differentiation

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Higher Unit 1Higher Unit 1

www.mathsrevision.comwww.mathsrevision.com

Finding the gradient for a polynomialDifferentiating a polynomial

Differentiating Negative Indices Differentiating Roots

Differentiating Brackets

Differentiating Fraction TermsDifferentiating with Leibniz Notation

Exam Type Questions

Equation of a Tangent Line

Increasing / Decreasing functionsMax / Min and Inflection Points

Curve Sketching

Max & Min Values on closed IntervalsOptimization

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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-pipe are very steep(S) but it is not very steep near the base(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 any point on a curve we need

to modify the gradient formula

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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 value We can improve the approximation by making the value of h

smallerThis 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 the value of h even smaller.

(x, f(x))

((x+h), f(x+h))True gradientApprox gradient

So the points are even closer together.


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Derivative

We have seen that on curves the gradient changes continually 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 a formula for the gradient.

This process is called DIFFERENTIATING

or FINDING THE DERIVATIVE

Differentiating

Finding the GRADIENT

Finding the rate of 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 4x 4x2 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

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Example 1 A curve has equation f(x) = 3x4

Its gradient is f '(x) = 12x3

At the point where x = 2 the gradient isf '(2) = 12 X 23 = 12 X 8 = 96

Example 2 A curve has equation f(x) = 3x2

Find the formula for its gradient and find the gradient when x = -4 .Its gradient is f '(x) = 6x

At the point where x = -4 the gradient isf '(-4) = 6 X -4 =-24


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g '(x) = 20x3 - 20x4 g '(2) = 20 X 23 - 20 X 24

= 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 X ax0

= a X 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 !!

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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 = 0

ie 3x(x - 2) = 0

ie 3x = 0 or x - 2 = 0

so x = 0 or x = 2

Now using original formula

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 = 1 xm

Consider ….. am X a-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 be in the form axn .

Bottom line terms get negative powers !!

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f(x) = 1 = x2

x -2

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

-2 x3

Example 10

g(x) = -3 x4

= -3x-4

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

12

x5

Negative IndicesHigher Outcome 3

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Example 11

h(y) = 4 3y3

= 4/3 y-3

So h '(y) =

-12/3 y-4 = -4 y4

Example 12

k(t) = 3 4t2/3

= 3/4 t-2/3

So k '(t) =- -6/12 t-5/3 = -1 2t5/3


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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) = 8 x4

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

-32 x5

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

= -32 -32

= 1


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Differentiating Roots

Fractional Indices

x X x = x

x1/2 X x1/2 = x1/2+1/2 = x1 = x

So it follows that …..

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

x = x1/2

In general nx = x1/n

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Also nxm or (nx)m = xm/n

Check the following using the power button on your calc.

641/2 = 64 = 8

641/3 = 364 = 4

25-1/2 = 1 = 1 or 0.2

25 5

163/4 = (416)3 = 23 = 8

1255/3 = (3125)5 = 55 = 3125

Fractional Powers

top line - power

bottom line - root

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so h'(y) = -7/3 y-10/3 = -7 33y10

Example 14

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

Example 15

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

2Example 16

h(y) = 1 3y7

= y-7/3

g(t) = t5

Differentiating RootsHigher Outcome 3

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Example 17Find the rate of change at the point where x = 4 on the curve with equation g(x) = 4 .

x g(x) = 4

x

= 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 .

Differentiating RootsHigher 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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Recall 17 7 7

Fractions

Reversing the above we get the following “rule” !

This can be used as follows …..

a + bc

a bc c

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Example 20

f(x) = 3x3 - x + 2 x2

= 3x - x-1 + 2x-2

f '(x) = 3 + x-2 - 4x-3

Fractions

= 3x3 - x + 2 x2 x2 x2

= 3 + 1 - 4 x2 x3

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Example 21 (tricky)

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

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

y X y1/2

= 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 to powers √=½

Brackets

Single fractions

Correct form

Indices

FractionsHigher Outcome 3

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= 1/4 - 1/16 - 1/16 = 1/8

= 1 - 1 - 1 2y1/2 2y3/2 y2

Also g'(4) = 1 - 1 - 1 2 X 41/2 2 X 43/2 42

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

FractionsHigher Outcome 3

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

Leibniz Notation

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

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

Example 22 Find dQ/dR

NB: Q = 9R2 - 15R-3

So dQ/dR = 18R + 45R-4 = 18R + 45 R4

Q = 9R2 - 15 R3

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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 NotationHigher Outcome 3

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Newton’s 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

so ds/dt = u + at

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

and this is Newton’s 1st Law of Motion

Real Life ExamplePhysics

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y = mx +c

y = f(x)

Equation of Tangents

tangent

NB: at A(a, b) gradient of line = gradient of curve

gradient of line = m (from y = mx + c )

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

it follows that m = f (a)

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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 point plus the gradient then we can use the formula y - b =

m(x - a) .

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

or y - 2 = x + 1

or y = x + 3

point is (-1,2)

m = 1


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Example 25

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

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 = -8 x3

when x = -2 then dy/dx = -8 (-2)3

= -8/-8 = 1m =

1

Equation of TangentsHigher Outcome 3

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

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

or y - 1 = x + 2

or 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)


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Example 26 - (other way round)

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

gradient of tangent = gradient of curve

dy/dx = 2x - 6

so 2x - 6 = 14

2x = 20 x = 10

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

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


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Increasing & Decreasing Functions and Stationary Points

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

X

y = f(x)

a b c d e f+

+

+

++

-

-

0

0

0

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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 negativeand 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.


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Example 27

For the function f(x) = 4x2 - 24x + 19 determine the intervals 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 > 24x > 3

Check: f (2) = 8 X 2 – 24 = -8

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


Higher Outcome 3

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Example 28

For the curve y = 6x – 5/x2 Determine if it is increasing or decreasing when x = 10.

= 6x - 5x-2

so dy/dx = 6 + 10x-3

when x = 10 dy/dx = 6 + 10/1000

= 6.01

Since dy/dx > 0 then the function is increasing.


y = 6x - 5 x2

= 6 + 10 x3

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Example 29Show that the function g(x) = 1/3x3 -3x2 + 9x -10

is never decreasing.

g (x) = x2 - 6x + 9

= (x - 3)(x - 3)= (x - 3)2

Since (x - 3)2 0 for all values of x

then g (x) can never be negative

so the function is never decreasing.

Squaring a negative or a positive value produces a positive value, while 02 = 0. So you will never

obtain a negative by squaring any real number.


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Example 30

Determine the intervals when the function

f(x) = 2x3 + 3x2 - 36x + 41

is (a) Stationary (b) Increasing (c) Decreasing.

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

= 6(x2 + x - 6)

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

Function is stationary when f (x) = 0

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


Higher Outcome 3

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determine when f (x) is positive & negative.

x -3 2

f’(x) + 0 - 0 +

Function increasing when f (x) > 0

ie x < -3 or x > 2

Function decreasing when f (x) < 0

ie -3 < x < 2


Higher Outcome 3

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Stationary Points and Their Nature

Consider this graph of y = f(x) again

X

y = f(x)

a b c+

+

+

+

+

-

-

0

0

0

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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 the gradient ( f(x) ) at either side of the stationary

value.


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Maximum Turning point

x af(x) + 0 -

Minimum Turning Point

x bf(x) - 0

+


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of Inflection

x c

f(x) + 0 +

Falling Point of Inflection

x d

f(x) - 0 -


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Example 31Find the co-ordinates of the stationary point on the curve y = 4x3 + 1 and determine its nature.

SP occurs when dy/dx = 0so 12x2 = 0

x2 = 0

x = 0

Using y = 4x3 + 1

if x = 0 then y = 1

SP is at (0,1)


Higher Outcome 3

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x 0

dy/dx + 0 +

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


dy/dx = 12x2

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Example 32

Find the co-ordinates of the stationary points on the curve 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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Nature Table

x 0 4

dy/dx - 0 - 0 +

So (0,24) is a Point of Infection

and (4,-232) is a minimum Turning Point


dy/dx=12x3 - 48x2

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Example 33Find the co-ordinates of the stationary points on the curve y = 1/2x4 - 4x2 + 2 and determine their nature.

SP occurs when dy/dx = 0

So 2x3 - 8x = 0

2x(x2 - 4) = 0

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

x = 0 or x = -2 or x = 2

Using y = 1/2x4 - 4x2 + 2if x = 0 then y = 2

if x = -2 then y = -6

SP’s at(-2,-6), (0,2) & (2,-6)

if x = 2 then y = -6


Higher Outcome 3

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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 includes important 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 their

nature 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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Dominant Terms

Suppose that f(x) = -2x3 + 6x2 + 56x - 99

As x +/- (ie for large positive/negative values) The formula is approximately the same as f(x) = -2x3

As x + then y -

As x - then y +

Graph roughly

Curve SketchingHigher 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)


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(b) Stationary Points occur where dy/dx = 0

so -6x + 12 = 0

6x = 12

x = 2

If x = 2

then y = -12 + 24 + 15 = 27

Nature Table

x 2

dy/dx + 0 -So (2,27)

is a Maximum Turning Point

Stationary Point is (2,27)


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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 15-1 5

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

15

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

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

y = -2x2 (x - 4) = -2x3 + 8x2

SPs occur where dy/dx = 0

so -6x2 + 16x = 0


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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) =512/27

naturex 0 8/3

dy/dx - 0 + 0 -


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

as x - then y +

Sketch

Xy = -2x2 (x – 4)

Curve Sketching

Cuts x – axis at 0 and 40 4

Max TP’s at (8/3, 512/27) (8/3, 512/27)

Y

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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) = 0So x = -2 or x = 2 but x2 -2


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

So 4x - 4x3 = 0 4x(1 - x2) = 0

4x(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)


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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 .


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Cuts y – axis at 8Cuts x – axis at -2 and 2

(c) Large valuesUsing y = -

x4as x + then y -

as x - then y -

Sketch is

X

Y

-2 28

(-1,9) (1,9)

y = 8 + 2x2 - x4

Max TP’s at (-1,9) (1,9)

Curve Sketching

Summarising

Higher Outcome 3

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Max & Min on Closed Intervals

In the previous section on curve sketching we dealt with the entire graph.

In this section we shall concentrate on the important details to be found in a small section of

graph.Suppose we consider any graph between the points

where x = a and x = b (i.e. a x b)

then the following graphs illustrate where we would expect 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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y =f(x)

xa b

c

(a, f(a))

(b, f(b))

(c, f(c))

max = f(b) end point

min = f(c) min TP

NB: a < c < b


Higher Outcome 3

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From the previous three diagrams we should be able to see that the maximum and minimum values of f(x) on the closed interval a x b can be found either at the end points or at a stationary point between the two 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 pointsdy/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 profits can 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 we would look for a min value or min TP.

Higher Outcome 3

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Higher Outcome 3Example 41

OptimizationQ. What is the maximum

volume

We can have for the given dimensions

A rectangular sheet of foil measuring 16cm X 10 cm has four small squares cut from each corner.

16cm

10cmx cm

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

This gives us a particular interval to consider !

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(16 - 2x) cm(10 - 2x) cm

x cm

The volume is now determined by the value of x so we can write

V(x) = x(16 - 2x)(10 - 2x)

= x(160 - 52x + 4x2)

= 4x3 - 52x2 +160x

We now try to maximize V(x) between 0 and 5

Optimization

By folding up the four flaps we get a small cuboid

Higher Outcome 3

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5End Points

V(0) = 0 X 16 X 10 = 0

V(5) = 5 X 6 X 0 = 0

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

= 4(3x2 - 26x + 40)

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

OptimizationHigher Outcome 3

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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 X 12 X 6 = 144

We now check gradient near x = 2


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x 2

V '(x) + 0 -

Hence max TP when x = 2

So max possible volume = 144cm3

Nature


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Example 42

When a company launches a new product its share of the market after x months is calculated by the formula

So after 5 months the share is

S(5) = 2/5 – 4/25 = 6/25

Find the maximum share of the market

that the company can achieve.

(x 2)2

2 4( )S x

x x


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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 22

2 4( ) 2 4 S x x x

x x

1 2 2 32 3

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

x x


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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


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x 4

S (x) + 0 -

S (3.9 ) = 0.00337…

S (4.1) = -0.0029…

Hence max TP at x = 4

And max share of market = S(4) = 2/4 – 4/16

= 1/2 – 1/4

= 1/4

Optimization

Nature

Higher Outcome 3

Differentiation

Higher Mathematics

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Technology

usingg x

andsoh x

formulag x

h approx gradient x

axnthen f x

3x4 f x

close point x

gradient of curve