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## calculus review for first exam

by: Samantha Notetaker

9

0

8

# calculus review for first exam MA 241

Samantha Notetaker
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basic list of things needed to know for the exam and explanations on how to do it
COURSE
calculus and analytical geometry I
PROF.
Keke Wang
TYPE
Study Guide
PAGES
8
WORDS
CONCEPTS
Math, Calculus, exam, review, study, guide, Derivatives, functions, Rules, Rules of Derivatives, Tangent, Limits
KARMA
50 ?

## Popular in Mathmatics

This 8 page Study Guide was uploaded by Samantha Notetaker on Thursday September 22, 2016. The Study Guide belongs to MA 241 at Embry-Riddle Aeronautical University - Prescott taught by Keke Wang in Fall 2016. Since its upload, it has received 9 views. For similar materials see calculus and analytical geometry I in Mathmatics at Embry-Riddle Aeronautical University - Prescott.

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Date Created: 09/22/16
MA 241 Study Guide Topics or skills to specifically study from homework and quizzes:  State values of the given graph and find domain and range (Homework 1)  Determine if graphs are functions or not (Homework 1)  Find domain with only given equation (Homework 1)  Match graphs with equations and vice versa (Homework 1)  Transitions: up, down, right, left, compress, stretch, reflect over x and y axis    (Homework 2)  Forming the equation given only transitions or graph (Homework 2)  Explain what the equation means (Homework 3)  Finding limit values from functions using limit laws (Homework 3)  Determine graphs that satisfy given conditions (Homework 3)  Adding, dividing, multiplying, subtracting limits (Homework 4)  Evaluate limits (Homework 4)  Identify problems with limit equations (Homework 4)  Find limits (Homework 4)  Discontinuities (Homework 5)  Jumps (Homework 5)  Explain discontinuities (Homework 5)  Removing discontinuities (Homework 5)  Derivate and derivative rules (Homework 6)  Tangent lines (notes) Examples of bulleted list above: (a) State value of f (1): 3 Look at where x=1 and go up to a point where it matches with the y axis (b) State value of f (3): 1 Do the same as part a  (c) State value of f (0): 1 (d) State domain and range of f   Domain: [­2,4] (we use the [ because these numbers are included in the domain and  range, if they weren’t you would use parenthesis)  Range: [­1,3] To find domain look at where the line on the graph starts at the x axis and follow from left to  right where it starts and stops tells you the domain. To find the range you simply start at the  bottom and go up looking at the y axis where its lowest point and highest point tells you the  range. Usually they will ask for it in interval notation. To determine if this graph is a function or not use the pencil rule as described in my first notes. The graph above is a function because it does not touch the y­axis at more than one point at a  time x3 x +2 x−24 f (x) = 5 ­3/  you set the denominator to zero and factor it out which leads you to (x+6) (x­4) because you take the product of the ­24 but its sum has to be equal to 2x. The only numbers that do that are ­6 and  4, so the domain is x is equal to all real numbers but x cannot equal 0. Therefore, the domain is:   (­∞, ­6) U (­6,4) U (4, ∞) The “U” means union which basically means there’s a whole between  those numbers. Matching equations:  6 f (g):  x           this would leave you with the last line in the graph 2 f (h): x    this is a quadratic function which means the graph is a parabola f (f):  x 3 this is a cubic function so it’s basically just a stretched parabola y = f (x) +3 this shifts the graph up 3 or whatever the constant seems to be y = f (x) ­3 this shifts the graph down 3 units  y = f (x+3): shifts 3 units to the left y = f (x­3) shifts 3 units to the right y = 3*f(x) stretches graph vertically by a factor of 3 y = (1/3) *f (x) compresses vertically graph by a factor of 3 y = f (3x) compresses graph horizontally by a factor of 3 y = f (1/3x) stretches horizontally graph by a factor of 3 y = ­f (x) reflects graph over the x axis y = f (­x) reflects graph over the y axis Say you have a transition of 3 units down, the graph is compressed horizontally by a factor of 2  and its reflected over the y axis. Your equation based on rules above: y = (­2x)­3 Equation: ­(3x+3)­2 This equation states that the graph is reflected over the x axis, compressed horizontally by a  factor of 3, shifted left 3 units and shifted down 2 units.  Limit laws: Lim[f(x) + or ­ g (x)] = lim f (x) + or – lim g (x) Lim [c*f(x)] = c* lim f (x) Lim [ f (x) * g (x)] = lim f (x) * g (x)  Lim f(x)/ g (x) = lim f(x)/ lim g (x) (g (x) cannot equal zero) n n Lim [ f(x) ¿  = [ lim f (x) ¿   ( N has to be a positive integer) Lim C = C  Lim  x n  where x approaches a =  a n  (n has to be a positive integer) Lim  √ x  where x approaches a =  √ a  (where n is a positive integer if n is even we assume  that a is greater than 0) n n Lim  √ f (x) =  √ lim ?f (x) (where n is a positive integer 2 of n is even we assume that lim  f(x) is greater or equal to 0)  Direct substitution property: =if f is a polynomial or a rational function and a is in the domain of  f.    (this should be your first result when trying to find a limit) Conjugate property: when the equation involves a root function you can cancel out the root by  multiplying it by its conjugate.  Graphs that satisfy given conditions: 2  f(x) = 4+x if x< ­2,  x  if ­2 ≤ x < 2, 6­x if x ≥ 2    the graph would be: Discontinuities and continuities: The limit of a function as x approaches a can often be found simply by calculating the value of  the function at a. Functions with this property is determined a continuous at a.  Requirements for this property: 1. Lim f(x) exists 2. Lim f(x) = f (a) 3.  f (a) is defined If you fail one of these conditions, then you need to remove that discontinuity by factoring out  the equation. However, you can have infinite discontinuity or a jump discontinuity  The Intermediate Value Theorem: suppose that f is continuous on [a, b], and let N be and number between f (a) and f (b) where f (a) does not equal f (b) then there exists a number c. Tangent lines: The tangent line to the curve y =f(x) at the point (p(a), f (a)) is the line through p with slope     m= lim as x approaches a f(x)­ f(a)/ x­a  provided that this limit exists the equation of the tangent line is y­ f(a)= m(x­a)  Derivatives and rules: The derivative of a f is denoted by f ‘(x) (prime of f)   d/d(x)= (C)=0 x ¿ xn−1  d/d(x)= ( = n  (Power Rule very important)  d/d(x)= (C* f(x)) = c* d/d(x) (f(x))  d/d(x)= [ f(x) +or – g(x)] = d/d(x) (f(x)) +or – d/d(x) (g(x)) The sine and cosine functions:  d/d(x) (sin(x)) = cos (x)  d/d(x) (cos(x)) = ­sin(x) Product Rule: [f(x)+g(x)]’= f’(x)* g(x) + f(x)* g’(x) Quotient rule: 2 [f(x)/g(x)]’ = f’(x)* g (x) – f(x)* g’(x)/  x

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