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Outcome 3. Higher. Using differentiation (Application). Higher Unit 1. Finding the gradient for a polynomial. Increasing / Decreasing functions. Max / Min and inflexion Points. Differentiating Brackets ( Type 1 ) . Curve Sketching. Differentiating Harder Terms (Type 2).

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Outcome 3

Higher

Using differentiation (Application)

Higher Unit 1

Finding the gradient for a polynomial

Increasing / Decreasing functions

Max / Min and inflexion Points

Differentiating Brackets ( Type 1 )

Curve Sketching

Differentiating Harder Terms (Type 2)

Max & Min Values on closed Intervals

Differentiating with Leibniz Notation

Optimization

Equation of a Tangent Line ( Type 3 )

Mind Map of Chapter

www.mathsrevision.com


Gradients & Curves

Outcome 3

Higher

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).

S

Demo

B


Gradients & Curves

Outcome 3

Higher

Gradient of tangent = gradient of curve at A

A

Demo

B

Gradient of tangent = gradient of curve at B


To find the gradient at any point on a curve we need to modify the gradient formula

Gradients & Curves

Outcome 3

Higher

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+h), f(x+h))

Approx gradient

(x, f(x))

True gradient


Gradients & Curves

Outcome 3

Higher

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 smaller

This means the two points are closer together.

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

Approx gradient

(x, f(x))

True gradient


Gradients & Curves

Outcome 3

Higher

We can improve upon this approximation by making the value of h even smaller.

So the points are even closer together.

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

Approx gradient

True gradient

(x, f(x))


Outcome 3

Higher

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.

Derivative

Finding the GRADIENT

Differentiating

The process of finding the gradient is called

Finding the rate of change

DIFFERENTIATING

or

FINDING THE DERIVATIVE (Gradient)


Derivative

Outcome 3

Higher

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)

2x24x

4x28x

Have guessed the rule yet !

5x1050x9

6x7 42x6

x3 3x2

x5 5x4

x99 99x98


Derivative

Rule for Differentiating

Outcome 3

Higher

It can be given by this simple flow diagram ...

multiply by the power

reduce the power by 1

If f(x) = axn

n

n

-1

ax

then f '(x) =

NB: the following terms & expressions mean the same

GRADIENT,

DERIVATIVE,

RATE OF CHANGE,

f '(x)


Derivative

Rule for Differentiating

Outcome 3

Higher

To be able to differentiate

it is VERY IMPORTANT that you are

comfortable using indices rules


Outcome 3

Higher

(I) f(x) = ax (Straight line function)

Special Points

Index Laws

x0 = 1

If f(x) = ax

= ax1

then f '(x) = 1 X ax0

= a X 1 = a

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

Also using y = mx + c

The line y = 12x has gradient 12,

and derivative = gradient !!


Outcome 3

Higher

Special Points

(II) f(x) = a, (Horizontal Line)

Index Laws

x0 = 1

If f(x) = a

= a X 1 = ax0

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

= 0

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

Also using formula y = c , (see outcome 1 !)

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


Differentiation techniques

Differentiation

=

Gradient

Differentiation

=

Rate of change

Name :


Calculus Revision

Differentiate


Calculus Revision

Differentiate


Calculus Revision

Differentiate


Derivative

Outcome 3

Higher

Example 1

A curve has equation f(x) = 3x4

Find the formula for its gradient and find the gradient when x = 2

Its gradient is f '(x) = 12x3

f '(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 is

f '(-4) = 6 X -4 =

-24


Derivative

Outcome 3

Higher

Example 3

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

g '(x) = 20x3 - 20x4

g '(2) = 20 X 23 - 20 X 24

= 160 - 320

= -160


Derivative

Outcome 3

Higher

Example 4

h(x) = 5x2 - 3x + 19

so h '(x) = 10x - 3

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

= -40 - 3 = -43

Example 5

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


Derivative

Outcome 3

Higher

Example 6 : Find the points on the curve

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

NB: gradient = derivative = f '(x)

Now using original formula

We need f '(x) = 2

ie 3x2 - 6x + 2 = 2

f(0) = 7

or 3x2 - 6x = 0

ie 3x(x - 2) = 0

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

ie 3x = 0 or x - 2 = 0

= 7

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

so x = 0 or x = 2


Calculus Revision

Differentiate


Calculus Revision

Differentiate

Straight line form

Differentiate


Calculus Revision

Differentiate

Straight line form

Differentiate


Calculus Revision

Differentiate

Straight line form

Chain Rule

Simplify


Calculus Revision

Differentiate

Straight line form

Differentiate


Calculus Revision

Differentiate

Straight line form

Differentiate


Calculus Revision

Differentiate

Straight line form

Differentiate


Outcome 3

Higher

Brackets

Basic Rule: Break brackets before you differentiate !

Example

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

= 2x(x2 - 9)

= 2x3 - 18x

So h'(x) = 6x2 -18


Calculus Revision

Differentiate

Multiply out

Differentiate


Calculus Revision

Differentiate

multiply out

differentiate


Calculus Revision

Differentiate

Straight line form

multiply out

Differentiate


Calculus Revision

Differentiate

multiply out

Differentiate


Calculus Revision

Differentiate

multiply out

Simplify

Straight line form

Differentiate


Calculus Revision

Differentiate

Multiply out

Straight line form

Differentiate


Outcome 3

Higher

Fractions

Reversing the above we get the following “rule” !

This can be used as follows …..


Fractions

Outcome 3

Higher

Example

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

= 3x3 - x + 2 x2 x2x2

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

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

= 3 + 1 - 4 x2 x3


Calculus Revision

Differentiate

Split up

Straight line form

Differentiate


Outcome 3

Higher

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

Leibniz Notation

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

eg y = 3x2 - 7x

so dy/dx = 6x - 7 .

Example 19

Q = 9R2 - 15 R3

Find dQ/dR

= 18R + 45 R4

NB: Q = 9R2 - 15R-3

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


Leibniz Notation

Outcome 3

Higher

Example 20

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


Real Life Example

Physics

Outcome 3

Higher

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


Equation of Tangents

y = mx +c

Outcome 3

Higher

y = f(x)

A(a,b)

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)


Straight line so we need a point plus the gradient then we can use the formula y - b = m(x - a) .

Equation of Tangents

Outcome 3

Higher

Example 21

Find the equation of the tangent line 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

= 2point is (-1,2)

Gradient:dy/dx = 3x2 - 2

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

m = 1

= 3 - 2 = 1


Equation of Tangents

Outcome 3

Higher

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

point is (-1,2)

m = 1

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

or y - 2 = x + 1

or y = x + 3


Equation of Tangents

Outcome 3

Higher

Example 22

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 = 1

m = 1


Equation of Tangents

Outcome 3

Higher

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)


Equation of Tangents

Outcome 3

Higher

Example 23 - (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

so2x - 6 = 14

2x = 20

x = 10

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

Point is (10,45)

Giving y = 100 - 60 + 5

= 45


Outcome 3

Higher

Increasing & Decreasing Functions and Stationary Points

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

y = f(x)

+

0

0

+

-

+

+

a

b

c

d

e

f

-

X

+

0


Increasing & Decreasing Functions and Stationary Points

Outcome 3

Higher

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 zero

and 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.


Increasing & Decreasing Functions and Stationary Points

Outcome 3

Higher

Example 24

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

f  (x) = 8x - 24

so 8x - 24 < 0

f(x) decreasing when f  (x) < 0

8x < 24

Check: f  (2) = 8 X 2 – 24 = -8

x < 3

f(x) increasing when f  (x) > 0

so 8x - 24 > 0

8x > 24

Check: f  (4) = 8 X 4 – 24 = 8

x > 3


Increasing & Decreasing Functions and Stationary Points

Outcome 3

Higher

Example 25

For the curve y = 6x – 5/x2

Determine if it is increasing or decreasing when x = 10.

y = 6x - 5 x2

= 6x - 5x-2

so dy/dx = 6 + 10x-3

= 6 + 10 x3

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

= 6.01

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


Increasing & Decreasing Functions and Stationary Points

Outcome 3

Higher

Example 26

Show 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

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.

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

then g (x) can never be negative

so the function is never decreasing.


Increasing & Decreasing Functions and Stationary Points

Outcome 3

Higher

Example 27

Determine the intervals when the function

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

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

f (x) = 6x2 + 6x - 36

Function is stationary when f (x) = 0

= 6(x2 + x - 6)

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

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

ie x = -3 or x = 2


Increasing & Decreasing Functions and Stationary Points

Outcome 3

Higher

We now use a special table of factors to determine when f (x) is positive & negative.

x

-3

2

-

+

+

0

0

f’(x)

Function increasing when f (x) > 0

ie x < -3 or x > 2

Function decreasing when f (x) < 0

ie -3 < x < 2


Outcome 3

Higher

y = f(x)

Stationary Points and Their Nature

Consider this graph of y = f(x) again

0

+

0

-

+

+

-

c

+

a

b

X

+

0


Stationary Points and Their Nature

Outcome 3

Higher

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 the gradient ( f(x) ) at either side of the stationary value.


Stationary Points and Their Nature

Outcome 3

Higher

Maximum Turning point

Minimum Turning Point

x

a

x

b

- 0 +

f(x)

f(x)

+ 0 -


Stationary Points and Their Nature

Outcome 3

Higher

Rising Point of inflexion

Other possible type of inflexion

x

c

x

d

f(x)

+ 0 +

f(x)

- 0 -


Stationary Points and Their Nature

Outcome 3

Higher

Example 28

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

SP occurs when dy/dx = 0

Using y = 4x3 + 1

so 12x2 = 0

if x = 0 then y = 1

x2 = 0

SP is at (0,1)

x = 0


Stationary Points and Their Nature

Outcome 3

Higher

Nature Table

x

0

+

+

dy/dx

0

dy/dx = 12x2

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


Stationary Points and Their Nature

Outcome 3

Higher

Example 29

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

Using y = 3x4 - 16x3 + 24

SP occurs when dy/dx = 0

So 12x3 - 48x2 = 0

if x = 0 then y = 24

12x2(x - 4) = 0

if x = 4 then y = -232

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

x = 0 or x = 4

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


Stationary Points and Their Nature

Outcome 3

Higher

Nature Table

4

x

0

dy/dx

- 0 - 0 +

dy/dx=12x3 - 48x2

So (0,24) is a Point of inflexion

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


Stationary Points and Their Nature

Outcome 3

Higher

Example 30

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

Using y = 1/2x4 - 4x2 + 2

SP occurs when dy/dx = 0

if x = 0 then y = 2

So 2x3 - 8x= 0

if x = -2 then y = -6

2x(x2 - 4) = 0

if x = 2 then y = -6

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

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

SP’s at(-2,-6), (0,2) & (2,-6)


Stationary Points and Their Nature

Outcome 3

Higher

Nature Table

x

-2

0

2

dy/dx

- 0 + 0 - 0 +

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

and (0,2) is a Maximum Turning Points


Outcome 3

Higher

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

Curve Sketching

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.


Curve Sketching

Outcome 3

Higher

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

Graph roughly

As x  + then y  -

As x  - then y  +


Curve Sketching

Outcome 3

Higher

Example 31

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 Sketching

Outcome 3

Higher

(b) Stationary Points

occur where dy/dx = 0

so -6x + 12 = 0

If x = 2

then y = -12 + 24 + 15 = 27

6x = 12

x = 2

Stationary Point is (2,27)

Nature Table

x

2

dy/dx

+ 0 -

So (2,27)

is a Maximum Turning Point


Curve Sketching

Outcome 3

Higher

Summarising

as x  + then y  -

(c) Large values

as x  - then y  -

using y = -3x2

Y

Sketching

5

Cuts x-axis at -1 and 5

-1

15

Cuts y-axis at 15

Max TP (2,27)

(2,27)

X

y = -3x2 + 12x + 15


Curve Sketching

Outcome 3

Higher

Example 32

Sketch 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

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

x = 0 or x = 4

(b) SPs

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

y = -2x2 (x - 4)

= -2x3 + 8x2

SPs occur where dy/dx = 0

so -6x2 + 16x = 0


Curve Sketching

Outcome 3

Higher

-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)2X (8/3 -4) =512/27

nature

x

0

8/3

-

+

-

0

0

dy/dx


Curve Sketching

Outcome 3

Higher

Summarising

(c) Large values

as x + then y -

using y = -2x3

as x - then y +

Y

Sketch

Cuts x – axis at 0 and 4

4

0

Max TP’s at (8/3, 512/27)

(8/3, 512/27)

X

y = -2x2 (x – 4)


Curve Sketching

Outcome 3

Higher

Example 33

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

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

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


Curve Sketching

Outcome 3

Higher

SPs occur where dy/dx = 0

(b) SPs

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)


Curve Sketching

Outcome 3

Higher

nature

x

-1

0

1

+

-

+

-

0

0

0

dy/dx

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


Curve Sketching

Outcome 3

Higher

Summarising

(c) Large values

Using y = - x4

Sketch is

as x + then y -

Y

as x - then y -

Cuts x – axis at -2 and 2

-2

2

Cuts y – axis at 8

8

(1,9)

Max TP’s at

(-1,9)

(-1,9)

(1,9)

X

y = 8 + 2x2 - x4


Outcome 3

Higher

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.


Max & Min on Closed Intervals

Outcome 3

Higher

y =f(x)

(b, f(b))

max = f(b) end point

(a, f(a))

min = f(a) end point

X

a b


Max & Min on Closed Intervals

Outcome 3

Higher

(c, f(c))

max = f(c ) max TP

y =f(x)

(b, f(b))

min = f(a) end point

(a, f(a))

x

a b

c

NB: a < c < b


Max & Min on Closed Intervals

Outcome 3

Higher

y =f(x)

max = f(b) end point

(b, f(b))

(a, f(a))

(c, f(c))

min = f(c) min TP

x

NB: a < c < b

c

a b


Max & Min on Closed Intervals

Outcome 3

Higher

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 34

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


Max & Min on Closed Intervals

Outcome 3

Higher

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

not in interval

in interval

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

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


Max & Min on Closed Intervals

Outcome 3

Higher

Extra bit

Using function notation we can say that

Domain = {xR: -1  x  2 }

Range = {yR: -20  y  0 }


Outcome 3

Higher

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.


Outcome 3

Higher

Optimization

Practical exercise on optimizing volume.

Graph

Problem


Q. What is the maximum volume

We can have for the given dimensions

Optimization

Outcome 3

Higher

Example 35

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

16cm

x cm

10cm

x cm

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

ie 0 < x < 5

This gives us a particular interval to consider !


Optimization

Outcome 3

Higher

By folding up the four flaps we get a small cuboid

x cm

(10 - 2x) cm

(16 - 2x) 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

Outcome 3

Higher

End Points

Considering the interval 0 < x < 5

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)


Optimization

Outcome 3

Higher

SPs occur when V '(x) = 0

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

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

ie x = 20/3or x = 2

not in interval

in interval

When x = 2 then

V(2) = 2 X 12 X 6 = 144

We now check gradient near x = 2


Optimization

Outcome 3

Higher

Nature

x

2

-

+

V '(x)

0

Hence max TP when x = 2

So max possible volume = 144cm3


Optimization

Outcome 3

Higher

Example 36

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

(x  2)

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.


Optimization

Outcome 3

Higher

End points

S(2) = 1 – 1 = 0

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

SPs occur where S (x) = 0


Optimization

Outcome 3

Higher

rearrange

8x2 = 2x3

8x2 - 2x3 = 0

2x2(4 – x) = 0

x = 0 or x = 4

In interval

Out with interval

We now check the gradients either side of 4


Optimization

Outcome 3

Higher

Nature

S (3.9 ) = 0.00337…

x  4 

S (4.1) = -0.0029…

-

+

0

S (x)

Hence max TP at x = 4

And max share of market = S(4)

= 2/4 – 4/16

= 1/2 – 1/4

= 1/4


Nature Table

Equation of tangent line

Leibniz Notation

x

-1

2

5

-

+

f’(x)

0

Straight Line

Theory

Max

Gradient at a point

f’(x)=0

Stationary Pts

Max. / Mini Pts

Inflection Pt

Graphs

f’(x)=0

Derivative

= gradient

= rate of change

Differentiation

of Polynomials

f(x) = axn

then f’x) = anxn-1


Outcome 3

Are you on Target !

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