Second derivatives: Concave and convex functions

​​In a nutshell

The second derivative can be used to determine whether a curve is concave (curving down) or convex (curving up) at any given point. The point at which a curve changes from being either concave or convex, to the other is called the point of inflection. This is at the point(s) where the second derivative is equal to zero.

Second derivative reminder

Differentiating a curve's equation twice gives the second derivative:

$$\begin{aligned}\text{curve: }&y=f(x)\\\text{first derivative: }&\frac{\text dy}{\text dx}=f'(x)\\\text{second derivative: }&\frac{\text d^2y}{\text dx^2}=f''(x)\end{aligned}$$​​​​

Concave

If a section of a curve is "concave", then it is curving downwards - that is to say the value of the gradient is getting lower. Thus, the second derivative, which is the measure of the rate of change of the gradient, is non-positive.

​$$\text{concave at }x\text{: } f''(x)\leq0$$​​

An example of a concave curve is given below. It is concave for all values of $$x$$.​

Some curves are only concave in particular intervals.

Example 1

Show that the following curve is concave for all positive values of $$x$$.

$$y=5-x^3$$​​

Find the second derivative by differentiating twice:

$$\begin{aligned}\frac{\text dy}{\text dx}&=-3x^2\\\frac{\text d^2y}{\text dx^2}&=-6x\end{aligned}$$​​​

This is negative for all positive values of $$x$$ and hence the curve is concave for all positive values of $$\underline{x}$$.

Convex

If a section of a curve is "convex" then it is curving upwards and thus the value of the gradient is increasing. It follows therefore that the second derivative is non-negative.

​$$\text{convex at }x\text{: } f''(x)\geq0$$​​

A curve that is convex for all $$x$$ is given below.

Curves can be convex in only some intervals of $$x$$.

Example 2

Find the interval in which the following curve is convex.

$$y=6x^2-\frac{x^4}{4}$$​​

You seek the interval(s) in which the second derivative is non-negative, so start by finding the second derivative:

$$\begin{aligned}\frac{\text dy}{\text dx}&=12x-x^3\\\frac{\text d^2y}{\text dx^2}&=12-3x^2\end{aligned}$$​​

Now set this to be non-negative and solve:

$$\begin{aligned}12-3x^2&\geq0\\12&\geq3x^2\\4&\geq x^2\end{aligned}$$​​

It follows that this curve is convex for $$\underline x$$ such that $$\underline{-2\leq x\leq2}$$.

Point of inflection

The point at which the second derivative changes sign is called a point of inflection. That is not to say that every point where the second derivative is zero is a point of inflection. 

Note: Showing that the second derivative is zero at a point is not enough to show that the point is a stationary point. You must also show that the sign changes on either side. For example, the point where $$x=0$$ on the curve $$y=x^4$$ has a zero second derivative, but is not a point of inflection.​

Points of inflection are often confused with stationary points, but points of inflection do not have to be stationary points. In particular, a point of inflection occurs on a curve when $$f''(x)=0$$ and the sign of $$f''(x)$$ is different to either side. It is the point at which the gradient goes from getting bigger to getting smaller, or vice versa. 

Points of inflection may not be easily spotted on a sketch of a curve, so stick to algebra to find them.

Example 3

Find the point of inflection on the curve $$y=f(x)$$ where $$f(x)=x^3+6x^2-10x-5$$.

One way to approach this would be to notice that because it is a cubic, there will be a point of inflection between two the stationary points (in the case that there is a single stationary point, this will also be the point of inflection). 

Though instead you can find the $$x$$ such that $$f''(x)=0$$ and check for a sign change of $$f''(x)$$​ on either side.

$$\begin{aligned}f(x)&=x^3+6x^2-10x-5\\f'(x)&=3x^2+12x-10\\f''(x)&=6x+12\end{aligned}$$​​

Setting this to zero and solving gives:

$$\begin{aligned}0&=6x+12\\-2&=x\end{aligned}$$​​

Now check for a change in sign of the second derivative to either side:

$$\begin{aligned}f''(-2.1)&=6(-2.1)+12=-0.6<0\\f''(-1.9)&=6(-1.9)+12=0.6>0\end{aligned}$$​​

Hence there is a change in sign, so there is a point of inflection when $$x=-2$$. The $$y$$-coordinate can be found by inserting this into $$f(x)$$:

$$f(-2)=(-2)^3+6(-2)^2-10(-2)-5=31$$​​

So the point of inflection is at $$\underline{(-2,31)}$$​.