Differentiation s5 Unit 1 Outcome 3CAPE
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Surds & Indicesxm . xn = xm+n
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Special Points(I) f(x) = ax (Straight line function)If f(x) = ax = ax1then f '(x) = 1 X ax0 = a X 1 = aIndex Laws x0 = 1So if g(x) = 12x then g '(x) = 12 Also using y = mx + cThe line y = 12x has gradient 12,and derivative = gradient !!HigherOutcome 3
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(II) f(x) = a, (Horizontal Line)If f(x) = a = a X 1 = ax0then f '(x) = 0 X ax-1= 0Index Laws x0 = 1So 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 !Special PointsHigherOutcome 3
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Name : Gradient=Rate of change=DifferentiationDifferentiationDifferentiation techniques
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Calculus RevisionDifferentiate
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Calculus RevisionDifferentiate
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Calculus RevisionDifferentiate
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Example 1A curve has equation f(x) = 3x4Its gradient is f '(x) = 12x3 f '(2) = 12 X 23 = 12 X 8 = 96Example 2A curve has equation f(x) = 3x2Find the formula for its gradient and find the gradient when x = 2Its gradient is f '(x) = 6xAt the point where x = -4 the gradient isf '(-4) = 6 X -4 =-24DerivativeHigherOutcome 3Find the formula for its gradient and find the gradient when x = -4
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Example 3If g(x) = 5x4 - 4x5 then find g '(2) .g '(x) = 20x3 - 20x4 g '(2) = 20 X 23 - 20 X 24 = 160 - 320 = -160DerivativeHigherOutcome 3
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Example 4h(x) = 5x2 - 3x + 19so h '(x) = 10x - 3and h '(-4) = 10 X (-4) - 3 = -40 - 3 = -43Example 5k(x) = 5x4 - 2x3 + 19x - 8, find k '(10) .k '(x) = 20x3 - 6x2 + 19So k '(10) = 20 X 1000 - 6 X 100 + 19 = 19419DerivativeHigherOutcome 3
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Example 6 : 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) = 2ie 3x2 - 6x + 2 = 2or 3x2 - 6x = 0 ie 3x(x - 2) = 0ie 3x = 0 or x - 2 = 0so x = 0 or x = 2Now using original formulaf(0) = 7f(2) = 8 -12 + 4 + 7 = 7Points are (0,7) & (2,7)DerivativeHigherOutcome 3
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Calculus RevisionDifferentiate
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Calculus RevisionDifferentiateStraight line formDifferentiate
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Calculus RevisionDifferentiateStraight line formDifferentiate
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Calculus RevisionDifferentiateChain RuleSimplifyStraight line form
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Calculus RevisionDifferentiateStraight line formDifferentiate
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Calculus RevisionDifferentiateDifferentiateStraight line form
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Calculus RevisionDifferentiateDifferentiateStraight line form
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Name : Gradient=Rate of change=DifferentiationDifferentiationDifferentiation techniques
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Calculus RevisionDifferentiateMultiply outDifferentiate
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Calculus RevisionDifferentiatemultiply outdifferentiate
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Calculus RevisionDifferentiateStraight line formmultiply outDifferentiate
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Calculus RevisionDifferentiatemultiply outDifferentiate
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Calculus RevisionDifferentiatemultiply outSimplifyDifferentiateStraight line form
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Calculus RevisionDifferentiateStraight line formMultiply outDifferentiate
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Calculus RevisionDifferentiateSplit upStraight line formDifferentiate
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If y is expressed in terms of x then the derivative is written as dy/dx .Leibniz NotationLeibniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc. eg y = 3x2 - 7x so dy/dx = 6x - 7 .Example 19Find dQ/dR NB: Q = 9R2 - 15R-3So dQ/dR = 18R + 45R-4HigherOutcome 3
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Example 20A curve has equation y = 5x3 - 4x2 + 7 . Find the gradient where x = -2 ( differentiate ! )gradient = dy/dx = 15x2 - 8xif x = -2 thengradient = 15 X (-2)2 - 8 X (-2)= 60 - (-16) = 76Leibniz NotationHigherOutcome 3
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Newtons 2ndLaw of Motions = ut + 1/2at2 where s = distance & t = time.Finding ds/dt means diff in dist diff in timeie speed or velocityso ds/dt = u + at but ds/dt = v so we getv = u + atand this is Newtons 1st Law of MotionReal Life ExamplePhysicsHigherOutcome 3
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A(a,b)y = mx +cy = f(x)Equation of TangentstangentNB: 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) HigherOutcome 3Demo
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HigherOutcome 3Equation of TangentsExample 21Find the equation of the tangent line to the curvey = 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 - 2when x = -1 dy/dx = 3 X (-1)2 - 2 = 3 - 2 = 1 m = 1Straight line so we need a point plus the gradient then we can use the formula y - b = m(x - a) .Demo
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Now using y - b = m(x - a) we get y - 2 = 1( x + 1)or y - 2 = x + 1or y = x + 3point is (-1,2)m = 1Equation of TangentsHigherOutcome 3
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Example 22Find 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 x3when x = -2 then dy/dx = -8 (-2)3= -8/-8 = 1m = 1Equation of TangentsHigherOutcome 3
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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 + 3AxesTangent cuts Y-axis when x = 0 so y = 0 + 3 = 3at point (0, 3)Tangent cuts X-axis when y = 0so 0 = x + 3 or x = -3at point (-3, 0)Equation of TangentsHigherOutcome 3
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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 - 6so2x - 6 = 14 2x = 20x = 10Put x = 10 into y = x2 - 6x + 5 Giving y = 100 - 60 + 5 = 45Point is (10,45)Equation of TangentsHigherOutcome 3
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Increasing & Decreasing Functions and Stationary PointsConsider the following graph of y = f(x) ..Xy = f(x)abcdef+++++--000HigherOutcome 3
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In the graph of y = f(x)The function is increasing if the gradient is positivei.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.Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3
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Example 24For the function f(x) = 4x2 - 24x + 19 determine the intervals when the function is decreasing and increasing.f (x) = 8x - 24f(x) decreasing when f (x) < 0so 8x - 24 < 0 8x < 24x < 3f(x) increasing when f (x) > 0
so 8x - 24 > 08x > 24x > 3Check: f (2) = 8 X 2 24 = -8 Check: f (4) = 8 X 4 24 = 8Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3
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Example 25For the curve y = 6x 5/x2 Determine if it is increasing or decreasing when x = 10. = 6x - 5x-2so dy/dx = 6 + 10x-3when x = 10 dy/dx = 6 + 10/1000 = 6.01Since dy/dx > 0 then the function is increasing. Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3
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Example 26Show that the function g(x) = 1/3x3 -3x2 + 9x -10 is never decreasing. g (x) = x2 - 6x + 9 = (x - 3)(x - 3)= (x - 3)2Since (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.Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3
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Example 27Determine the intervals when the functionf(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 stationary when f (x) = 0ie 6(x + 3)(x - 2) = 0 ie x = -3 or x = 2Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3
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We now use a special table of factors to determine when f (x) is positive & negative. x-32f(x)
+Function increasing when f (x) > 0ie x < -3 or x > 2Function decreasing when f (x) < 0ie -3 < x < 2Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3-0 +0
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Stationary Points and Their NatureConsider this graph of y = f(x) againXy = f(x)abc+++++--000HigherOutcome 3
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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 NatureHigherOutcome 3
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Maximum Turning pointxaf(x) + 0 -Minimum Turning Pointxbf(x) - 0 +Stationary Points and Their NatureHigherOutcome 3
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Rising Point of inflexion xcf(x) + 0 +Other possible type of inflexion xdf(x) - 0 -Stationary Points and Their NatureHigherOutcome 3
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Example 28Find 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 = 0Using y = 4x3 + 1 if x = 0 then y = 1SP is at (0,1)Stationary Points and Their NatureHigherOutcome 3
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Nature Tablex 0dy/dx +So (0,1) is a rising point of inflexion.Stationary Points and Their Naturedy/dx = 12x2HigherOutcome 3 + 0
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Example 29Find the co-ordinates of the stationary points on the curve y = 3x4 - 16x3 + 24 and determine their nature.SP occurs when dy/dx = 0So 12x3 - 48x2 = 0 12x2(x - 4) = 012x2 = 0 or (x - 4) = 0x = 0 or x = 4Using y = 3x4 - 16x3 + 24 if x = 0 then y = 24if x = 4 then y = -232SPs at (0,24) & (4,-232)Stationary Points and Their NatureHigherOutcome 3
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Nature Tablex04dy/dx - 0 - 0 +So (0,24) is a Point of inflexion and (4,-232) is a minimum Turning PointStationary Points and Their Naturedy/dx=12x3 - 48x2HigherOutcome 3
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Example 30Find the co-ordinates of the stationary points on the curve y = 1/2x4 - 4x2 + 2 and determine their nature.SP occurs when dy/dx = 0So 2x3 - 8x = 0 2x(x2 - 4) = 02x(x + 2)(x - 2) = 0x = 0 or x = -2 or x = 2Using y = 1/2x4 - 4x2 + 2if x = 0 then y = 2if x = -2 then y = -6SPs at(-2,-6), (0,2) & (2,-6)if x = 2 then y = -6Stationary Points and Their NatureHigherOutcome 3
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Nature Tablex0dy/dx-22 - 0 + 0 - 0 +So (-2,-6) and (2,-6) are Minimum Turning Pointsand (0,2) is a Maximum Turning PointsStationary Points and Their NatureHigherOutcome 3
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Curve SketchingNote: 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 = 0for 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.HigherOutcome 3
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Dominant TermsSuppose 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 roughlyCurve SketchingHigherOutcome 3
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Example 31Sketch the graph of y = -3x2 + 12x + 15(a) AxesIf x = 0 then y = 15If y = 0 then -3x2 + 12x + 15 = 0( -3)x2 - 4x - 5 = 0(x + 1)(x - 5) = 0x = -1 or x = 5Graph cuts axes at (0,15) , (-1,0) and (5,0)Curve SketchingHigherOutcome 3
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(b) Stationary Points occur where dy/dx = 0so -6x + 12 = 06x = 12x = 2If x = 2 then y = -12 + 24 + 15 = 27Nature Tablex2dy/dx + 0 -So (2,27) is a Maximum Turning PointStationary Point is (2,27)Curve SketchingHigherOutcome 3
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(c) Large valuesusing y = -3x2as x + then y - as x - then y -SketchingXYy = -3x2 + 12x + 15Curve SketchingCuts x-axis at -1 and 5SummarisingCuts y-axis at 15-15Max TP (2,27)(2,27)15HigherOutcome 3
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Example 32Sketch the graph of y = -2x2 (x - 4)(a) AxesIf x = 0 then y = 0 X (-4) = 0If y = 0 then -2x2 (x - 4) = 0x = 0 or x = 4Graph cuts axes at (0,0) and (4,0) .-2x2 = 0 or (x - 4) = 0(b) SPsy = -2x2 (x - 4)= -2x3 + 8x2SPs occur where dy/dx = 0
so -6x2 + 16x = 0Curve SketchingHigherOutcome 3
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-2x(3x - 8) = 0-2x = 0 or (3x - 8) = 0x = 0 or x = 8/3If x = 0 then y = 0 (see part (a) ) If x = 8/3 then y = -2 X (8/3)2 X (8/3 -4) =512/27naturex08/3dy/dx - Curve SketchingHigherOutcome 3 - + 0 0
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Summarising(c) Large valuesusing y = -2x3as x + then y -as x - then y +SketchXy = -2x2 (x 4)Curve SketchingCuts x axis at 0 and 404Max TPs at (8/3, 512/27) (8/3, 512/27)YHigherOutcome 3
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Example 33Sketch the graph of y = 8 + 2x2 - x4(a) AxesIf 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) = 0or (2 + x) (2 - x)(2 + x2) = 0So x = -2 or x = 2 but x2 -2Curve SketchingHigherOutcome 3
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(b) SPsSPs occur where dy/dx = 0So 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 - x4when x = 0 then y = 8when x = -1 then y = 8 + 2 - 1 = 9 (-1,9) when x = 1 then y = 8 + 2 - 1 = 9 (1,9)Curve SketchingHigherOutcome 3
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naturex-101dy/dx+So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .Curve SketchingHigherOutcome 3+--000
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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 isXY-228(-1,9)(1,9)y = 8 + 2x2 - x4Max TPs at (-1,9)(1,9)Curve SketchingSummarisingHigherOutcome 3
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Max & Min on Closed IntervalsIn 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.HigherOutcome 3
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y =f(x)Xa b(a, f(a))(b, f(b))max = f(b) end pointmin = f(a) end pointMax & Min on Closed IntervalsHigherOutcome 3
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xy =f(x)(b, f(b))(a, f(a))max = f(c ) max TP min = f(a) end pointa b(c, f(c))cNB: a < c < bMax & Min on Closed IntervalsHigherOutcome 3
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y =f(x)xa b c(a, f(a))(b, f(b))
(c, f(c))max = f(b) end pointmin = f(c) min TPNB: a < c < bMax & Min on Closed IntervalsHigherOutcome 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 pointsExample 34Find the max & min values of y = 2x3 - 9x2 in the interval where -1 x 2. End pointsIf x = -1 then y = -2 - 9 = -11If x = 2 then y = 16 - 36 = -20 Max & Min on Closed IntervalsHigherOutcome 3
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Stationary pointsdy/dx = 6x2 - 18x = 6x(x - 3)SPs occur where dy/dx = 0 6x(x - 3) = 06x = 0 or x - 3 = 0x = 0 or x = 3in interval not in intervalIf x = 0 then y = 0 - 0 = 0Hence for -1 x 2 , max = 0 & min = -20Max & Min on Closed IntervalsHigherOutcome 3
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Extra bitUsing function notation we can say thatDomain = {xR: -1 x 2 } Range = {yR: -20 y 0 }Max & Min on Closed IntervalsHigherOutcome 3
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Derivative GraphsHigherOutcome 3Demo
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OptimizationNote: 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.HigherOutcome 3
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OptimizationHigherOutcome 3ProblemPractical exercise on optimizing volume. Graph
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Example 35HigherOutcome 3OptimizationQ. What is the maximum volumeWe can have for the given dimensionsA rectangular sheet of foil measuring 16cm X 10 cm has four small squares each x cm cut from each corner. 16cm10cmx cmNB: x > 0 but 2x < 10 or x < 5ie 0 < x < 5This gives us a particular interval to consider ! x cm
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(16 - 2x) cm(10 - 2x) cmx cmThe volume is now determined by the value of x so we can writeV(x) = x(16 - 2x)(10 - 2x)= x(160 - 52x + 4x2)= 4x3 - 52x2 +160xWe now try to maximize V(x) between 0 and 5OptimizationBy folding up the four flaps we get a small cuboid HigherOutcome 3
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Considering the interval 0 < x < 5End PointsV(0) = 0 X 16 X 10 = 0V(5) = 5 X 6 X 0 = 0SPsV '(x) = 12x2 - 104x + 160 = 4(3x2 - 26x + 40)= 4(3x - 20)(x - 2)OptimizationHigherOutcome 3
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SPs occur when V '(x) = 0 ie 4(3x - 20)(x - 2) = 03x - 20 = 0 or x - 2 = 0ie x = 20/3 or x = 2not in intervalin intervalWhen x = 2 thenV(2) = 2 X 12 X 6 = 144We now check gradient near x = 2OptimizationHigherOutcome 3
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x2V '(x)+ Hence max TP when x = 2So max possible volume = 144cm3NatureOptimizationHigherOutcome 3- 0
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Example 36When a company launches a new product its share of the market after x months is calculated by the formulaSo after 5 months the share isS(5) = 2/5 4/25 = 6/25Find the maximum share of the market that the company can achieve.(x 2)OptimizationHigherOutcome 3
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End pointsS(2) = 1 1 = 0There is no upper limit but as x S(x) 0. SPs occur where S (x) = 0OptimizationHigherOutcome 3
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8x2 = 2x38x2 - 2x3 = 02x2(4 x) = 0x = 0 or x = 4Out with intervalIn intervalWe now check the gradients either side of 4rearrangeOptimizationHigherOutcome 3
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x 4 S (x)S (3.9 ) = 0.00337S (4.1) = -0.0029Hence max TP at x = 4And max share of market = S(4)= 2/4 4/16= 1/2 1/4= 1/4OptimizationNatureHigherOutcome 3+-0
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Differentiationof Polynomialsf(x) = axnthen fx) = anxn-1Derivative = gradient = rate of change Graphsf(x)=0f(x)=0Stationary PtsMax. / Mini PtsInflection PtNature TableGradient at a pointEquation of tangent lineStraight LineTheoryLeibniz Notation
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Are you on Target ! Update you log book Make sure you complete and correct ALL of the Differentiation 1questions in the past paper booklet.Outcome 3
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