optimisation using derivatives i

Optimisation Using Optimisation Using Derivatives IDerivatives I

Positive and negative values of the derivative as an indication of the points at which the function is increasing or decreasing

zero values of the derivative as an indication of stationary points

Greatest and least values of functions

relationship between the graph of a function and the graph of its derivative

methods of determining the nature of stationary points

greatest and least values of functions in a given interval

If the gradient is positive, the curve slopes upwards from left to If the gradient is positive, the curve slopes upwards from left to rightright

If the gradient is negative, the curve slopes downwards from left If the gradient is negative, the curve slopes downwards from left to rightto right

Page Page 501 Ex 21.1501 Ex 21.11-21-2 (orally)(orally)

State the values of x for which the curve is State the values of x for which the curve is (a) increasing(a) increasing(b) decreasing(b) decreasing(c) stationary(c) stationary

State the values of x for which the curve is State the values of x for which the curve is (a) increasing x<3, x>2(a) increasing x<3, x>2(b) decreasing(b) decreasing(c) stationary(c) stationary

State the values of x for which the curve is State the values of x for which the curve is (a) increasing x<3, x>2(a) increasing x<3, x>2(b) decreasing -3<x<2(b) decreasing -3<x<2(c) stationary(c) stationary

State the values of x for which the curve is State the values of x for which the curve is (a) increasing x<3, x>2(b) decreasing -3<x<2(c) stationary x=-3, x=2

Positive and negative values of the Positive and negative values of the derivative as an indication of the points derivative as an indication of the points

at which the function is increasing or at which the function is increasing or decreasingdecreasing

Consider the following functions and Consider the following functions and their derivatives:their derivatives:

y = xy = x22

y = xy = x33

y = xy = x33 - 2x - 2x22 – 2x + 1 – 2x + 1

Page Page 501 Ex 21.1501 Ex 21.133

Greatest and Least ValuesGreatest and Least Values

Find the greatest and least values of the Find the greatest and least values of the function y = xfunction y = x33 + 2x + 2x22 - 4x + 5 - 4x + 5

for -3 for -3 x x 2 2

Check bounds

When x = -3, y= 8 When x = 2, y = 13

Check stationary values

y = xy = x3 3 + 2x+ 2x22 - 4x + 5 - 4x + 5

y’ = 3xy’ = 3x2 2 + 4x – 4 = (3x-2)(x+2)+ 4x – 4 = (3x-2)(x+2)

S.P. when y’=0 S.P. when y’=0 x= x= ⅔⅔ or x = -2 or x = -2

When x = When x = ⅔⅔, y = , y = 3.5193.519 (3dp) When x= -2, y = (3dp) When x= -2, y = 1313

Greatest value is 13 ; Least value is 3.519 (3dp)Greatest value is 13 ; Least value is 3.519 (3dp)

y = xy = x33 + 2x + 2x22 - 4x + 5 - 4x + 5

Page Page 505 Ex 21.2505 Ex 21.211

Model : Find the stationary points on the curveModel : Find the stationary points on the curve y = x y = x33+2x+2x2 2 – 4x +2– 4x +2

2

0)2)(23(

0443

0

)2)(23(

443

242

32

2

2

23

xorx

xx

xx

whenoccurSP

xx

xx

xxxy

dxdy

dxdy

When x = ⅔, y = 0.519 (3dp)

When x = -2, y = 10

X = -2X = -2 X = X = ⅔⅔

dydy = = dxdx

When x = ⅔, y = 0.519 (3dp)

When x = -2, y = 10

X < -2X < -2 X = -2X = -2-2 < x < -2 < x <

⅔⅔ X = X = ⅔⅔ X > X > ⅔⅔

dydy = = dxdx

When x = ⅔, y = 0.519 (3dp)

When x = -2, y = 10

X < -2X < -2 X = -2X = -2-2 < x < -2 < x <

⅔⅔ X = X = ⅔⅔ X > X > ⅔⅔

00 00dydy = = dxdx

When x = ⅔, y = 0.519 (3dp)

When x = -2, y = 10

X < -2X < -2 X = -2X = -2-2 < x < -2 < x <

⅔⅔ X = X = ⅔⅔ X > X > ⅔⅔

++ 00 -- 00 ++dydy = = dxdx

When x = ⅔, y = 0.519 (3dp)

When x = -2, y = 10

X < -2X < -2 X = -2X = -2-2 < x < -2 < x <

⅔⅔ X = X = ⅔⅔ X > X > ⅔⅔

++ 00 -- 00 ++dydy = = dxdx

Max at (-2, 10) Min at (⅔ ,0.519)

Now check your answer on your GC

Read bottom of 509 - 510Read bottom of 509 - 510

Page Page 510 Ex 21.3510 Ex 21.31 a,d,e,f1 a,d,e,f4-6, 9, 34-6, 9, 3

The The SecondSecond Derivative Derivative

The derivative of is called the second The derivative of is called the second

derivative derivative

At a S.P.,At a S.P.,

if is positive then that point is a minimum if is positive then that point is a minimum

if is negative then that point is a maximum if is negative then that point is a maximum

dx

dy

2

2

dx

yd

2

2

dx

yd

2

2

dx

yd

Model Model Find the nature of the stationary Find the nature of the stationary pointspoints

on the curve y = on the curve y = xx33+2x+2x2 2 – 4x +2– 4x +2

2

0)2)(23(

0443

0

)2)(23(

443

242

32

2

2

23

xorx

xx

xx

whenoccurSP

xx

xx

xxxy

dxdy

dxdy

)10,2(max

,2

)519.0,(min

,

46

2

2

2

2

2

2

32

32

atimum

negxWhen

atimum

posxWhen

x

dx

yd

dx

yd

dx

yd

Model Model Draw the graph of y = 2 sin 3x for 0Draw the graph of y = 2 sin 3x for 0 < x < x < 2< 2ΠΠ

y = 2 sin 3xy = 2 sin 3xdy/dx = 6 cos 3xdy/dx = 6 cos 3xdd22y/dxy/dx22 = -18 sin 3x = -18 sin 3x

S.P. when dy/dx = 0 S.P. when dy/dx = 0 6 cos 3x = 06 cos 3x = 0 cos 3x = 0cos 3x = 0 3x = (2n+1)3x = (2n+1)ΠΠ/2/2 x = x = (2n+1)(2n+1)Π/Π/66 x = … x = … ΠΠ/6, 3/6, 3ΠΠ/6, 5/6, 5ΠΠ/6, 7/6, 7ΠΠ/6, …/6, …

xx dy/dy/dxdx

dd22y/dxy/dx2 2

(-18 sin 3x)(-18 sin 3x)

ΠΠ/6/6 00 --

ΠΠ/2/2 00 ++

55ΠΠ/6/6 00 --

77ΠΠ/6/6 00 ++

33ΠΠ/2/2 00 --

Max at (Max at (ΠΠ/6,2)/6,2)

Min at (Min at (ΠΠ/2,-2)/2,-2)

Max at (5Max at (5ΠΠ/6,2)/6,2)

Min at (7Min at (7ΠΠ/6,-2)/6,-2)

Max at (3Max at (3ΠΠ/2,2)/2,2)

1111ΠΠ/6/6 00 ++ Min at (11Min at (11ΠΠ/6,-2)/6,-2)

Check on GC

Page Page 510 Ex 21.3510 Ex 21.344

Repeat

using the second derivative to determine the nature of stationary points

+ Page + Page 518 Ex 21.4518 Ex 21.422

PerformanceFaith Honour

St Luke’s Anglican School