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# Calculus I MAC 2311

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This 7 page Class Notes was uploaded by Connor Abernathy II on Monday October 12, 2015. The Class Notes belongs to MAC 2311 at Florida International University taught by Robert Storfer in Fall. Since its upload, it has received 13 views. For similar materials see /class/221712/mac-2311-florida-international-university in Calculus and Pre Calculus at Florida International University.

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Date Created: 10/12/15

MAC 2311 ANALIZING THE GRAPH OF A FUNCTION y f x 1 Find the domain of the function f Some functions and their domains are given below FUNCTION DOMAIN y 13x Where P The set of Real Numbers is annlvnnmial 00 00 PM lex 0 rx where r Is a Rational Function fx Sin x The set Effoeiljmnbers f x COS x The set of Real Numbers w fxtanx x1 cosx 0 x1 x k g k an integer fx secx colsx x1 cosx 0 x1 x k g k an integer fx cotx 005x xl sinx 0 xl xi kn39 k an integer sinx fxcscx 1 xlsinx 0xlx k7rkaninteger sinx fx 81C The set of Real Numbers 0000 fx1nx The interval0oo f xlnx oo0u0oo fx sin 1 x arcsin x The interval 11 fx tan 1 x arctan x The set of Real Numbers new fx sec391 x arc sec x 00 1 U 1 00 fx V where n is an even The interval 0 00 integer n 2 2 The set of Real Numbers fx where n is an odd waw integer n 2 3 2 Find any intercepts off The Xintercepts are all points 97 for which f 97 0 and the yintercept there will be at most one otherwise you do not have a function is the unique number f 0 If x 0 is not in the domain off then f has no yintercept Functions in the above table with no yintercept include fx csc x fx cot x fx In x fx lnlx andfx sec391 x Two very common functions with no y L x2 intercepts include fx i and fx x 3 Look for any vertical andor horizontal asymptotes of the graph Vertical asymptotes occur where lim f x ioo lim f x ice or lim f x ice and are the lines x c for all such c Horizontal asymptotes are any lines y L for which lim f x L In the case of a rational function where the degree of the numerator is exactly one more than that of the denominator then the function will have the oblique asymptote y ax b a 7i 0 ax b a 7i 0 is the quotient resulting from dividing the numerator by the denominator 4 Using f 39 nd all intervals of increasing and decreasing Locate any critical points stationary points andor points of nondifferentiability 5 Using the rst or second derivative tests nd all relative extrema Remember the existence of a critical point does not guarantee a relative extrema at that point but if f were to have a relative extrema it would do so at a critical point In other words the existence of a critical point is a necessary but not suf cient condition for a function to have a relative extrema 6 Using f quot determine all intervals of concave up and concave down Be sure to nd all in ection points 7 Check for vertical tangent lines and comer points Vertical tangent lines are consequences of the points of non differentiability c f due to in nite behavior of the rst derivative near c If we can write lim f39x 00 or oo then there will be no relative extrema at x c a good example here is fx x 3 at x 0 But if f 39 exhibits in nite behavior near 0 but the onesided limits do not agree that is lim f x oo but lim f x 00 or vica versa then you have a relative extrema at x c and graphically the extrema is a cusp Think of the function f x x 3 x Z at x 0to help with this idea Comer points occur at points of nondifferentiability x cfor which all that can be said is lim f 39x does not exist due to disagreement of onesided limits of f 39 The function 1 x lt 0 then f39x We proved this in class 1 x gt 0 earlier in the semester and is something that should be known Comer points are very good candidates for the occurrence of relative extrema fx M with x 0 is a classic example ofthis If fx Ix MAC 2311 CHAPTER FOUR DIFFERENTIATION FORI1ULAS For the chain rule forms assume it u x is a differentiable function ofx Logarithmic functions 1 fx lnx d 1 BASIC FORM 1n x dx x 2 fx10gb x 1 BASIC FORM ilogb x dx x In b Nate lhaifarmula one is a special case affaimula two Exponential Functions 3 f x e d x x BASIC FORM 6 e dx 4 fx b d x x BASIC FORM b J b lnb dx Nate Ihalfarmula three is a special case affarmulafaur Inverse Trigonom etric Functions 5 fx sin391 x 2 BASIC FORM isin 1 x dx 1 x 6 fx tan391 x d 1 1 BASIC FORM tan x 2 dx 1 x 7 fx sec391 x BASIC FORM sec 1 x 1 llexZ l d 1 du CHAIN RULE FORM 1nu dx u dx d 1 du CHAIN RULE FORM 10gb u dx it In b dx d u u du CHAIN RULE FORM 6 e dx dx CHAIN RULE FORM 1amp3 b 1n b dx dx d 1 du CHAIN RULE FORM s1n u dx 1 2 dx d 1 1 du CHAIN RULE FORM tan u 2 dx 1u dx 1 du 61 1 CHAIN RULE FORM sec u dx lulxju2 1 Also cos391 x sin391 x cot391 x tan391 x and csc391 x lsec391 x MAC 23 l l GRAPHING POLYNOMIALS An nth degree polynomial functionP x anx anilxH ch do n a whole number an at 0 can be graphed using the following steps 1 Find the yintercept P 0 a0 2 Find all the zeros of the polynomial The zeros are the numbers x0 for which Px0 0 Graphically x0 is an x intercept of P Indicate the zeros on the your graph 3 Use the multiplicity of x0 to decide whether the graph of P touches or crosses the xaxis at x0 4 Determine the intervals of increasing and decreasing of P and nd all the relative extremum using the First or Second Derivative Tests Indicate the relative extremum on the graph Since polynomials have no points of nondifferentiability then any relative extrema will occur at stationary points 5 Determine the intervals of concave up and concave down Find and label any in ection points 6 Extend your graph to the left and right by nding the end behavior of P The end behavior is completely determined by lim anxquot and lim anxquot NOTES 1 Polynomials are defined continuous and in nitely differentiable on the Set of Real Numbers 00 co In nite differentiability implies the other two features 2 Polynomials have no vertical asymptotes This is because polynomials are continuous everywhere Also non constant polynomials have no horizontal asymptotes because their limits at ioo do not exist as nite values 3 Please keep in mind that if a function has a relative extremum then the extremum will occur at a critical point But the converse of this statement is not true In other words being a critical point is necessary but not suf cient condition for en extremum to occur Existence of a critical point does not guarantee you will have an extremum there A classic example ofthis is the function fx x3 f x 3x2 3 x 0 is a critical point but no sign change of the derivative occurs at x 0 Therefore there is no relative maximum or minimum off at x 0 or anywhere else because x 0 is the only critical point In fact fx xquot is increasing on oooowhy 4 Since the derivative of an nth degree polynomial is a polynomial of degree n l and the maximum number of zeros a polynomial can have is equal to its degree then a polynomial of degree 71 cannot have more than n 1 critical points We can now conclude that an nth degree polynomial can have at most n l relative extrema why 5 Using similar reasoning as in 4 above no polynomial can have more than n 2 in ection points MAC 2311 GRAPHING RATIONAL FUNCTIONS P x A Rational Function Rx where Q is not the zero polynomial can be graphed using the following steps Q06 1 Find the domain of R The domain of R is the set of all real numbers for which Qx 0 2 If x 0 belongs to the domain of R then nd the yintercept R0 If x is not in the domain of R then the graph of l x R has no yintercept eg the function f given by f x 2 has no yintercept x x 3 Find all the zeros of the polynomial The zeros are the numbers x0 in the domain of R for which the numeratorP x0 0 Graphically x0 is an Xintercept of P Indicate the zeros on your graph 4 Find and draw any horizontal asymptotes you will have at most one on your graph The horizontal asymptote is the line y L for which lim L If the degree of the numeratorP is greater than that of the denominator Q then there is no horizontal asymptote why 5 Find and indicate any points where the graph crosses the horizontal asymptote if there is such a line 6 Find and then draw all vertical asymptotes Remember if both P a and Qa are zero then x a is not a vertical asymptote Rather a is a removable discontinuity To account for the removable discontinuity plot a hole where the point aRa would be as if the function was continuous there For example x 4 is a removable 416 x 4x4 2 l6 2 d1scont1nu1ty ofthe function rx x 4 For x at 4 rx x 4 4 x 4 That makes your x x x graph of r be the line y x 4 with a hole at the point 4 8 7 Determine the intervals of increasing and decreasing of R and find all the relative extremum using the First or Second Derivative Tests Indicate the relative extremum on the graph 8 Determine the intervals of concave up and concave down Find and label any in ection points 9 In the special situation where the degree of the numerator is one more than that of the denominator your graph will have an oblique asymptote y ax b where ax b is the quotient that results from dividing P by Q Find draw and label the oblique asymptote NOTE An oblique asymptote is a special case of the general polynomial asymptote that occurs when the degree of the numerator P which we will call n is larger than that of Q which we will refer to as m When that is the case your graph will approach a polynomial of degree n m This is because of the Division Algorithm which tells us that Px Rx S x where S 1s a polynomial of degree n m and the degree of R 1s less than that Q Q06 Qx Theniigggg xliIEoSx2EZJLIELSXX1LIILE and gt0 asx gtioo

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