Math Differentiation W11

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Maximum, minimum and saddle point

A point on a curve at which the gradient is zero, i.e. where dydx

= 0 , is called a stationary point. At a

stationary point, the tangent to the curve is horizontal and the curve is “flat”.

There are three types of stationary points. We are going to have a look at these three types and find a

way to distinguish one from another.

Minimum point

In this case, the gradient of the curve is negative to the left of the point P . To the right of point P , the

gradient of the curve is positive.

to the left of P at point P to the right of P

dydx

< 0 dydx

= 0 dydx

> 0

Maximum point

In this case, the gradient of the curve is positive to the left of point P . To the right of point P , the

gradient of the curve is negative.

to the left of P at point P to the right of P

dydx

> 0 dydx

= 0 dydx

< 0

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Saddle point

In this case, the gradient has the same sign each side of the stationary point P .

to the left of P at point P to the right of P

dydx

> 0 dydx

= 0 dydx

> 0

to the left of P at point P to the right of P

dydx

< 0 dydx

= 0 dydx

< 0

Example: Find the coordinates of the stationary points on the curve y = x3 + 3x2 +1 and determine

their nature. Sketch the curve.

Solution: When y = x3 + 3x2 +1 , then dydx

= 3x2 + 6x . At a stationary point, dydx

= 0 . Therefore,

3x2 + 6x = 03x x + 2( ) = 0⇒ x1 = 0; x2 = −2

When x = 0⇒ y = 1⇒ point 0 /1( )

When x = −2⇒ y = 5⇒ point −2 / 5( )

To determine the nature of each stationary point, we must examine the gradient each

side of each point. For point 0 /1( ) :

x −1 0 1

dxdy

−3 0 9

negative positive \ _ /

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Note: When choosing x values each side of x = 0 , the chosen interval must not

include any other stationary point.

The stationary point 0 /1( ) is a minimum.

For the point −2 / 5( ) :

x −3 −2 −1

dxdy

9 0 −3

positive negative / _ \

The stationary point −2 / 5( ) is a maximum.

There are two stationary points on the curve: 0 /1( ) a minimum and −2 / 5( ) a maximum.

To sketch the curve, plot the stationary points and notice that the curve y = x3 + 3x2 +1

cuts the y -axis at the point 0 /1( ) .

Second derivative and stationary points Consider the graph of some function y = f x( ) which possesses a maximum, a minimum and a saddle

point. Consider also the graph of the derived function as shown below.

• At the maximum point P , the gradient of the derived function is negative. That is, d2ydx2

< 0 at a

maximum.

• At the saddle point Q , the gradient of the derived function is zero. That is, d2ydx2

= 0 .

• At the minimum point R , the gradient of the derived function is positive. That is, d2ydx2

> 0 .

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Note: Saddle point: If d2ydx2

= 0 , we generally speak of a point of inflexion. Such a point of inflexion is

called a saddle point, if dydx

= 0 as in the picture shown below.

This provides an easier method for classifying stationary points, as the following example illustrates.

Example: Find the coordinates of the stationary points on the curve f x( ) = x3 − 6x2 −15x +1 , and

using the second derivative to determine their nature.

Solution: When f x( ) = x3 − 6x2 −15x +1 , then f ' x( ) = 3x2 −12x −15 .

At stationary points, f ' x( ) = 0 . That is,

3x2 −12x −15 = 0

3 x2 − 4x − 5( ) = 03 x − 5( ) x +1( ) = 0

Solving gives x1 = 5 or x2 = −1 .

x1 = 5⇒ y1 = −99⇒ P1 5 / −99( )

x2 = −1⇒ y2 = 9⇒ P2 −1 / 9( )

The second derivative is given by f '' x( ) = 6x −12 . At point P1 5 / −99( ) , f '' 5( ) = 18 > 0 .

Therefore, the stationary point P1 5 / −99( ) is a minimum.

At the point P2 −1 / 9( ) , f '' −1( ) = −18 < 0 . Therefore, the stationary point P2 −1 / 9( ) is a

maximum.

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Exercise

1. Find the coordinates of the points on each of the following curves at which the gradient is zero.

a) y = 2x2 − 3x +1

b) y = x4 − 32x + 3

c) y = 5 + 3x2 − x3

2. Find the coordinates of the stationary points on the following curves. In each case, determine

their nature and sketch the curve.

a) y = x3 − 3x + 3

b) y = 8x3 − x4

c) y = x4 − 8x2 − 9

3. Determine the intervals of increase and decrease for each of the following functions.

a) f x( ) = x2 − 8x + 3

b) f x( ) = x + 4x

c) f x( ) = x3 − 3x2 − 9x + 4

4. By investigating the stationary points of f x( ) = x3 + 3x2 + 6x − 30 and sketching the curve

y = f x( ) , show that the equation f x( ) = 0 has only one real solution.