Convex and Concave Curves

Example 1: Characterising Graphs

Say we have a graph of the function $f(x) = x(x^2 + 1)$.

Find the parts of the graph where the function is convex or concave, and find the point(s) of inflexion.

[3 marks]

$f(x) = x(x^2 + 1) = x^3 + x$ gives

$f''(x) = 6x$

$f''(x) = 0$, when $x = 0$

$f''(x) textcolor{red}{< 0}$ when $x<0$. Here we have a concave section.

$f''(x) textcolor{purple}{> 0}$ when $x>0$. Here we have a convex section.

When $6x = 0$, i.e. $x=0$ , we have a point of inflexion, since the curve changes from concave to convex at this point.

So, the function is concave for $x < 0$, has a point of inflexion at the origin, and is convex for $x > 0$.

Example 2: Stationary Points of Inflexion

Show that the curve $y = x^3 – 6x^2 + 12x$ has a stationary point of inflexion at $x = 2$

[3 marks]

Find $dfrac{dy}{dx}$ and $dfrac{d^2 y}{dx^2}$:

$begin{aligned} dfrac{dy}{dx} &= 3x^2 – 12x + 12 [1.2em] dfrac{d^2 y}{dx^2} &= 6x – 12 end{aligned}$

When $x = 2$, $dfrac{dy}{dx} = 3(2)^2 – 12(2) + 12 = 0$, so there is a stationary point at $x = 2$

When $x = 2$, $dfrac{d^2 y}{dx^2} = 6(2) – 12 = 0$

When $x<2$, $dfrac{d^2 y}{dx^2} textcolor{red}{< 0}$ and when $x>2$, $dfrac{d^2 y}{dx^2} textcolor{blue}{> 0}$, so there is a point of inflexion at $x=2$.

Hence, the curve has a stationary point of inflexion at $x=2$.