# Convex and Concave Curves

## Convex and Concave Curves Revision

**Convex and Concave Curves**

We define a curve f(x) as **convex** or **concave** when we observe the behaviour of f''(x).

Make sure you are happy with the following topics before continuing.

**Distinguishing Between Convex and Concave**

**Convex curves** curve downwards and **concave curves** curve upwards.

That doesn’t sound particularly mathematical, though…

- When f''(x) \textcolor{purple}{> 0}, we have a portion of the graph where the gradient is
**increasing**, so the graph is**convex**at this section. - When f''(x) \textcolor{red}{< 0}, we have a portion of the graph where the gradient is
**decreasing**, so the graph is**concave**at this section.

An easy way to test for both is to connect two points on the curve with a straight line.

- If the line is above the curve, the graph is
**convex**. - If the line is below the curve, the graph is
**concave**.

**Points of Inflexion**

A **point of inflexion** occurs when the curve transitions from **convex** to **concave** or vice versa.

We’re looking for sections of the graph where f''(x) = 0.

**Note:Â **While all points of inflexion have f''(x) = 0, not all points where f''(x) = 0 are points of inflexion. We have to check the curve actually changes from **convex** to **concave** or vice versa by seeing what happens on either side of the point.

For example, for f(x) = x^3 - 3x^2 + 2x + 1

\begin{aligned} f'(x) &= 3x^2 - 6x + 2 \\ f''(x) &= 6x - 6 \end{aligned}

So,

f''(x) = 0 when x = 1

Then,

f''(x) \textcolor{red}{< 0} for x < 1

f''(x) \textcolor{purple}{> 0} for x > 1

The curve changes from **concave** to **convex** at x=1, so there is a point of inflexion at x = 1.

**Note:** If there is a point of inflexion that is also a stationary point (i.e. f'(x) = 0 also), then it is called a **stationary point of inflexion**.

**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.

## Convex and Concave Curves Example Questions

**Question 1:** Show that x^4 + 2x^2 has no points of inflexion.

**[3 marks]**

Let f(x) = x^4 + 2x^2. Then

f'(x) = 4x^3 + 4x and f''(x) = 12x^2 + 4

For any points of inflexion, f''(x) = 0.

Equating 12x^2 + 4 =0, we have x^2 = - \dfrac{1}{3}. We are unable to find a (real) root for this, so conclude that there are no points of inflexion.

**Question 2:** Find the point(s) of inflexion of \sin x, for 0 \leq x \leq 2\pi measured in radians.

**[3 marks]**

f(x) = \sin x gives

f'(x) = \cos x and f''(x) = -\sin x

f''(x) = 0 means that x = 0, \pi, 2\pi.

So, we have points (0, 0), (\pi, 0) and (2\pi, 0).

**Question 3:** The function f(x) = 2x^4 - 3x^3 has two stationary points. Show that only one of them is a point of inflexion.

**[4 marks]**

f(x) = 2x^4 - 3x^3 gives

f'(x) = 8x^3 - 9x^2 and f''(x) = 24x^2 - 18x = 6x(4x - 3)

For f''(x) = 0, we either require

x = 0 or x = \dfrac{3}{4}

Using these values in f'(x), we have

f'(x) = 0 when x = 0

and

f'(x) = - \dfrac{27}{16} when x = \dfrac{3}{4}

Therefore, we have one stationary point of inflexion when x = 0.

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