Calculus - Santowski 10/8/20151Calculus - Santowski.

Calculus - Santowski

04/21/23 1Calculus - 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


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


(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


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) = 3x5 – 5x3

04/21/23Calculus - Santowski 6

(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 = x3(3 – 5x2) So x = 0 and 3 = 5x2 x = + √(5/3)


(c) symmetry: evaluate f(-x) = 3(-x)5 – 5(-x)3

So f(-x) = -(3x5–5x3) = -f(x) odd symmetry

(d) Asymptotes: polynomials have no vertical asymptotes, so we do not need to calculate limxc

(d) Asymptotes: polynomials have no horizontal asymptotes, so we do not need to calculate limx+∞


(e) intervals of increase and decrease d/dx f(x) = 15x4 – 15x2 = 15x2(x2-1) = 15x2(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) = +180 f’(-0.5) = -45/16 f’(0.5) = -45/16 f’(2) = +180

Thus f(x) increases on (-∞,-1) and also on (1,∞) and decreases on (-1,1)


(f) extrema: f(x) has: x= -1 is a max, x = 0 is a stationary point, x = 1 is a min

f(-1) = -3 + 5 = 2 point (-1,2) f(0) = 0 point (0,0) f(1) = 3 – 5 = -2 point (1,-2)


(g) concavity & inflection points: Here f”(x) = 60x3 – 30x = 30x(2x2-1) 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) = -30 f”(-0.5) =15/2 So f”(0.5) = -15/2 f”(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),∞)


(h) now we sketch the function, paying attention to the x- and y-coordinates of key features


We can repeat the process for other functions:


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f (x) =1

1+ x 2

g(x) =x 2

1− x 2

h(x) = x − cos x

k(x) = e−x 2

m(x) = ln x 2 −1( )

Sketch the graph of where k is any positive integer


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f (x) =k − x

k 2 + x 2

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) = 2x6 + 3x5 + 3x3 – 2x2


We will estimate all max/min points and intervals of concavity of f(x) = 2x6 + 3x5 + 3x3 – 2x2

We will use the window x? and yF2 ZOOMFIT

Here is the graph

Which really doesn’t give us much info other than end behaviour


f(x) = 2x6 + 3x5 + 3x3 – 2x2

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 0.5


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


First Derivative Graphs Second Derivative graph


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, 0.19


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 = -1.23 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


Repeat the same graphic based exercise for the following two functions:


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f (x) =x 2 + 7x + 3

x 2

g(x) =x 2 x +1( )

3

x − 2( )2x − 4( )

4

h(x) = sin x + sin2x( ) on 0,π[ ]

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)


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g(x) =1

x 2 + 2x + c

Handout #1(S5.5, Q1-12; S7.2, Q6; S8.2, Q12; S8.4, Q9)

Handout #2 (p300-301) Q1-10 & Q21-25


Calculus - Santowski 10/8/20151Calculus - Santowski.

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