Calculus - Santowski 10/8/20151Calculus - Santowski

10/8/20152Calculus - Santowski

 NOTE: This lesson is intended as a CONSOLIDATION of a variety of skills/concepts taught before Christmas break  (1) Review fundamental graph features  (2) Review fundamental algebra skills  (3) Apply calculus based concepts of test for concavity, test for increasing and decreasing functions and the concepts of limits to help us sketch graphs and describe the behaviour of functions 10/8/20153Calculus - Santowski

 One simple method that can be used to sketch the graph of a curve is to determine ordered pairs (i.e. in Algebra I,II)  Why do we NOT do this in Calculus?  So we will REVIEW calculus based methods for determining the general shapes/features of curves 10/8/20154Calculus - Santowski

 (a) domain  (b) intercepts  (c) symmetry (even & odd)  (d) asymptotes (we’ll use limits)  (e) intervals of increase/decrease  (f) local extrema  (g) concavity and inflection points 10/8/20155Calculus - Santowski

 This example is intended to be SIMPLE and basic, but illustrate the thinking process involved in developing and presenting a solution  Use Calculus based methods to sketch the curve f(x) = 3x 5 – 5x 3 10/8/2015Calculus - Santowski6

 (a) domain: the domain of all polynomial functions is xεR  (b) intercepts: to find the y-intercept, evaluate f(0)  here f(0) = 0  (b) intercepts: to find the x-intercept(s), solve f(x) = 0  So 0 = x 3 (3 – 5x 2 )  So x = 0 and 3 = 5x2  x = + √(5/3) 10/8/2015Calculus - Santowski7

 (c) symmetry: evaluate f(-x) = 3(-x) 5 – 5(-x) 3  So f(-x) = -(3x 5 –5x 3 ) = -f(x)  odd symmetry  (d) Asymptotes: polynomials have no vertical asymptotes, so we do not need to calculate lim x  c  (d) Asymptotes: polynomials have no horizontal asymptotes, so we do not need to calculate lim x  +∞ 10/8/2015Calculus - Santowski8

 (e) intervals of increase and decrease  d/dx f(x) = 15x 4 – 15x 2 = 15x 2 (x 2 -1) = 15x 2 (x+1)(x- 1) which is differentiable on xεR  So critical points are at f’(x) = 0, so at x = -1,0,1  To find intervals of increase/decrease, we can use a sign chart or the first derivative test  f’(-2) = +180f’(-0.5) = -45/16  f’(0.5) = -45/16f’(2) = +180  Thus f(x) increases on (-∞,-1) and also on (1,∞) and decreases on (-1,1) 10/8/2015Calculus - Santowski9

 (f) extrema: f(x) has: x= -1 is a max, x = 0 is a stationary point, x = 1 is a min  f(-1) = = 2  point (-1,2)  f(0) = 0  point (0,0)  f(1) = 3 – 5 = -2  point (1,-2) 10/8/2015Calculus - Santowski10

 (g) concavity & inflection points:  Here f”(x) = 60x 3 – 30x = 30x(2x 2 -1) which is differentiable on xεR  So IP exist where f”(x)=0, so at x = 0 and x = +√(0.5)  To test for intervals of concavity, we can use a sign chart or the second derivative test  So f”(-1) = -30f”(-0.5) =15/2  So f”(0.5) = -15/2f”(1) = 30  So f(x) is con down on (-∞,-√(0.5)) and on (0,√(0.5)) and con up on (-√(0.5),0) and (√(0.5),∞) 10/8/2015Calculus - Santowski11

 (h) now we sketch the function, paying attention to the x- and y-coordinates of key features 10/8/2015Calculus - Santowski12

We can repeat the process for other functions: 10/8/2015Calculus - Santowski13

 Sketch the graph ofwhere k is any positive integer 10/8/2015Calculus - Santowski14

 In the following exercises, we will NOT use the calculus options from the homescreen, nor will we use any calculus based Math options from the graph menu  We will use calculus to reveal all important aspects of the graph  We will estimate all max/min points and intervals of concavity of f(x) = 2x 6 + 3x 5 + 3x 3 – 2x 2 10/8/2015Calculus - Santowski15

 We will estimate all max/min points and intervals of concavity of f(x) = 2x 6 + 3x 5 + 3x 3 – 2x 2  We will use the window x  ? and y  F2 ZOOMFIT  Here is the graph  Which really doesn’t give us much info other than end behaviour 10/8/2015Calculus - Santowski16

 f(x) = 2x 6 + 3x 5 + 3x 3 – 2x 2  So we will adjust our view window to [-3,3] and [-50,100], we can see some clearer details  The graph is showing  A minimum between -2 & -1 and a horizontal tangent and maybe inflection points say at - 1 and 0 ands /8/2015Calculus - Santowski17

 So, if I cannot use the calculus options on the graphing menu to find max/min and intervals of concavity, what can I do to find the relevant points (critical points & inflection points)  Recall that the max/mins of a function can be found using the zeroes of the derivative  Recall that the max/mins of the derivative function can be found using the zeroes of the second derivative 10/8/2015Calculus - Santowski18

First Derivative GraphsSecond Derivative graph 10/8/2015Calculus - Santowski19

 So let’s use the graph to find the various zeroes of the first and second derivative graphs  f’(x) = 0 when x = -1.62, 0, 0.35  f”(x) = 0 when x = -1.23, /8/2015Calculus - Santowski20

 So our conclusions would be that f(x) has an absolute minimum at x = -1.62, a local max at x = 0 and a local min at x = 0.35  The inflection points are at x = and x = 0.19  The intervals of concavity are then [-∞,-1.23] and [0.19,∞] for con up and [-1.23,0.19[ for con down 10/8/2015Calculus - Santowski21

 Repeat the same graphic based exercise for the following two functions: 10/8/2015Calculus - Santowski22

 How does the graph of vary as the value of c changes?  Develop your solution using appropriate algebraic and graphic evidences.  (Consider relevant graphic features like max/min, inflection points, discontinuities) 10/8/2015Calculus - Santowski23

 Handout #1(S5.5, Q1-12; S7.2, Q6; S8.2, Q12; S8.4, Q9)  Handout #2 (p )  Q1-10 & Q /8/2015Calculus - Santowski24