Types of Stationary Point

If x_sp is the stationary point, then if we consider points either side of x_sp, there are 4 types of behaviour of the gradient.

	x<xsp	x = xsp	x>xsp	curve
(i)	+ve	zero	-ve	max
(ii)	-ve	zero	+ve	min
(iii)	+ve	zero	+ve	inflection
(iv)	-ve	zero	-ve	inflection

Stationary points, like (iii) and (iv), where the gradient doesn't change sign produce S-shaped curves, and the stationary points are called points of inflection.


How to determine if a stationary point is a max, min or point of inflection.

The rate of change of the slope either side of a turning point reveals its type.

But a rate of change is a differential. So all we need to do is differentiate the slope, dy/dx, with respect to x. In other words we need the 2nd differential, or

(dy/dx), more usually called (dee 2 y by dee x squared)

Examples

1. y(x) = 9x² - 2

= 18x

= 18

2. y(x) = 4x⁵ -

= 20x⁴ +

= 80x³ -

3. p = 3q³ - 4q² + 6

= 9q² - 8q

= 18q - 8

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Rules for stationary points
i) At a local maximum, = -ve

ii) At a local minimum, = +ve

iii) At a point of inflexion, = 0, and we must examine the gradient either side of the turning point to find out if the curve is a +ve or -ve p.o.i.

Examples

1. Taking the same example as we used before:

y(x) = x³ - 3x + 1

= 3x² - 3, giving stationary points at (-1,3) and (1,-1)

= 6x

At stationary point (-1,3), x = -1, so = -6, so it's a maximum.

At stationary point (1,-1), x = +1, so = +6, so it's a minimum

So we can finally sketch the curve:




2. y = x³ + 8

= 3x², which is equal to zero at the stationary point.

If 3x² = 0, x = 0, and so y = +8, so the stationary point is at (0,8).

= 6x

So, at the stationary point (0,8), = 0, so we have a point of inflexion. But dy/dx is +ve either side of this point (e.g. at x = +1, dy/dx = +3, at x = -1, dy/dx = +3), so the curve has a positive point of inflection.


3. Where are the turning point(s), and does it (or they) indicate a max or min in the function p(q) = 4 - 2q - 3q²?

= -2 - 6q, which at the turning point = 0
so -2 - 6q = 0, 6q = -2, q = -, and p = 4

We have one turning point at (-, 4).

= -6, so the turning point is a maximum.
(This is consistent with what we said earlier, that for quadratics if the x² term is -ve, we have a maximum).

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More complicated Differentiation Rules
A. Product Rule

If a function y(x) can be written as the product of two other functions, say u(x) and v(x), then the differential of y(x) is given by the product rule:

e.g. if y(x) = u(x) v(x)
then

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= u. + v.

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Examples

1. y(x) = (x² + 2)(x + 1)

let u = x² + 2, so that du/dx = 2x
let v = x + 1, so that dv/dx = 1

so = u. + v = (x² + 2).1 + (x + 1).2x

= 3x² + 2x + 2

(Note: we can check this by expanding out the brackets)

y(x) = x³ + x² + 2x + 1, dy/dx = 3x² + 2x + 2

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2. y(x) = x³(3 - + 3x²)
Let u = x³, and v = 3 - + 3x²
= x³( + 6x) + (3 - + 3x²).3x²