3$. Indeed, for such $x$ we have $$f(x)-f(3)=(x-3)f'(\xi)=(x-3)\cdot kx(\xi-3)^3 $$ where $\xi$ is between $x$ and $3$. In particular, $x-3$ and $\xi-3$ have the same sign. Hence $f(x)-f(3)$ is $kx$ times the product of four non-zero numbers of same sign, i.e, $kx$ times something positive, i.e., something negative (because $k$ is negative and $x$ is positive). Hence $f(x)<f(3)$ as claimed.