CALCULUS 1: LIMITS, AVERAGE GRADIENT AND FIRST PRINCIPLES DERIVATIVES

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1 6 LESSON CALCULUS 1: LIMITS, AVERAGE GRADIENT AND FIRST PRINCIPLES DERIVATIVES Learning Outcome : Functions and Algebra Assessment Standard 1..7 (a) In this section: The limit concept and solving for limits Function notation (brief revision section) Finding the rate of change of a function between two points or average gradient Calculating the derivative of a function from first principles Limit Concept Background The concept of finding a limit looks at what happens to a function value (-value) of a curve, as we get close to a specific -value on the curve. One wa to think of it is: if ou start walking in a certain direction and find ourself heading towards a big boulder, this boulder will be our limit. Eample Eample 1 Let s start b looking at the function ƒ() = shown below: f() 6 Now let s consider the lim (we sa: the limit of ƒ() as tends to zero). So, 0 we want to know what the -value is equal to when approaching = 0. Page 88

2 We sa: lim ƒ() = lim = 3(0) + 6 Substitute in = 0, and remove lim 0 = 6 Eample In the case of ƒ() = _ 16 (shown below) 4 we can use a limit to find what value ƒ() tends to as tends to four. 8 f() Eample But if we substitute = 4 into the limit, we will get lim ƒ() = _ 0, but we know that this 4 0 is wrong. So how can we solve this algebraicall? Well, let s see if we can simplif ƒ() at all: ƒ() = _ The numerator is a difference of two squares, so we can factorise ( 4)( + 4) = The ( 4) in the numerator and denominator will cancel 4 = + 4 Now let s solve for the limit of ƒ(): lim ƒ() = lim 4 4 = lim 4 _ Now we can substitute in for = 4 = 8 this means that ƒ() tends to 8 but does not equal 8 at the value of 4 Note that if substituting for the limit produces a zero denominator, factorise and cancel first! Activit 1 Activit 1) lim 5 + ) lim ) lim 1_ + + 1_ Page 1 Lesson 1 Algebra Page 89

3 Revision of Function Notation Eample Eample Eample 1 If ƒ() = 3 +, find the value of: 1. ƒ() ƒ() tells us that in the formula ƒ(), we must make the substitution = ƒ() = () 3() + = = 0 this tells us that on the curve ƒ(), at the point =, ƒ() = 0. In actual fact we have found the co-ordinates (; 0) of a point on the curve ƒ().. ƒ( 1) ƒ( 1) tells us to make the substitution = 1 in ƒ() ƒ( 1) = ( 1) 3( 1) + = = 6 3. ƒ( + ) ƒ( + ) we substitute = + into ƒ() ƒ( + ) = ( + ) 3( + ) + multipl out = simplif = + 4. ƒ( + h) in this case we must substitute = + h into f(). ƒ( + h) = ( + h) 3( + h) + multipl out = + h + h 3 3h + this cannot be simplified further and is therefore our final answer Eample If ƒ() = ƒ( + h) ƒ() +, determine : h ƒ( + h) ƒ() h = ( + h) + ( + h) ( + ) Substitute = + h for ƒ( + h) h = + h + h + + h Remember to distribute the negative sign h h( + h + 1) = Factorise and take a common factor of h h = + h + 1 This cannot be simplified further and is our final answer Page 90

4 Average Gradient Background When dealing with straight line graphs, it is eas to determine the slope of the same graph b calculating the change in -value between two points and the change in -value between the same two points. Then we substitute these into change in the formula for gradient =. However, when dealing with curves the change in gradient changes at ever point on the curve, therefore we need to work with the average gradient. The average gradient between two points is the gradient of a straight line drawn between the two points. Eample Given the equation ƒ() = 6, find the average gradient between = 1 and = 4. First we need to find the value of ƒ() at = 1. This is the -value of ƒ() at = 1. ƒ(1) = (1) 1 6 (making the substitution = 1) = 7 = 5 Net we must find the value of ƒ() at = 4 ƒ(4) = (4) 4 6 = (16) 10 = Now we can substitute these values into the average gradient formula: _ B A B = ( 5) A 4 1 = _ 7 3 = 9 Eample The Derivative (First Principles) Background We will now investigate the average gradient between points. B calculation we can show that this is equivalent to: ƒ( + h) ƒ() h f() f() ( ; f()) f( + h) ( + h ; f( + h)) + h 009 Page 1 Lesson 1 Algebra Page 91

5 Now think about it: as h, the distance between our -values, gets smaller our points get closer together. Lets look at an eample of this: let = for the graph below. ( h = 3, since 5 = 3) for h = 3: f() f() f( + h) h = 3 h 5 g() for h = 1: f() f() f( + h) h = 1 h 3 g() for h = 0.5 f() f() f( + h) h.5 g() h = 0.5 Page 9 In each of the graphs above, g() is a straight line with the same gradient as the average gradient of the curve between the points and + h. Can ou see that as h decreases the two points on the ais get closer together? Can ou also see that as the two poins on the -ais get closer together, the average gradient changes? So as the two points get closer together (as h becomes 0) the average gradient between the two points becomes the actual gradient at the point. However, h

6 cannot equal 0 as that would give us a 0 denominator, so we must use limits. The actual gradient at a point is called the derivative. To find the limit of the gradient as h tends to zero, we use the formula: lim ƒ( + h) f(). We call this the derivative of the function and use the h 0 h notation ƒ () This method is called finding the derivative from first principles. Eample Find the derivative of the function ƒ() = 3 from first principles. Solution: ƒ( + h) = 3( + h) = 3( + h + h ) = 3 + 6h + 3h Start b finding f( + h), which is the first part of the formula Now substitute the epressions for ƒ() and ƒ( + h) into the first principles formula: Eample Solution ƒ( + h) ƒ() ƒ () = lim h h 0 (3 + 6h + 3h ) (3 ) h 0 h 3 + 6h + 3h 3 h 0 h = lim = lim 6h + 3h = lim h h 0 ƒ () = lim h(6 + 3h) h 0 h Dont forget to change the signs of the term in the second bracket This can be simplified further If we were to evaluate this now, we would get ƒ () = _ 0. But we 0 have a common factor of h in the numerator, so let s factorise. The h in the numerator will cancel out the h in the denominator = lim (6 + 3h) h 0 = 6 +3(0) Substitute in for h = 0, and remove lim h 0 = 6 This is the derivative of ƒ() = 3 Activit Activit 1. For the curve ƒ() = - + 9, find: 1.1 the average gradient between = 0 and = 6 1. the derivative, from first principles, at = 3. From first principle, find ƒ () for the following functions:.1 ƒ() = Page 1 Lesson 1 Algebra Page 93

7 . ƒ() = Calculus : General Rules for Differentiation, and Tangents Learning Outcome : Functions and Algebra Assessment Standard 1..7 (a) In this section we will: determine the general rules for differentiation determine the gradient and equation of a tangent to a function at a desired point General Rules for Differentiation Note: there are man different was of writing ƒ (), we can use: D [ƒ()] we sa: the derivative of ƒ() with respect to _ d d we sa: the derivative of with respect to or f we sa: the derivative of or f() Derivative Laws I am sure ou will agree that finding the derivative from first principles is a rather time-consuming process and that a more time-efficient method is needed. Luckil for ou, there is one! A set of rules have been created in order to speed up the process, and these are: 1. If ƒ() =k n where k is a constant, then ƒ () = k n n 1 If ƒ() = 6 3, then ƒ() = 6 3 = 18. If ƒ() = k, where k is a constant, then ƒ () = 0 If the ƒ() is equal to a constant, then the derivative of the function is zero. If ƒ() = 3, then ƒ () = 0 3. If ƒ() = g() + h(), then ƒ () = g () + h () The derivative of a sum is the sum of the derivatives Ifƒ() = , then ƒ () = Similarl, if ƒ() = g() h(), then ƒ () = g () h () The derivative of a difference, is the difference of the derivatives Ifƒ() = 6 3 6, then ƒ () = Page 94

8 Remember: 1_ = If ou have in the denominator, take it to the numerator and change the sign of the eponent 3 _ = 1_ 3 If ou have a surd, change it to power form. 4 + = _ If ou have man terms over a common denominator, with a single term in the denominator, separate the terms out into several fractions Eamples: Find the derivatives of the following functions 1.1 ƒ() = f ()= ƒ() = _ + 4 f () = _ 1.3 ƒ() = 3 _ = 3 - Take the to the numerator b changing the sign of the eponent ƒ () = 3( 3 ) = 6 3 = _ ƒ() = _ = _ ƒ () = = 4 + _ 4 Activit 3 Eample Activit Find the derivatives of the following functions: _ 1. ƒ() = + 1 _. = 3 ( 3) 3. = Page 1 Lesson 1 Algebra Page 95

10 m Tangent = (4) + = 10 Therefore, the gradient of the tangent toƒ() at the point = 4 is 10. If we want to find the equation of the tangent, we need a point on the curve. Since we onl know the -value in this case, we need to find the -value at = 4, so we use the original curve to get this -value ƒ (4) = (4) + (4) + 1 making the substitution = 4 into ƒ() = = 5 so we have the point on ƒ(): (4, 5) We know that the general formula for a straight line graph is = m + c. Now that we have found a point on the graph and we know the gradient, we can solve for c: 5 = 10(4) + c substituting in = 5, = 4 and m = 10 5 = 40 + c 5 40 = c c = 15 Therefore, the equation for the tangent to ƒ() at = 4 is = Eample Consider ƒ() = Sketch: ƒ() 1. Determine: ƒ () 1.3 Determine: ƒ (1) 1.4 What does this number represent? 1.5 Hence determine the equation of the tangent to ƒ() at = Determine the equation of the tangent at =. Eample Solution 1.1 Sketch: ƒ() First find intercepts with the aes: Let = 0 to find the -intercept, so = 8 and the -intercept is at (0, 8) To find the -intercepts let = 0, so 0 = ( + )( + 4) and the -intercepts are at ( ; 0) and ( 4 ; 0) Ais of smmetr: = _ b a = _ ( ) ( 1) = _ = 1 For the turning point we must find the -value at the ais of smmetr: = ( 1) ( 1) + 8 = = Solution Page 1 Lesson 1 Algebra Page 97

11 Turning Point ( 1; 9) Now we can sketch ƒ(): ( 1; 9) (0 ; 8) ( 4 ; 0) ( ; 0) 1. Determine: ƒ () ƒ () = 1.3 Determine: ƒ (1) ƒ (1) = (1) substitute = 1 into ƒ () = What does this number represent? The gradient of the tangent at = Hence determine the equation of the tangent to ƒ() at = 1 We have alread found the gradient of the tangent at = 1. Now we need to find a point on ƒ() and solve for the c from = m + c. We need to find the -value of the curve at = 1, so we need to find ƒ(1): ƒ(1) = (1) (1) + 8 = = 5 (1; 5) is the point on ƒ() at which we are tring to find the tangent Now we can use this to solve for c: 5 = (-4)(1) + c substitute in = 5, m = -4 and = 1 5 = -4 + c c = c = 9 = is the tangent to ƒ() at = 1 Page 98

12 1.6 Determine the equation of the tangent at = First we need to find ƒ ( ) ƒ ( ) = ( ) = 4 = This is the gradient of the tangent of ƒ() at = Now we need to find the corresponding -value at this point ƒ( ) = ( ) ( ) + 8 = = 8 ( ; 8) is the point on ƒ() at which we are tring to find the tangent We have everthing we need to solve for c in the equation (substitute the point ( ;8)) = m + c: = + c 8 = 4 + c c = 1 = + 1 is the tangent to ƒ() at = Activit 4 Activit 1. Find the equation of the tangent to the curve ƒ() = ( 1) at the point where the graph meets the -ais. Find the equation of the tangent to the curve ƒ() = + + at the point (1 ; ) 3. The line = is a tangent to the curve of ƒ() = Determine the point of contact. 009 Page 1 Lesson 1 Algebra Page 99

13 4. Find the equation of the tangent to the curve of ƒ() = _ 3 at the point = 1 Calculus 3: Sketching and Analsing Graphs Using Calculus Learning Outcome : Functions and Algebra Assessment Standard 1..7 (a) In this section: Sketching a cubic graph Finding the equation of a cubic graph How to work with points of Inflections Application of maima and minima in real-life scenarios Calculating the rate of change in real-life scenarios Cubic Graphs Background The cubic graph has a general formula: ƒ() = a 3 + b + c + d. Let s first look at some new concepts and terminolog that must be learnt: Stationar point: a point on the function where the gradient is zero. There are three kinds of stationar points: 1. Local maimum: f () = 0 local maimum positive gradient negative gradient. Local Minimum: negative gradient positive gradient local minimum f () = 0 Page 100

14 3. Point of Inflection: f BE CAREFUL: Not all points of inflection are stationar points! The gradient of the curve MUST BE ZERO at a stationar point. As with curve sketching in the past: For -intercept, put = 0 then solve for, i.e. = d. Intercept at (0; d) For -intercepts, put = 0, i.e. solve a 3 + b + c + d = 0. There could be one, two or three real roots. For stationar points, put ƒ () = 0 and solve for, i.e. 3a + b + c = 0, then use these values to find the -value of the curve at these points. Find the points of inflection b setting the second derivative equal to zero. Points of inflection, f () = 0 There could be: TWO stationar points, in which case the possible graphs for a>0 are: A B C 3 real roots turning points A B real roots turning points A 1 real root turning points And for a < 0: A B C 3 real roots turning points A real roots turning points B 1 real root turning points ONE stationar point, in which case the possible graphs are: A 009 Page 1 Lesson 1 Algebra Page 101

15 NO stationar points: in this case ou ma have a point of inflection for which the gradient is not zero. Eample Eample 1 (Curve Sketching) ƒ() = Sketch ƒ(), clearl showing all intercepts with the aes and stationar points. Stationar Points: A good wa to start is to find the stationar points. In order to do this, we must find ƒ () = 0 since at the minimums and maimums, the gradient of the curve is 0. ƒ () = = 0 factorise (3 + 1)( 1) = 0 = _ 1 3 OR = 1 Now we need to find the -values of the curve at the stationar points, so we must find ƒ ( _ 1 3) and ƒ(1). Substitute = _ 1 into ƒ() Substitute = 1 into ƒ() 3 f( _ 1 3 ) = ( _ 1 3) 3 ( _ 1 3) ( _ 1 3) + 10 ƒ(1) = = _ 1 7 _ _ = 9 = (1; 9) 7 = _ 75 7 = 10 _ , ( _ 1 3 ; 10 5_ 7) Page 10

16 These are the two points on ƒ() at which we have stationar points. From the stationar points, we can now make a rough sketch of the graph: Points of Inflection: to find if there are an points of inflection, we need to solve for ƒ () = 0 ƒ () = 6-6 = 0 6 = = _ 1 3 \ there is a point of inflection at = _ 1 3. We need to find f( _ 1 3 ) f( _ 1 3 ) = ( _ 1 3 )3 (_ 1 3 ) (_ 1 3 ) + 10 = _ 1 7 _ 1 9 _ = = _ 59 7 = (_ 1 3 ; _ 59 7 ) -intercepts: Net we must find the -intercepts b making ƒ() = 0 Cubic functions are more difficult to factorize than quadratic functions. Using the factor theorem is often the easiest wa. REMEMBER: when using the factor theorem, if we find ƒ(a) = 0 for some value, a, then ( a) is a factor of ƒ(). Use trial and error to get our first factor. Tr ƒ( 1) ƒ(1), ƒ(-), and ƒ() until ou get one equal to zero ƒ(1); ƒ( 1) 0; ƒ() 0 but ƒ( ) = = 0 + is a factor. We usuall get the other factor(s) b doing long division, snthetic division or the inspection method. B looking at the rough sketch above can ou see wh there is onl 1 -intercept -intercept: to find the -intercept, let = 0 = (0) 3 (0) = Page 1 Lesson 1 Algebra Page 103

17 Now we can sketch the graph b joining all the points: note that if coefficient of 3 in the original equation is positive, the graph has shape and if the coefficient of 3 is negative,. In this case we have 1 3 so the shape is as follows. ( _ 1 3 ;_ 75 7 ) 10 ( _ 1 3 ;_ 59 7 ) f() (1; 9) _1 _ Eample Eample (Finding the Equation of a Curve) Find the equation of ƒ(): (0 ; 16) f ( ; 0) ( ; 0) Use = a( 1 ) ( ) ( 3 ) Where 1, and 3 are -intercepts This is another form of the general formula of a cubic graph. This form is normall used when finding the equation of a cubic graph as opposed to sketching the graph. The -intercepts are = and =, but at = the graph does not cut the aes, so here we have two -intercepts that are the same. Making this substitution in the general form, we have: = a( + ) ( ) ( ) We are also given the point (0; 16) which is the -intercept. We can substitute this into out general form to find the value of a: 16 = a(0 + )(0 ) (0 ) 16 = a()( )( ) 16 = 8a a = = ( + )( ) Substitute a = and multipl out. = ( + 4)( 4 + 4) = Page 104

18 = Eample 3 (Point of Inflection). Find the points of inflection and stationar points in = If we were to look at the function = and tr to solve for the stationar points of this function we would get: _ d d = = 0 (solve for the stationar points) This cannot be factorised easil, so we use the quadratic formula: = b ± b 4ac a = ± ( ) 4(3)(1) (3) _ = ± _ Eample Now we have a problem, there _ is no solution for as we have 8 in the quadratic equation. Therefore we can sa that there are no stationar points on this curve. However we can find the point of inflection b setting the second point of inflection ( 1 ; ) 3 f() = + + derivative equal to zero. Activit 5 Activit 1. The figure represents the graphs of the functions f() = 4 + and g() = g() F B D 0 A f() C E Determine: 1.1 The lengths of DA, OB, and CF. 009 Page 1 Lesson 1 Algebra Page 105

19 1. The coordinates of the turning point E. 1.3 The value(s) of for which the gradients of ƒ() and g() are equal.. ƒ() = Use a calculator to find the coordinates of the stationar points of ƒ(), correct to one decimal digit.. Use the remainder theorem to find the roots of ƒ() = 0.3 Sketch the curve of ƒ(), showing all the important points clearl. 0.4 Determine the gradient of ƒ() at each of the -intercepts. 3. Find an equation for ƒ() using the sketch: f Page 106

20 4. Calculate all the stationar points and points of Inflection in the curve = Maima and Minima in Calculus Real Life Application There are man situations in life where we need to find the maimum or minimum of some quantit. Often businesses need to find was to minimise costs and maimise productivit. This is where maima and minima come in ver useful. First, let s note some basics that will be useful: often we will get asked to maimise or minimise a certain area, volume, or cost function etc. r Rectangle Circle Perimeter = + Perimeter (circumference) = πr Area = Area = πr h h r Bo Clinder Volume = h Volume = πr h Surface Area = + h + h...(closed) Surface Area = πrh + πr... (closed) Surface Area = + h + h...(open) Surface Area = πrh + πr (open) Background In order to find the maimum or minimum of a function, the derivative of the function must be found and made to equal zero. This is similar to the cubic graphs section, where if we wanted to find the local maimum or minimum we did the same thing. The difference here is that instead of finding a local maimum or minimum of a cubic graph, we ma be finding the maimum or minimum volume, area, cost, profit etc. 009 Page 1 Lesson 1 Algebra Page 107

21 Eample Eample Eample 1 A factor has emploees and makes a profit of P rand per week. The relationship between the profit and number of emploees is epressed in the formula P() = , where is the number of emploees Calculate: 1. The number of emploees needed for the factor to make a maimum profit. First we need to find the derivative of P() P () = Net, to find the maimum profit, we need to make P () = 0 and then solve for (the number of emploees needed to achieve this profit). P () = 0 0 = = 600 = 100 = ±10 Remember that (10) = ( 10), so we must include both the positive and negative answers But it is not possible to have a negative number of people! = 10. The maimum profit Now, all we have to do is substitute = 10 back into the formula for P() P(10) = (10) (10) P(10) = (1000) P(10) = P(10) = 5000 Ma Profit is R5000 Eample You have bought 16m of fencing and would like to use it to enclose a field. If length = h and width =, find h in terms of. What is the maimum area possible for the field? h Perimeter=( length) + ( breadth) so we can use the 16 m to get a first equation. 16 = h + 8 = h + Page 108

22 h = 8 Now that we have epressed h in terms of, we can get an equation for the area that does not contain h. When we form the equation that must be differentiated we must alwas epress it in terms of one unknown (usuall ). Area = length breadth (this is knowledge ou have from grade 9 and 10) = h but we know h = 8 = (8 ) = 8 For maimum area, we need to find the derivative of the area, then make darea _ = 0, and solve for : d darea _ = 8 Now, make this = 0 d 0 = 8 = 8 = 4 Therefore, for the maimum area the width of the field must be 4m. Now substitute this into the equation for the length : h = 8 h= 4 For the maimum possible area of the field, the field must be a square with both a length and a width of 4 m, ie the area = 16 m. Activit 6 Activit 1. The sum of one number and three times another number is 1. Find the two numbers if the sum of their squares is a minimum, and give this minimum value.. A rectangular bo without a lid is made of card board of negligible thickness. The sides of the base are cm b 3 cm, and the height is h cm..1 If the total eternal surface area is 00 cm, show that h = _ 0 _ 3 5. Find the dimensions of the bo for which the volume is a maimum 3. ABCD is a square of side 30 cm. PQRS is a rectangle drawn within the square so that BP = BQ = DS = DR = cm. A P B 3.1 Prove that the area (A) of PQRS is given b A = 60 cm Q 3. What is the maimum possible area of PQRS? S D R C 009 Page 1 Lesson 1 Algebra Page 109

23 Rates of Change Real Life Application In an applications that rel on the speed of cars, planes, or even the speed a ball travels through the air, rates of change can be used to calculate the velocit (rate of change of displacement with time) and acceleration (rate of change of velocit with time) of a moving object. So net time ou re watching the F1, motorbike racing, or even our favourite sportsman or woman in action, ou re watching calculus in motion. Velocit/speed = the derivative of displacement/distance with respect to time Acceleration = the derivative of velocit/speed with respect to time Eample Solution Eample Eample 1 A stone is dropped from the top of a tower. The distance, S metres, which it has fallen after t seconds is given b S = 5t. Calculate the average speed of the stone between t = and t = 4 seconds. Solution: As the question asked for average speed, we need to find the average gradient of the curve defined b S = 5t. So first we need to find the value of S at the two points t = and t = 4: S = 5() S = 5(4) = 5(4) = 5(16) = 0 = 80 S(4) S() Average speed = m/s = _ 4 m/s = _ 60 = 30m/s Eample An object is projected verticall upwards from the ground and its movement is described b S(t) = 11t 16t, where S = distance from the starting point in metres and t = time in seconds. 1. Determine its initial velocit The initial velocit of the object is its velocit at t = 0. This is also the rate of change of distance at t = 0. Therefore we need to find the formula for the derivative of S(t) and evaluate it at t = 0: v(t) = ds _ = 11 3t. Remember velocit is the derivative of distance/ dt displacement. v(0) = 11 3(0) substitute = 0 =11 m/s this is the initial velocit Page 110

24 . Determine its velocit after 3 seconds Now we substitute t = 3 into the formula we have found for the velocit v(3) = 11 3(3) = = 16 m/s this is the velocit after 3 seconds 3. Find the maimum height the object reaches This will be the maimum displacement of the object which, from maima and minima, we can find b putting s (t) = 0 and solving for t: S (t) = 11 3t 11 3t = 0 put equal to zero to find t at maimums 11 = 3t t = _ 11 3 = 7_ = 3.5 seconds S (_ 7 ) = 11 (_ 7 ) 16 (_ 7 ) substitute t = 3.5 into S(t) to find maimum height = = 196 It reaches a height of 196m 4. When will it be more than 96m above the ground? Now we need to create an inequalit for S(t)>96m and solve for t: 11t 16t > 96 0 > 16t 11t + 96 move everthing to the same side of the inequalit sign t 7t + 6 < 0 factorise and simplif (t 6)(t 1) < < t < 6 It is above 96 m from 1 to 6 seconds. 5. What is the acceleration of the object? Acceleration is the change in velocit with respect to time. So we need to find the derivative of the velocit: v(t) = 11 3t a(t) = v (t) a(t) = 3 m.s - Activit 7 Activit 1. A large tank is being filled with water through certain pipes. The volume of water in the tank at an given time is given b V(t) = t t, where the volume is measured in cubic metres and the time in minutes. 1.1 Find the rate at which the volume is increasing when t = 1 minute 009 Page 1 Lesson 1 Algebra Page 111

25 1. At what time does the water start running out of the tank?. A cricket ball is thrown verticall up into the air. After seconds, its height is metres where = Determine:.1 the velocit of the ball after 3 seconds. the maimum height reached b the ball..3 the acceleration of the ball..4 the total distance travelled b the ball when it returns to the ground. Activit Activit 8 1. Differentiate the following with respect to : 1.1 ( _ 1 )( + _ 1 ) _ _ 4. Determine: lim ( ) _ 3. If =, determine _ d. Give the answer with positive eponents. d 4. Evaluate: lim _ Determine ƒ () from first principles if ƒ() = 3 6. Find the average gradient of g() between = 1 and = 3 if g() = 7. Determine _ d d if: 7.1 = ( 3 1) _ 7. = _ = ( 1 ) = 7.5 = If ƒ() = _ + _ +, find ƒ () and hence evaluate ƒ (-) Page 11

26 Activit 9 1. The figure shows the curve given b ƒ() = a Point P ( ; 9) lies on the curve of ƒ(). Activit P(; 9) f() 1.1 Determine the value of a. 1. Determine the equation of the tangent at point P. 1.3 Determine the average gradient of the curve between = and = 5. Given: ƒ () = 0 at = 3 and = 1 onl and that ƒ( 5) = 0; ƒ( ) = 0; ƒ(1) = 0; ƒ(0) = 1. Draw a rough sketch of a graph of ƒ() assuming it is a cubic polnomial. 3. ƒ() = b + c + d. A and B are turning points with B(6; 0) 0 B(6; 0) A T f() Find: 3.1 the values of b, c and d. 3. the coordinates of A 3.3 T has an -coordinates of 3, find the -coordinate of T 3.4 Equation of the tangent to the curve at T 4.1 Show that the gradient of the tangent at (; 4) to the curve = _ 4 is double that of the gradient at the point (4; 7) 4. Determine the co-ordinates of the point on the curve at which the gradient is Page 1 Lesson 1 Algebra Page 113

27 Solutions to Activities Activit lim ( + 1) = _ 4 Activit f(6) 8(0) 1.1 gradient = = _ = 6 1. = 3: f (n) = lim f(3 + h) f(3) h 0 h lim (3 + h) + 9 ( 3) + 9) h 0 h lim 9 6h h h 0 h lim h( 6 h) h 0 h 6 0 = 6 (.1 ƒ () = lim + h) + ( + ) h 0 h lim + h + h + h 0 h lim _ h + h h 0 h lim h( + h) h 0 h lim ( + h) = h 0. ƒ () = lim 3( + h) + 3( + h) - (3 + 3) h 0 h 3 lim + 6h + 3h h 3 3 h 0 h lim h(6 + 3h + 3) h 0 h (6 + 3h + 3) = lim h 0 Activt 3 1. ƒ () = _ 1 _ 1 _ 1 _ 3. = 3 ( ) 3. Activit _ d d = = + 1 d _ d = 1 1. f() = + 1 into: (0; 1) f () = gradient of tangent : f (0) = (0) = tangent: = c (sub(0; 1)) = n + 1 Page 114. ƒ() = - + +

28 ƒ () = + 1 at = 1: ƒ (1) = (1) + 1 = 1 \ m of tangent = 1 \ tangent: = + c sub (1 ; ) = 1 + c \ 3 = c \ tangent: = = is a tangent to ƒ() = \ m = 11 and gradient: ƒ () = 4 5 \ 4 5 = 11 4 = 16 \ = 4 to find, sub = -4 into either equation: = 11( 4) + 44 or ƒ( 4) = ( 4) 5( 4) + 1 = 0 = 0 \ point of contact: (-4 ; 0) 4. ƒ() = _ 3 = 3 1 ƒ () = 3 = 3 _ at = -1: f ( 1) = 3 _ ( 1) = 3 _ = 3 1 \ m of tangent = -3 to find the -value: f( 1) = _ 3 1 = 3 \ point of contact: ( 1 ; 3) \ tangent: = 3 + c sub ( 1 ; 3) \ 3 = 3( 1) + c 6 = c \ tangent: = 3 6 Activit 5 1. Finding -intercepts of g(): ( = 0) \ = 0 ( + 1)( 1)( + 1) = 0 \ = 1 or = 1 \ B( 1; 0) and A(1; 0) -intercept of g(): 1 \ C(0; 1) Finding -intercept of f(): ( = 0) 4 + = 0 \ = 1 \ D ( 1 ; 0 ) 009 Page 1 Lesson 1 Algebra Page 115

29 1.1 DA = 1 + _ 1 = 3 OB = 1 CF = 1 + = 3 1. g () = = 0 (3 1)( + 1) = 0 \ = _ 1 3 or = 1 At E, = 1 \ = g 3 ( 1 3) = 3 7 \ E ( 1 3 ; 3 7 ) 1.3 Gradient of f() = 4 (from m = 4) \ g () = = 4 \ = 0 (3 + 5)( 1) = 0 = _ 5 or = 1 (values of for which the gradient of f() and g() are 3 equal.).1 f () = = 0 =,577 or = 1,4 \ =,6 or = 1,4 \ f(,6) = 0,4 f(1,4) = (0,4) (,6; 0,4) (1,4; 0,4). ( 1)( )( 3) = 0 (-intercepts).3 \ = 1 or = or = (1,4; 0,4) ( ; 0) (,6; 0,4) Page = 1; m = = ; m = 1 = 3; m =.5 = _ 1 + c (sub(1; 0)) Note: a normal is a line to the tangent m of tangent = m of normal = _ 1 = _ 1 + _ 1 3. = a ( + 1)( + 1)( 3) sub(0; 9) 9 = a(1)(1)( 3) 9 = 3a a = 3 = 3( + 1)( + 1)( 3) =

30 = No stationar points since derivative = 0 has no solution. However the second derivative equal to zero does have an answer: = 0 so = _ 9 ; = _ Activit 6 1. The numbers are _ 18 5 and 6_ 5 ; the minimum is 14 _ 5. Dimensions are _ 0 cm; 4cm; 10cm 3 _ 3.1 Using Pthag: SR = Note: RC = QC = 30 QR = (30 ) + (30 ) (30 ) _ QR = (30 ) Area PQRS = ( n) ( (30 )) A = (30 ) A = 60 n cm 3. ma Area: A = 60 4 = 0 = 15 ma area A(15) = 60(15) (5) 450 cm Activit Rate of increase: V (t) = 10 t at t = 1: V (1) = 8m 3 /minute 1. ma volume: V (t) = 10 t = 0 t = 5 min.1 velocit = _ d d _ d = d = 3: velocit = = 0 m/s. ma height: _ d = = 0 d = 5 Ma height at = 5: = 50(5) 5(5) = 15m.3 Acceleration = _ d = 10 m/s d.4 Double the maimum = 50 m Activit _ 1 = \ derivative = + _ 3 ( )( 1. 1) = ( ) 1 \ derivative = 1.3 derivative = _ \ derivative = _ 1 = _ _ lim. ( )( + 1) ( ) 009 = lim 1_ ( + 1) = ( + 1) = 6 Page 1 Lesson 1 Algebra Page 117

31 _ 3. = Make the subject of the formula. = _ _ \ = 1 1_ \_ d d = _ _ 1 1_ 4. lim ( )( + + 4) = + () + 4 = 3 ( )( + ) + f( + h) f() = lim 3( + h) ( 3 ) h 0 h h 0 h 3 6h 3h + 3 h 0 h 5. f () = lim = lim = 6 3(0) = 6 g(3) g(1) 6. m = = _ 16 0 = = \_ d d = = + 3 = + 1_ \_ d d = + _ 1 1_ 7.3 = + \_ d d = _ = _ 8 = = _ 1 _ + 1 \_ d d = _ 8 = 1 + \_ d d = _ + _ 3 _ 3 5_ 8. ƒ() = _ \ f () = _ \ ƒ ( ) = _ = _ _ ƒ () = 7 3_ _ Activit sub (; 9) : 9 = a() \ a = 1 1. gradient at = : f () = 3 \ f () = 1 \ tangent: = 1 + c sub (; 9) \ c = 15 and = 1 15 f(5) f() 1.3 m = = _ = 39. Note: f ( 3) = 0 and f ( 1) = 0 \ these are stationar points. f( 5) = 0; f( ) = 0; f(1) = 0 \ these are the -intercepts f(0) = 1 ie. 1 is the -intercept value Page 118

32 3. d = 0 (graph passes through the origin) 3.1 to find b and c ( unknowns) we solve simultaneousl. from B f(6) = 0 \ (6) 3 + 6(6) + c(6) + 0 = 0 6b + c = 36 \ c = 36 6b since B is a stationar point f (6) = 0 : f () = 3 + b + c = 0 f (6) = 3(6) + b(6) + c = 0 (from eqn 1: c = 36 6b) \ b + c = 0 6b = 7 b = 1 \ b b = 0 \ c = eqn: = \_ d d = at A (stationar point) : = = 0 ( 6)( ) = 0 \ A = f() = 3 \ A(; 3) 3.3 f(3) = 7 \ T = m of tangent: f (3) = 3(3) + 4(3) 36 = 9 \ = 9 + c sub T(3; 7) \ C = 54 \ = _ d d = 3 _ 1 gradient at = : 3 _ 1 () = gradient at = 4: 3 _ 1 (4) = 1 \ gradient at (; 4) is double that at point (4; 7) 4.1 (8; 7) 6.1 OP = units; OQ = 1 unit; OR = 6 units; ON = 1 units 6. S(4; 36) 6.3 OT = 30 units; TN = 3 units 6.4 = Z(-1; 14) 7. p = _ (1; s7) 009 Page 1 Lesson 1 Algebra Page 119

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