differentiation-140202103726-phpapp01.ppt

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Higher Unit 1

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Finding the gradient for a polynomial

Differentiating a polynomial

Differentiating Negative Indices

Differentiating Roots

Differentiating Brackets

Differentiating Fraction Terms

Differentiating with LeibnizNotation

Exam Type Questions

Equation of a Tangent Line

Increasing / Decreasing functions

Max / Min and Inflection Points

Curve Sketching

Max & Min Values on closed Intervals

Optimization

Higher Outcome 3


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On a straight line the gradient remains constant,however with curves the gradient changes continually,and the gradient at any point is in fact the same as

the gradient of the tangent at that point.

The sides of the half-pipeare very steep(S) but it is

not very steep near thebase(B).

B

S

Gradients & CurvesHigher Outcome 3


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A

Gradient of tangent = gradient of curve at A

B

Gradient of tangent = gradient of curve at B



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Gradients & Curves

To find the gradient at anypoint on a curve we need to

modify the gradient formulaHigher Outcome 3

For the function y = f(x) we do this by taking the point (x, f(x))

and another very close point ((x+h), f(x+h)).

Then we find the gradient between the two.

(x, f(x))

((x+h), f(x+h))

True gradient

Approx gradient

2 1

2 1

-

-

y ym

x x


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The gradient is not exactly the same but is

quite close to the actual valueWe can improve the approximation by making the value of hsmaller

This means the two points are closer together.

(x, f(x))

((x+h), f(x+h))

True gradient

Approx gradient



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We can improve upon this approximation by making thevalue of heven smaller.

(x, f(x))

((x+h), f(x+h))

True gradientApprox gradient

So the points are even closer together.



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Higher Outcome 3

Derivative

We have seen that on curves the gradient changescontinually and is dependant on the position on the

curve. ie the x-value of the given point.

We can use the formula for the curve to produce aformula for the gradient.

This process is called DIFFERENTIATING

or FINDING THE DERIVATIVE

DifferentiatingFinding theGRADIENT

Finding the rateof change


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If the formula/equation of the curve is given by f(x)

Then the derivative is called f '(x) - f dash x

There is a simple way

of finding f '(x) from f(x).

f(x) f '(x)

2x2 4x4x2 8x

5x10

50x9

6x7 42x6

x3 3x2

x5 5x4

x99

99x98

DerivativeHigher Outcome 3


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Rules for Differentiating

These can be given by the simple flow diagram ...

multiply by

the power

reduce the

power by 1

Or

If f(x) = axn

then f '(x) = naxn-1

NB: the following terms & expressions mean the same

GRADIENT, DERIVATIVE, RATE OF CHANGE, f '(x)

Derivative

Higher Outcome 3


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www.math

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com Example 5 If g(x) = 5x4- 4x5 then find g '(2) .

g '(x) = 20x3- 20x4

g '(2) = 20 X23- 20 X24

= 160 - 320

= -160



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Your turn

Page 91 ex 6D ( odd numbers )

Page 92-93 ex. 6E ( even numbers )


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Special Points

(I) f(x) = ax, where a is any real no.

If f(x) = ax = ax1

then f '(x) = 1 Xax0 = aX 1 = a

Index Laws

x0= 1

So if g(x) = 12x then g '(x) = 12

Also using y = mx + c

The line y = 12x has gradient 12,

and derivative = gradient !!

Higher Outcome 3


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com (II) f(x) = a, where a is any real no.

If f(x) = a = a X 1 = ax0

then f '(x) = 0 X ax--1 = 0

Index Laws

x0= 1

So if g(x) = -2 then g '(x) = 0

Also using formula y = c , (see outcome1 !)

The line y = -2 is horizontal so has gradient 0 !

Special PointsHigher Outcome 3


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Example 6

h(x) = 5x2- 3x + 19 so h '(x) = 10x - 3

and h '(-4) = 10 X (-4) - 3 = -40 - 3 = -43

Example 7

k(x) = 5x4- 2x3+ 19x - 8, find k '(10) .

k '(x) = 20x3 - 6x2 + 19

So k '(10) = 20 X 1000 - 6 X 100 + 19 = 19419



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Find the points on the curve

f(x) = x3- 3x2+ 2x + 7 where the gradient is 2.

NB: gradient = derivative = f '(x)

We need f '(x) = 2

ie 3x2 - 6x + 2 = 2

or 3x2

- 6x = 0ie 3x(x - 2) = 0

ie 3x = 0 or x - 2 = 0

so x = 0 or x = 2

Now using originalformula

f(0) = 7

f(2) = 8 -12 + 4 + 7

= 7

Points are (0,7) & (2,7)



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Negative Indices

Index Law x-m

= 1xm

Consider .. am Xa-m = am+(-m) = a0 = 1

also am X 1 =

am

1

This gives us the following

NB: Before we can differentiate a term it must bein the form axn .

Bottom line terms get negative powers !!

Higher Outcome 3


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Example 9f(x) = 1 =

x2x -2

So f '(x) = -2x-3 =-2x3

Example 10

g(x) = -3x4

= -3x-4

So g '(x) = 12x-5 = 12

x5

Negative IndicesHigher Outcome 3


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Example 13

The equation of a curve is f(x) = 8 (x 0)

x4

Find the gradient at the point where x = -2 .

f(x) = 8x4 = 8x-4 so f '(x) = -32x-5 =-32x5

Required gradient = f '(-2) = -32(-2)5

= -32-32

= 1

Negative IndicesHigher Outcome 3


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Fractional Indices

x Xx = x

x1/2 X x1/2 = x1/2+1/2 = x1 = xSo it follows that ..

Similarly 3x = x1/3 and 4x = x1/3

x = x1/2

In general nx = x1/n

Higher Outcome 3


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so h'(y) = -7/3y-10/3 = -7

33y10

Example 14

f(x) = x = x1/2 so f '(x) = 1/2x-1/2 = 1

2x

Example 15

= t5/2so g'(t) = 5/2t3/2 = 5t3

2

Example 16

h(y) = 13y7

= y-7/3

g(t) = t5


Higher Outcome 3


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Example 17

Find the rate of change at the point where x = 4 on thecurve with equation g(x) = 4 .

xg(x) = 4x

= 4x-1/2

NB: rate of change = gradient = g'(x) .

g'(x) = -2x-3/2 = -2(x)3

so g'(4) = -2(4)3 =

-2

/8 =-1

/4 .


Higher Outcome 3


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Brackets

Basic Rule: Break brackets before you differentiate !

Example 18 h(x) = 2x(x + 3)(x -3)

= 2x(x2- 9)

= 2x3 - 18x

So h'(x) = 6x2-18

Also h'(-2) = 6 X(-2)2-18 = 24 - 18 = 6

Higher Outcome 3


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www.math

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Recall 1

7 7 7

Fractions

Reversing the above we get the following rule !

This can be used as follows ..

a + b

c

a b

c c

Higher Outcome 3


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Example 21 (tricky)

g(y) = (y + y)(y + 1)yy

= (y + y1/2)(y + 1)

y Xy1/2

= y2 + y3/2+ y + y1/2

y3/2

= y2 + y3/2 + y + y1/2

= y1/2 + 1 + y-1/2 + y-1

y3/2 y3/2 y3/2 y3/2

Change topowers =

Brackets

Singlefractions

Correct form

Indices

Fractions

Higher Outcome 3


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www.math

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1/2y-1/2 - 1/2y-3/2 - y-2

= 1/4 - 1/

16 - 1/

16 = 1/

8

= 1 - 1 - 1

2y1/2 2y3/2 y2

Also g'(4) = 1 - 1 - 12 X 41/2 2 X43/2 42

= 1 - 1 - 12 X4 2 X(4)3 16

Fractions

Higher Outcome 3


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If y is expressed in terms of x then the derivative iswritten as dy/dx .

Leibniz Notation

Leibniz Notation is an alternative way of expressingderivatives to f'(x) , g'(x) , etc.

eg y = 3x2- 7x so dy/dx = 6x - 7 .

Example 22Find

dQ

/dR

NB: Q = 9R2- 15R-3

So dQ/dR

= 18R + 45R-4 = 18R + 45R4

Q = 9R2

- 15R3

Higher Outcome 3


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Example 23

A curve has equation y = 5x3- 4x2+ 7 .

Find the gradient where x = -2 ( differentiate ! )

gradient = dy/dx = 15x2- 8x

if x = -2 then

gradient = 15 X(-2)2- 8 X(-2)

= 60 - (-16) = 76

Leibniz Notation

Higher Outcome 3


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Newtons 2ndLaw of Motion

s = ut + 1/2at2 where s = distance & t = time.

Finding ds/dt means diff in dist diff in time

ie speed or velocity

sods

/dt = u + atbut ds/dt = v so we get v = u + at

and this is Newtons 1st Law of Motion

Real Life ExamplePhysics

Higher Outcome 3


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www.math

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y = mx +c

y = f(x)

Equation of Tangents

tangent

NB: at A(a, b) gradient of line = gradient of curvegradient of line = m (from y = mx + c )

gradient of curve at (a, b) = f(a)

it follows that m = f(a)

Higher Outcome 3


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Higher Outcome 3


Example 24

Find the equation of the tangent to the curve

y = x3

- 2x + 1 at the point where x = -1.Point: if x = -1 then y = (-1)3- (2 X-1) + 1

= -1 - (-2) + 1

= 2 point is (-1,2)

Gradient: dy/dx= 3x2- 2

when x = -1 dy/dx= 3 X(-1)2- 2

= 3 - 2 = 1 m = 1

Tangent is line so we need a pointplus the gradient then we can use

the formula y - b = m(x - a) .


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Now using y - b = m(x - a)

we get y - 2 = 1( x + 1)

or y - 2 = x + 1

or y = x + 3

point is (-1,2)

m = 1


Higher Outcome 3


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Example 25

Find the equation of the tangent to the curve y = 4x2

at the point where x = -2. (x 0)Also find where the tangent cuts the X-axis and Y-axis.

Point:when x = -2 then y = 4(-2)2

= 4/4 = 1

point is (-2, 1)

Gradient: y = 4x-2 so dy/dx = -8x-3 = -8x3

when x = -2 then dy/dx = -8

(-2)3=-8

/-8= 1 m = 1


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Now using y - b = m(x - a)

we get y - 1 = 1( x + 2)

or y - 1 = x + 2or y = x + 3

Axes Tangent cuts Y-axis when x = 0

so y = 0 + 3 = 3 at point (0, 3)

Tangent cuts X-axis when y = 0

so 0 = x + 3 or x = -3 at point (-3, 0)


Higher Outcome 3


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Example 26 - (other way round)

Find the point on the curve y = x2- 6x + 5 wherethe gradient of the tangent is 14.

gradient of tangent = gradient of curve

dy/dx= 2x - 6

so 2x - 6 = 142x = 20 x = 10

Put x = 10 into y = x2- 6x + 5

Giving y = 100 - 60 + 5= 45 Point is (10,45)


Higher Outcome 3


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Increasing & Decreasing Functionsand Stationary Points

Consider the following graph of y = f(x) ..

X

y = f(x)

a b c d e f+

+

+

+

+

--

0

0

0

Higher Outcome 3

I i & D i F i


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In the graph of y = f(x)

The function is increasing if the gradient is positive

i.e. f (x) > 0 when x < b or d < x < f or x > f .The function is decreasing if the gradient is negative

and f (x) < 0 when b < x < d .

The function is stationary if the gradient is zeroand f (x) = 0 when x = b or x = d or x = f .These are called STATIONARY POINTS.

At x = a, x = c and x = e

the curve is simply crossing the X-axis.


Higher Outcome 3


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Example 27

For the function f(x) = 4x2- 24x + 19 determine theintervals when the function is decreasing and increasing.

f (x) = 8x - 24

f(x) decreasing when f (x) < 0 so 8x - 24 < 0

8x < 24

x < 3

f(x) increasing when f (x) > 0 so 8x - 24 > 0

8x > 24

x > 3

Check: f (2) = 8 X2 24 = -8

Check: f (4) = 8 X4 - 24 = 8


Higher Outcome 3


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Example 30

Determine the intervals when the function

f(x) = 2x3

+ 3x2

- 36x + 41is (a) Stationary (b) Increasing (c) Decreasing.

f (x) = 6x2+ 6x - 36

= 6(x2+ x - 6)

= 6(x + 3)(x - 2)

Function is stationarywhen f (x) = 0

ie 6(x + 3)(x - 2) = 0

ie x = -3 or x = 2


Higher Outcome 3


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www.mathsrevision.com We now use a special table of factors to

determine when f (x) is positive & negative.

x -3 2

f(x) + 0 - 0 +

Function increasingwhen f (x) > 0 ie x < -3 or x > 2

Function decreasingwhen f (x) < 0 ie -3 < x < 2


Higher Outcome 3

St ti P i ts


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Stationary Pointsand Their Nature

Consider this graph of y = f(x) again

X

y = f(x)

a b c+

+

+

+

+

-

-

0

0

0

Higher Outcome 3

St ti P i t


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www.mathsrevision.com This curve y = f(x) has three types of stationary point.

When x = a we have a maximum turning point (max TP)

When x = b we have a minimum turning point (min TP)

When x = c we have a point of inflexion (PI)

Each type of stationary point is determined by thegradient ( f(x) ) at either side of the stationary value.


Higher Outcome 3

St ti P i t


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Maximum Turning point

x a

f(x) + 0 -

Minimum Turning Point

x bf(x) - 0 +


Higher Outcome 3

St ti P i t


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www.mathsrevision.com Rising Point

of Inflection

x cf(x) + 0 +

Falling Pointof Inflection

x df(x) - 0 -


Higher Outcome 3

St ti P i t


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Example 31

Find the co-ordinates of the stationary point on thecurve y = 4x3+ 1 and determine its nature.

SP occurs when dy/dx = 0

so 12x2= 0

x2= 0

x = 0

Using y = 4x3+ 1

if x = 0 then y = 1

SP is at (0,1)


Higher Outcome 3

St ti P i ts


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www.mathsrevision

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x 0

dy/dx + 0 +

So (0,1) is a rising point of inflexion.


dy/dx = 12x2

Higher Outcome 3

St ti n P ints


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Example 32

Find the co-ordinates of the stationary points on thecurve y = 3x4- 16x3 + 24 and determine their nature.

SP occurs when dy/dx = 0

So 12x3- 48x2 = 0

12x2(x - 4) = 0

12x2 = 0 or (x - 4) = 0

x = 0 or x = 4

Using y = 3x4- 16x3 + 24

if x = 0 then y = 24

if x = 4 then y = -232

SPs at (0,24) & (4,-232)


Higher Outcome 3


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Stationary Points


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Nature Table

x 0

dy/dx

-2 2

- 0 + 0 - 0 +

So (-2,-6) and (2,-6) are Minimum Turning Points

and (0,2) is a Maximum Turning Points


Higher Outcome 3


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Curve Sketching

Note: A sketch is a rough drawing which includesimportant details. It is not an accurate scale drawing.

Process

(a) Find where the curve cuts the co-ordinate axes.

for Y-axis put x = 0

for X-axis put y = 0 then solve.

(b) Find the stationary points & determine theirnature as done in previous section.

(c) Check what happens as x +/- .

This comes automatically if (a) & (b) are correct.

Higher Outcome 3


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Example 34

Sketch the graph of y = -3x2+ 12x + 15

(a) Axes If x = 0 then y = 15

If y = 0 then -3x2+ 12x + 15 = 0 ( -3)

x2- 4x - 5 = 0

(x + 1)(x - 5) = 0

x = -1 or x = 5

Graph cuts axes at (0,15) , (-1,0) and (5,0)

Curve SketchingHigher

Outcome 3


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www.mathsrevision

.com (c) Large values

using y = -3x2

as x + then y -

as x - then y -

Sketching

X

Y

y = -3x2+ 12x + 15

Curve Sketching

Cuts x-axis at -1 and 5

Summarising

Cuts y-axis at 15-15

Max TP (2,27)(2,27)

15

Higher

Outcome 3


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Example 35Sketch the graph of y = -2x2(x - 4)(a) Axes

If x = 0 then y = 0 X -4 = 0

If y = 0 then -2x2(x - 4) = 0

x = 0 or x = 4

Graph cuts axes at (0,0) and (4,0) .

-2x2= 0 or (x - 4) = 0

(b) SPsy = -2x2(x - 4) = -2x3+ 8x2

SPs occur where dy/dx = 0

so -6x2+ 16x = 0


Outcome 3


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-2x(3x - 8) = 0

-2x = 0 or (3x - 8) = 0

x = 0 or x = 8/3

If x = 0 then y = 0 (see part (a) )If x = 8/3 then y = -2 X(8/3)2 X(8/3 -4) =512/27nature

x 0 8/3

dy/dx - 0 + 0 -


Outcome 3


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Summarising(c) Large valuesusing y = -2x3 as x + then y -

as x - then y +

Sketch

X

y = -2x2(x 4)

Curve Sketching

Cuts x axis at 0 and 40 4

Max TPs at (8/3, 512/27)(8/3, 512/27)

Y

Higher

Outcome 3


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Example 36 Sketch the graph of y = 8 + 2x2- x4

(a) Axes If x = 0 then y = 8 (0,8)

If y = 0 then 8 + 2x2- x4= 0

Graph cuts axes at (0,8) , (-2,0) and (2,0)

Let u = x2 so u2= x4

Equation is now 8 + 2u - u2= 0

(4 - u)(2 + u) = 0

(4 - x2)(2 + x2) = 0

or (2 + x) (2 - x)(2 + x2) = 0

So x = -2 or x = 2 but x2 -2


Outcome 3


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(b) SPs SPs occur where dy/dx = 0

So 4x - 4x3 = 0

4x(1 - x2

) = 04x(1 - x)(1 + x) = 0

x = 0 or x =1 or x = -1

Using y = 8 + 2x2- x4

when x = 0 then y = 8

when x = -1 then y = 8 + 2 - 1 = 9 (-1,9)

when x = 1 then y = 8 + 2 - 1 = 9 (1,9)


Outcome 3


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www.mathsrevision

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x -1 0 1

dy/dx + 0 - 0 + 0 -

So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .


Outcome 3


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Max & Min on Closed Intervals

In the previous section on curve sketching we dealtwith the entire graph.

In this section we shall concentrate on the importantdetails to be found in a small section of graph.

Suppose we consider any graph between the pointswhere x = a and x = b (i.e. a x b)

then the following graphs illustrate where we wouldexpect to find the maximum & minimum values.

Higher

Outcome 3


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y =f(x)

Xa b

(a, f(a))

(b, f(b)) max = f(b) end point

min = f(a) end point


Higher

Outcome 3


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x

y =f(x)(b, f(b))

(a, f(a))

max = f(c ) max TP

min = f(a) end point

a b

(c, f(c))

c NB: a < c < b


Higher

Outcome 3


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From the previous three diagrams we should be able tosee that the maximum and minimum values of f(x) onthe closed interval a x b can be found either at

the end points or at a stationary point between thetwo end points

Example 37

Find the max & min values of y = 2x3- 9x2 in the

interval where -1 x 2.

End points If x = -1 then y = -2 - 9 = -11

If x = 2 then y = 16 - 36 = -20


Higher

Outcome 3


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Stationary points

dy/dx = 6x2- 18x = 6x(x - 3)

SPs occur where dy/dx = 0

6x(x - 3) = 0

6x = 0 or x - 3 = 0

x = 0 or x = 3

in interval not in interval

If x = 0 then y = 0 - 0 = 0

Hence for -1 x 2 , max = 0 & min = -20


Higher

Outcome 3


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Using function notation we can say that

Domain = {xR: -1 x 2 }

Range = {yR: -20 y 0 }


Higher

Outcome 3


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Optimization

Note: Optimum basically means the best possible.

In commerce or industry production costs and profitscan often be given by a mathematical formula.

Optimum profit is as high as possible so we would look

for a max value or max TP.

Optimum production cost is as low as possible so wewould look for a min value or min TP.

Higher

Outcome 3

Q. What is the maximum volume


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Higher

Outcome 3Example 41

OptimizationQ. What is the maximum volume

We can have for the givendimensions

A rectangular sheet of foil measuring 16cm X10 cm hasfour small squares cut from each corner.

16cm

10cmx cm

NB: x > 0 but 2x < 10 or x < 5

ie 0 < x < 5

This gives us a particular interval to consider !


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www.mat

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.com Considering the interval 0 < x < 5End Points

V(0) = 0 X16 X10 = 0

V(5) = 5X 6 X0 = 0

SPs V '(x) = 12x2- 104x + 160

= 4(3x2- 26x + 40)

= 4(3x - 20)(x - 2)

OptimizationHigher

Outcome 3


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www.mat

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.com SPs occur when V '(x) = 0

ie 4(3x - 20)(x - 2) = 0

3x - 20 = 0 or x - 2 = 0

ie x = 20/3 or x = 2

not in intervalin interval

When x = 2 then V(2) = 2 X12 X6 = 144

We now check gradient near x = 2

OptimizationHigher

Outcome 3


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End points S(2) = 1 1 = 0

There is no upper limit but as x S(x) 0.

SPs occur where S (x) = 0

3 2

8 2'( ) 0S x

x x

1 2

2

2 4( ) 2 4 S x x x

x x

1 2 2 3

2 3

2 8'( ) 2 4 2 8S x x x x x

x x

OptimizationHigher

Outcome 3

O


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8x2

= 2x3

8x2- 2x3 = 0

2x2(4 x) = 0

x = 0 or x = 4

Out with interval In interval

We now check the gradients either side of 4

3 2

8 2'( ) 0S x

x xrearrange

OptimizationHigher

Outcome 3


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Calculus


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Differentiate24 3 7x x

8 3x

Calculus


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Differentiate

3 26 3 9x x x

x

3 26 3 9x x x

x x x x

2 16 3 9x x x

Split up

Straight line form

22 6 9x x Differentiate

Calculus


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Differentiate

3 1

2 2

2 5x x

1 3

2 23 1

2 22 5x x

1 3

2 25

23x x

Calculus


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Differentiate2

3 1x

x

1 1

2 23 2 1x x

Straight line form

Differentiate

1 3

2 21

2

32

2x x

1 3

2 23

2x x

Calculus


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Differentiate (3 5)( 2)x x

23 6 5 10x x x Multiply out

Differentiate 6 1x

23 10x x


91/124

Calculus


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Differentiate2( 2 )x x x

multiply out3 22x x

differentiate23 4x x

Calculus


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Differentiate

3

1 1

x x

1 3x x Straight line form

Differentiate2 4( 3)x x

2 43x x

Calculus


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Differentiate

2x x x

1

22x x xStraight line form

multiply out

Differentiate

5 3

2 2x x

3 1

2 25 3

2 2x x

Calculus


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Differentiate 2 4x x

multiply out

Simplify

4 8 2x x x

Differentiate

6 8x x

Straight line form

1

26 8x x 1

23 1x

Calculus


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Differentiate5( 2)x

Chain rule45( 2) 1x

45( 2)xSimplify

Calculus


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Differentiate3(5 1)x

Chain Rule23(5 1) 5x

2

15(5 1)xSimplify

Calculus

R i i


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Differentiate2 4(5 3 2)x x

Chain Rule 2 34(5 3 2) 10 3x x x

Calculus

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Differentiate

5

2(7 1)x

Chain Rule

Simplify

3

25

2(7 1) 7x

3235

2 (7 1)x

Calculus

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Differentiate

1

2(2 5)x

Chain Rule

Simplify

3

21

2(2 5) 2x

3

2(2 5)x

Calculus

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Differentiate 3 1x

Chain Rule

Simplify

Straight line form 1

23 1x

1

21

2 3 1 3x

1

23

2 3 1x

Calculus

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Differentiate

4

2

5

3x

Chain Rule

Simplify

Straight line form2 45( 3)x

2 520( 3) 2x x

2 540 ( 3)x x

Calculus

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Differentiate1

2 1x

Chain Rule

Simplify

Straight line form

1

2(2 1)x

3

21

2(2 1) 2x

3

2

(2 1)x

Calculus

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Differentiate2

2 cos sin3

x x

22sin cos

3

x x

Calculus

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Differentiate 3cosx

3sinx

Calculus

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Differentiate cos 4 2sin 2x x

4sin 4 4 cos 2x x

Calculus

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Differentiate4cos x

34cos sinx x

Calculus

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Differentiate sinx

Chain Rule

Simplify

Straight line form

1

2(sin )x

1

21

2(sin ) cosx x

1

21

2 cos (sin )x x

1

2

cos

sin

x

x

Calculus

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Differentiate sin(3 2 )x

Chain Rule

Simplify

cos(3 2 ) ( 2)x

2cos(3 2 )x

Calculus

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Differentiatecos5

2

x

Chain Rule

Simplify

1

2cos5xStraight line form

1

2sin 5 5x

5

2sin5x

Calculus

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Differentiate ( ) cos(2 ) 3sin(4 )f x x x

( ) 2 sin(2 ) 12 cos(4 )f x x x

Calculus

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Differentiate2 43200( )A x x

x

Chain Rule

Simplify

Straight line form2 1( ) 43200A x x x

2( ) 2 43200A x x x

2

43200( ) 2A x x

x

Calculus

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Differentiate 22

( )f x x

x

Differentiate

Straight line form

1

22( ) 2f x x x

1 321

2( ) 4f x x x


114/124

Calculus

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Differentiate

62siny x

Chain Rule

Simplify

62 cos 1dy

xdx

62cosdy

xdx

Calculus

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Differentiate3

(8 )

4

A a a

multiply out

Differentiate

2364

A a a

36

2A a

Calculus

Revision


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Differentiate

1

3 2( ) (8 )f x x

Chain Rule

Simplify

1

3 221

2( ) (8 ) ( 2 )f x x x

1

2 3 2( ) (8 )f x x x

Calculus

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Differentiate16

, 0y x x

x

Differentiate

Straight line form

1

216y x x

3

21 8dy

xdx

Calculus

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Differentiate23 3 16( )

2

A x x

x

Straight line form

Multiply out23 3 3 3 16( )

2 2A x x

x

Differentiate

2 13 3( ) 24 32

A x x x

2

( ) 3 3 24 3A x x x

Calculus

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Differentiate 1

2( ) 5 4f x x

Chain Rule

Simplify

1

21

2( ) 5 4 5f x x

1

25

2( ) 5 4f x x

Calculus

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Differentiate216( ) 240

3

A x x x

32( ) 240

3A x x

Calculus

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Differentiate 2( ) 3 (2 1)f x x x

Multiply out

Differentiate

3 2( ) 6 3f x x x

2( ) 18 6f x x x


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