Further Differentiation

Transcription

1 Worksheet 39 Further Differentiation Section Discriminant Recall that the epression a + b + c is called a quadratic, or a polnomial of degree The graph of a quadratic is called a parabola, and looks like one of the following: The are smmetrical about a stationar point which is either a local minimum or maimum Parabolas do not have points of inflection In the quadratic = a + b + c, if the co-efficient of is greater than zero, the parabola is concave up; if a is negative, the parabola is concave down The option a = 0 is precluded as this would result in a linear polnomial which is a straight line when graphed The quadratic formula found b solving a + b + c = 0 is given b = b ± b 4ac The a value under the square root sign, b 4ac, is called the discriminant, and we denote this b We can tell quite a lot about the curve = a + b + c just b evaluating = b 4ac If > 0 If = 0 If < 0 then the graph will cut the -ais in two places, ie there are two -intercepts the graph will touch the -ais in one place onl the graph will sit wholl above or below the -ais depending on the sign of a There are no -intercepts Using this information and the information gained from the derivative, we can sketch the graph of an quadratic Eample : Sketch the graph of = We have a =, b = 3, and c = Since a > 0, the stationar point is a minimum Further, b 4ac > 0

2 so there are two -intercepts, which are found in this case through factorization: = ( + )( + ) So = 0 when ( + )( + ) = 0, or when = or We find where the stationar point is b setting d = 0 This gives d d d = + 3 = 0 which has the solution = 3 When = 3, = ( 3 ) + 3( 3 ) + = 4 The stationar point is at ( 3, ) When = 0, =, so the -intercept 4 is With all this information at our disposal, we can now draw the graph of = : 4 Eample : Sketch = + This can be rewritten as = + so that a =, b =, and c = We have a < 0 so the stationar point is a maimum We find the -coordinate of the stationar point b setting d d = 0 d d = + = 0 which has the solution =, and the corresponding value is = ( ) + = 7 4 The stationar point has the coordinates (, 7 ) The discriminant is 4 b 4ac = 7 < 0 so there are no -intercepts The -intercept is given when = 0, so = The graph of = + then looks like:

3 Eercises: Sketch the graph of each of the following using the method studied in section (a) = (b) = (c) = (d) = 6 (e) = 5 3 Section Second Derivatives Recall from the last worksheet the discussion on concavit To determine whether a stationar point was a maimum, minimum, or a point of inflection, we looked at the changes in slope as we moved from one side of the stationar point to the other There is an easier method of determining the concavit of a graph at an point (not just a critical point) The method is based upon the notion that concavit is a measure of the change of a slope As we viewed the derivative as a measure in the change of the height of a function we can view the derivative of the derivative as a change in the slope of the graph The derivative of the derivative is called the second derivative, and it is denoted b one of the following: f (),, depending on how the function is defined d d, or D (f) Eample : Find the second derivative of = d d = 0 d d = 0 So the second derivative of = is 0 Eample : Find the second derivative of f() = f () = f () =

4 The second derivative is a help with curve sketching as it tells about the concavit of the graph at an point - this is most useful at critical points If f () = 0, the concavit is changing, so that the critical point is a point of inflection If f () > 0, then the graph is concave up at If f () < 0, then the graph is concave down at This means () If f () = 0 and f () > 0 there is a minimum turning point at () If f () = 0 and f () < 0 there is a maimum turning point at Eample 3 : For the function f() = , find the stationar points and describe their important characteristics We first find the solutions of f () = 0 f () = = 3( + + ) = 3( + )( + ) = 0 which has the solution = Also f( ) = ( ) 3 + 3( ) + 3( ) = So the critical point is at (, ) Now, f () = At =, f ( ) = 6( ) + 6 = 0 Therefore the concavit is changing at =, so the point (, ) is a point of inflection Eample 4 : Find the stationar points of the function f() = 4 +, and describe their properties We first find solutions of f () = 0 f () = 4 3 = ( ) When f () = 0, ( ) = 0 which has the solutions = 0 and = ± These are the -coordinates of the critical points The second derivative is f () = At = 0, f (0) = < 0 so at = 0 we have a local maimum At =, f ( ) = 4 > 0 so at = 0 we have a local minimum At =, f ( ) = 4 > 0 so at = 0 we have a local minimum 4

5 Eercises: Find the stationar points of each of the following and describe their important characteristics (a) f() = 3 4 (b) f() = (c) f() = (d) f() = (e) f() = Section 3 Further Sketching We now have a comprehensive range of tools that help us sketch curves, especiall those of polnomial functions We can find intercepts, eamine what happens to the function at certain values of, find critical points, and find properties of critical points Let s put these tools to use Eample : From the information in eample 4 in section, sketch the function = f() = 4 + The critical points were at = 0, and at = ± At = 0, f(0) =, and this is a local maimum At =, f() = 3 4 At =, f() = 3 4 and this point was a local minima and this point was a local minima The -intercept is (0, ) We leave the question of the -intercepts for the moment As ±, f() From all this information, we can now draw some of f() = 4 + : 3 4 5

6 Since the three critical points shown are all the critical points, there can be no other changes in direction, so the graph can be completed as follows: (, 3 4 ) (, 3 4 ) Eample : In section, eample 3, we looked at the function f() = Sketch this function There was onl one critical point, (, ), and it was a point of inflection The -intercept is found b letting = 0, which gives f(0) = 0 The -ais intercepts are given b the solutions to f() = 0: = ( ) = 0 So one solution is = 0 The solutions to = 0 are given b = 3 ± 9 = 3 ± 3 which has no real solutions So = 0 is the onl -intercept As, f() and as, f() The graph of f() = then looks like: - Eample 3 : Sketch the function f() = f () = 3 which is zero when either = 0 or = ± f () = 6 We have f (0) = 0 f () = minimum at (, 3) f (0) = 0 maimum at (, 9) There is an inflection point at (0, 3) The graph of f() is shown: 6

7 Eercises: Using the method outlined in section 3, sketch the following curves (a) f() = (b) f() = + 8 (c) f() = (d) f() = (e) f() =

8 Eercises 39 Further Differentiation For each of the following quadratic functions, find the sign of the discriminant to determine if there are 0,, or roots (a) f() = (b) f() = 4 + (c) f() = 4 (d) f() = (e) f() = For each of the following, sketch a function which satisfies the given conditions (a) f () > 0 and f () > 0 (b) f () > 0 and f () < 0 (c) f () < 0 and f () > 0 (d) f () < 0 and f () < 0 (e) f() = 4, f () = 0, f () < 0 for <, and f () > 0 for > 3 (a) Sketch the function f() = , labelling all intercepts and critical points Determine the nature of the critical points (b) An economist stated: Although our current account deficit is increasing, the government policies to reduce it seem to be taking effect If D represents the current account deficit, and t represents time, what can be determined about dd and d D? dt dt 8

9 Answers 39 (a) > 0, roots (b) < 0, no roots (c) > 0, roots (d) = 0, root (e) < 0, no roots (a) Concave up and increasing (b) Concave down and increasing (c) Concave up and decreasing (d) Concave down and decreasing (e) Change of concavit at (, 4) 4 3 (a) f() = ( 3) ( + 3) Intercepts at = 3, 3 f () = 3( 3)( + ) Stationar points at = 3, f () = 6 6 f (3) > 0 so there is a minimum point at (3, 0) f ( ) < 0 so there is a maimum point at (, 3) 3 4 (b) dd dt > 0 and d D dt < 0 9