# This second derivative is showing a point of inflection rather than a minimum point

For the curve $y=(2x-1)^4$, the derivative $8(2x-1)^3$ shows that the only stationary point is $(0.5, 0)$, which means it just touches the $x$-axis at that point. However, when determining the nature of the stationary point, the second derivative is $48(2x-1)^2$. Substituting $x=0.5$ into the second derivative gives $\frac{d^2y}{dx^2}=0$, which implies the stationary point is a point of inflection. However, when drawing the graph, it is actually a minimum point.

I also checked it by substituting $x=0.49$ and $x=0.51$ into the first derivative, which gives a negative and positive gradient, respectively, showing a minimum point.

So why does the second derivative show a point of inflection?

Shouldn't it be positive, which shows a minimum point?

Is it possibly to do with the fact that the curve does not actually cross the $x$-axis but only touches it, or is that irrelevant?

• I don't understand. May 28 '17 at 13:11
• It's because the third derivative is also $0$. So the second derivative being $0$ doesn't mean it's an inflection point. May 28 '17 at 13:11
• So does that mean to be absolutely sure about inflection points, you also have to check the third derivative? May 28 '17 at 13:12
• Yes. Or better yet, just check the sign of the derivative on either side of the stationary point. Much quicker, especially for more complicated functions. May 28 '17 at 13:13
• This is an A-level chain rule maths question from C3, so we haven't learnt about the third derivative. May 28 '17 at 13:13

$\dfrac{d^2y}{dx^2}=0$ does not imply that there is a zone of inflection. As an example, $y=x^4$. This has a stationary point at $x=0$, and the second derivative here is $12x^2=12\cdot0^2=0$. But this is not a zone of inflection, it is a minimum. The point is, second derivative being $0$ is not a definitive way of checking if there is a zone of inflection. It is true that at any zone of inflection, the second derivative will be $0$, but the converse is not true, as illustrated by my example above.

An alternative method is that you can check values either side of the stationary point to see what the true nature is. (Alternatively you could check higher derivatives, but that is sometimes tedious).

In this case, $$y=(2x-1)^4\\y'=8(2x-1)^3$$ You know there is a stationary point at $\frac12$, so substitute in $x=\frac12\pm\epsilon$ where $\epsilon$ is small. Then $$y'=8(\pm2\epsilon)^3=\pm16\epsilon^3$$ So is $\epsilon<0$ then $y'<0$, and if $\epsilon>0$ then $y'>0$. This means it is a minimum, since the function is decreasing just before the stationary point (i.e. for $\epsilon<0$), and is increasing after (i.e. for $\epsilon>0$).

• Ok that makes sense, I usually check my answers using this method. But I don't fully understand the point about the second derivative being 0 not necessarily meaning a point of inflection. May 28 '17 at 13:25
• If $\frac{d^2y}{dx^2}=0$ does not always mean a point of inflection then what else does it mean? May 28 '17 at 13:26
• The second derivative being zero simply means that the gradient function has a stationary point at that point ($\frac12$) in your example. May 28 '17 at 13:27
• So is it only when the second derivative is positive or negative that you can definitely be sure that it is a minimum/maximum point? May 28 '17 at 13:30
• @kashveyron Yes, exactly :) The rules are $y''>0\iff \text{minimum}$, $y''<0\iff \text{maximum}$, $y''=0\impliedby \text{inflection}$, which means $y''=0$ doesn't guarantee anything about the point. May 28 '17 at 13:32

Just like $f'(x_0)=0$ doesn't mean $x_0$ is a point of maximum or minimum, $f''(x_0)$ doesn't mean $x_0$ is a point of inflection.

The usual example for the first case is $f(x)=x^3$: $f'(0)=0$, but $0$ is a point of inflection and neither a point of maximum nor a point of minimum.

The condition that $f''(x_0)=0$ when $x_0$ is a point of inflection is only necessary and not sufficient (for points where the second derivative exists).

Your function is essentially the same as $f(x)=x^4$, which obviously has a minimum at $0$.

For “well behaved” functions (more precisely, analytic, but the condition can be relaxed), a point of inflection is where the first derivative changes from increasing to decreasing or conversely: so they're the points where the first derivative has a maximum or minimum.

For $f(x)=x^4$, we have $f'(x)=4x^3$, and this function has neither a maximum nor a minimum at $0$.

When you're looking for the points of inflection, you certainly find the points where the second derivative vanishes. Among them there are the inflection points and you have to check for the sign of the second derivative at either side of the point.

A different method is to look at the lowest $n$ such that $f^{(n)}(x_0)\ne0$: if $n$ is even the point $x_0$ is either a maximum or a minimum; if $n$ is odd the point $x_0$ is a point of inflection (assuming $f''(x_0)=0$ to begin with).

However, this method might not be conclusive. Consider $$f(x)=\begin{cases} 0 & x=0 \\ \exp(-1/x^2) & x\ne0 \end{cases}$$ Then $f^{(n)}(0)=0$ for every $n$; $0$ is obviously a point of minimum.

The function $$F(x)=\int_{0}^x f(t)\,dt$$ has the same property: $F^{(n)}(0)=0$ for all $n$, and $F$ has a point of inflection at $0$.

On the other hand, the function $$f(x)=\begin{cases} 0 & x=0 \\ x^4\sin(1/x) & x\ne0 \end{cases}$$ has $f''(0)=0$, but the second derivative takes positive and negative values in every left and right neighborhood of $0$, so $0$ is not a point of inflection, but it's not a point of maximum or minimum either.

There can be other points of inflection, though: the function $f(x)=\sqrt{x}$ has an inflection at $0$, where it is not differentiable.

• +1 for the very nice explanation as well as the counter-example at the end of the answer. May 31 '17 at 8:11

you must use higher derivatives: $$y'=4(2x-1)^3\cdot 2$$ $$y''=12(2x-1)^2\cdot4$$ $$y'''=24(2x-1)\cdot 8$$ $$y^{iv}=24\cdot 16>0$$ therefore we have a minimum at $$x=\frac{1}{2}$$

• the order of the derivative must be odd May 28 '17 at 13:15
• Oh I think I get it now, is that because it is a quartic? May 28 '17 at 13:15
• yes thats right May 28 '17 at 13:16
• I don't fully understand what you mean by 'the order of the derivative must be odd'. May 28 '17 at 13:17
• You may have to check your derivatives : you forgot the "$2$" factor in $(2x-1)$. May 28 '17 at 13:17