Convex and concave curves and points of inflectionEdexcel A-Level Maths: Revision notes

The second derivative describes how the gradient changes.

• Where $f''(x)>0$ the gradient is increasing and the curve is convex (it bends upwards, like a bowl, and lies above its tangents).
• Where $f''(x)<0$ the gradient is decreasing and the curve is concave (it bends downwards and lies below its tangents). Example: $y=x^{3}-6x^{2}+2$ has $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=6x-12$, so it is concave for $x<2$ and convex for $x>2$.

A point of inflection is a point where the curve changes from convex to concave or vice versa. At such a point $f''(x)=0$ and $f''(x)$ changes sign. To find one: solve $f''(x)=0$, then show the sign of $f''$ is different either side, then find $y$. For $y=x^{3}-6x^{2}+2$, $f''(x)=6x-12=0$ gives $x=2$, with $f''<0$ before and $f''>0$ after; $y=-14$, so the point is $(2,-14)$. A point of inflection need not be stationary: the gradient there is usually non-zero.

At a stationary point, $f''>0$ gives a minimum and $f''<0$ a maximum, but if $f''(x)=0$ the test fails: the point may be a minimum, a maximum or a point of inflection. Check the sign of $f'(x)$ on either side, or the sign change of $f''$. Take $y=x^{n}$ with $n>2$: $f'(0)=0$ and $f''(0)=0$. If $n$ is even (e.g. $x^{4}$) the origin is a minimum: $f'$ goes from negative to positive and $f''=12x^{2}$ does not change sign. If $n$ is odd (e.g. $x^{3}$) the origin is a stationary point of inflection: $f''=6x$ changes sign and $f'=3x^{2}\geq0$ either side.

Use the second derivative with other features: intercepts, stationary points and their nature, and the intervals where the curve is convex or concave. Example: $y=x^{4}-8x^{3}+18x^{2}$ has $f''=12(x-1)(x-3)$, so inflections at $(1,11)$ and $(3,27)$ and concave for $1<x<3$. Also $f'=4x(x-3)^{2}$ is zero at $x=0$ (minimum) and $x=3$ (stationary inflection). Between the inflections the curve bends downwards; outside them it bends upwards.

If $N$ is a quantity changing with time, $\frac{\mathrm{d}N}{\mathrm{d}t}$ is its rate of growth and $\frac{\mathrm{d}^{2}N}{\mathrm{d}t^{2}}$ is how fast that rate is changing. A point of inflection on an $N$-$t$ curve is where the rate of growth is greatest or least. Example: $N=3t^{2}-\frac13t^{3}$ has $\frac{\mathrm{d}N}{\mathrm{d}t}=6t-t^{2}$ and $\frac{\mathrm{d}^{2}N}{\mathrm{d}t^{2}}=6-2t$. The inflection at $t=3$ is where the rate of growth is greatest ($9$ thousand users per week).