Using differentiation (Application) Outcome 3 Using differentiation (Application) Finding the gradient for a polynomial Increasing / Decreasing functions Differentiating Brackets ( Type 1 ) Max / Min and inflexion Points Differentiating Harder Terms (Type 2) Curve Sketching Differentiating with Leibniz Notation Max & Min Values on closed Intervals Equation of a Tangent Line ( Type 3 ) Optimization Mind Map of Chapter

Gradients & Curves Demo 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 A B Outcome 3 Higher Gradient of tangent = gradient of curve at A A 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 ((x+h), f(x+h)) Approx gradient (x, f(x)) 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

So the points are even closer together. 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))

Derivative 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. 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) 2x2 4x 4x2 8x Have guessed the rule yet ! 5x10 50x9 6x7 42x6 x3 3x2 x5 5x4 x99 99x98

Derivative If f(x) = axn n n -1 ax then f '(x) = 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

Special Points Outcome 3 (I) f(x) = ax (Straight line function) Higher (I) f(x) = ax (Straight line function) 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 !!

Special Points Outcome 3 (II) f(x) = a, (Horizontal Line) Index Laws Higher (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 = Gradient Differentiation = Rate of change Name :

Calculus Revision Differentiate

Calculus Revision Differentiate

Calculus Revision Differentiate

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

If g(x) = 5x4 - 4x5 then find g '(2) . 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 Example 4 h(x) = 5x2 - 3x + 19 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 Example 6 : Find the points on the curve 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 so x = 0 or x = 2 Points are (0,7) & (2,7)

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

Basic Rule: Break brackets before you differentiate ! Outcome 3 Higher 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

Calculus Revision Differentiate multiply out Differentiate

Calculus Revision Differentiate multiply out Simplify Straight line form Differentiate

Calculus Revision Differentiate Multiply out Straight line form

Fractions Outcome 3 Reversing the above we get the following “rule” ! Higher Reversing the above we get the following “rule” ! This can be used as follows …..

Fractions f(x) = 3x3 - x + 2 x2 = 3x3 - x + 2 x2 x2 x2 Outcome 3 Higher Example f(x) = 3x3 - x + 2 x2 = 3x3 - x + 2 x2 x2 x2 = 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

Leibniz Notation Outcome 3 Higher Leibniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc. 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 NB: Q = 9R2 - 15R-3 So dQ/dR = 18R + 45R-4 = 18R + 45 R4

Leibniz Notation = 60 - (-16) = 76 Outcome 3 Example 20 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 Newton’s 2ndLaw of Motion 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 y = f(x) A(a,b) tangent 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 = 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

Equation of Tangents Outcome 3 Now using y - b = m(x - a) 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 m = 1 Outcome 3 Example 22 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 Now using y - b = m(x - a) 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 Example 23 - (other way round) 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 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)

Increasing & Decreasing Functions and Stationary Points Outcome 3 Higher Consider the following graph of y = f(x) ….. y = f(x) + + - + + a b c - d e f X +

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 f(x) decreasing when f  (x) < 0 so 8x - 24 < 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 - + + f’(x) Function increasing when f (x) > 0 ie x < -3 or x > 2 Function decreasing when f (x) < 0 ie -3 < x < 2

Stationary Points and Their Nature Outcome 3 Higher y = f(x) Consider this graph of y = f(x) again + - + + + a - b c X +

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) + 0 - f(x)

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 Nature Table x dy/dx Higher Nature Table x + + dy/dx 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. SP occurs when dy/dx = 0 Using y = 3x4 - 16x3 + 24 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 x 4 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. SP occurs when dy/dx = 0 Using y = 1/2x4 - 4x2 + 2 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 2 dy/dx - 0 + 0 - 0 + So (-2,-6) and (2,-6) are Minimum Turning Points and (0,2) is a Maximum Turning Points

Curve Sketching Outcome 3 Higher 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.

Suppose that f(x) = -2x3 + 6x2 + 56x - 99 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 As x  + then y  - Graph roughly As x  - then y  +

Curve Sketching Outcome 3 Example 31 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)

is a Maximum Turning Point 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 Summarising Outcome 3 (c) Large values Higher Summarising (c) Large values as x  + then y  - using y = -3x2 as x  - then y  - Y Sketching Cuts x-axis at -1 and 5 -1 5 Cuts y-axis at 15 15 Max TP (2,27) (2,27) X y = -3x2 + 12x + 15

Graph cuts axes at (0,0) and (4,0) . 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 -2x(3x - 8) = 0 -2x = 0 or (3x - 8) = 0 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)2 X (8/3 -4) =512/27 nature x 8/3 - + - dy/dx

Curve Sketching Summarising Outcome 3 (c) Large values 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 Max TP’s at (8/3, 512/27) (8/3, 512/27) X y = -2x2 (x – 4)

Curve Sketching Outcome 3 Example 33 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 (b) SPs SPs occur where dy/dx = 0 Higher (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)

So (0,8) is a min TP while (-1,9) & (1,9) are max TPs . Curve Sketching Outcome 3 Higher nature x -1 1 + - + - dy/dx So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .

Curve Sketching Summarising Outcome 3 (c) Large values Using y = - x4 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 Max TP’s at (-1,9) (-1,9) (1,9) (1,9) X y = 8 + 2x2 - x4

Max & Min on Closed Intervals Outcome 3 Higher 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)) (a, f(a)) min = f(a) end point 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 a b c

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

Extra bit 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 }

Note: Optimum basically means the best possible. Optimization Outcome 3 Higher 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.

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

Considering the interval 0 < x < 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)

We now check gradient near 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 = 20/3 or 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

So max possible volume = 144cm3 Optimization Outcome 3 Higher Nature x 2 + - V '(x) Hence max TP when x = 2 So max possible volume = 144cm3

Optimization S(5) = 2/5 – 4/25 = 6/25 Outcome 3 Example 36 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.

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

We now check the gradients either side of 4 Optimization Outcome 3 Higher rearrange 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

And max share of market = S(4) Optimization Outcome 3 Higher Nature S (3.9 ) = 0.00337… x  4  S (4.1) = -0.0029… + - 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

Equation of tangent line Nature Table Equation of tangent line Leibniz Notation -1 2 5 + - x f’(x) Max Straight Line Theory 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