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by: CMDChiimeh

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# MATH 1151 EXAM I - Study Guide Math 1151

Marketplace > Ohio State University > Mathematics (M) > Math 1151 > MATH 1151 EXAM I Study Guide
CMDChiimeh
OSU
GPA 2.71
Calculus I
Dr. Bobby Ramsey

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Functions, Limits, Laws and Derivatives
COURSE
Calculus I
PROF.
Dr. Bobby Ramsey
TYPE
Study Guide
PAGES
4
WORDS
CONCEPTS
Calc, functions
KARMA
50 ?

## Popular in Mathematics (M)

This 4 page Study Guide was uploaded by CMDChiimeh on Friday January 30, 2015. The Study Guide belongs to Math 1151 at Ohio State University taught by Dr. Bobby Ramsey in Spring2015. Since its upload, it has received 189 views. For similar materials see Calculus I in Mathematics (M) at Ohio State University.

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Date Created: 01/30/15
SPRING 2015 MIDTERM 1 STUDY GUIDE MATH 1151 3 FEBRUARY CALCULUS I INTRODUCTION TO FUNCTIONS Transformations of Functions Vertical Shifts y fx Suppose c gt 0 I y fx I c shifts the graph of y fx a distance of c units I I y fx I c shifts the graph of y fx a distance of c units Horizontal Shifts x Suppose c gt 0 I y fx I c shifts the graph of y fx a distance of c units to the I y fx I c shifts the graph of y fx a distance of c units to the Vertical Stretches y fx Suppose c gt1 I y fx I c the graph ofy fx by a factor of c units I y fx Ic the graph of y fx by a factor of c units Horizontal Stretches x Suppose c gt1 I y fx Ic the graph ofy fx by a factor of c units I y fx I c the graph ofy fx by a factor of c units Graph Re ections I y Ifx re ects the graph of y fx about the I y fIx re ects the graph of y fx about the Types of Functions I Polynomial 0 Linear 9PX mx b o Quadratics always a parabola 9Px ax2 bx c o Cubics 9Px ax3 bx2 cx d I Power 0 fx xa where a is a constant I Exponentials 0 fx as where base a is a positive constant I Rationals O ffx PX QX I Algebraics SPRING 2015 MIDTERM 1 STUDY GUIDE MATH 1151 3 FEBRUARY CALCULUS 0 fx x2 1 I Logarithms 0 fx logax I Trigonometrics 0 fx sin x IDEA OF LIMITS Lim Ha fx L This means that fx quotgoes to 9 L as X quotgoes to 9 a I LeftHand Limit Lim x9afx L As x approaches afrom the fx goes to L I RightHand Limit Lim Hal fx L As x approaches afrom the fx goes to L Suppose that c is a constant and limits lim x9e fx and lim x93 gx exist 1 The limit of a sum is the of limits a limxsa W I gm 1imxsa mxn Inmxsa gm 2 The limit of a difference is the of the limits a limxsa fx lgx1 1imxsa mxn Inmxsa gm 3 The limit of a function is the of the function a limxsa Ifx I1imx a fx 4 The limit of a product is the of the limits a 11mm W I gm 1imxsa mxn Inmxsa gm 5 The limit of a quotient is the of the limits only if the denominator 0 a 11mm fx lgx1 1imxsa fxI1imx a gx1 Infinite Limits Letf be a function de ned on an open interval with the number a except at a Then hm xsa fx I w SPRING 2015 MIDTERM 1 STUDY GUIDE MATH 1151 3 FEBRUARY CALCULUS I A whole function is continuous when there aren t any holes jumps or gaps of any kind in its graph Drawing entire graph without pick pencil off page Three Requirements of Continuous Functions 1 fa exist 2 x a CXiSt 3 x a Continuous Functions FUNCTION EXAMPLE WHERE Polynomials Lim xsl 2392 I 3 2I2 I 3 Everywhere 8 239 I Rational Lim x93 x2 9 X 3 Lim x93 x 91 1 idem Lim xam 6 Radicals Limxeo x Nx 9 3 39 SW31 Nx 9 I91 Mx 9 I 9 Nx Everywhere 1n doma1n Limxeo IMx 9 9 I Lim HI MI 9 3 II93 6 I Exponential Limxsoo e4X equot2X j e Ix e2X 3e x Limxsoq 6 e 8 e39 3e 5X Limx l 6 e6 8 e2l 3e5l Everywhere 6 8 39 4 I Logarithmic Limxs Ioga x Everywhere I Trigonometric Limeso 1 cosG 62 Limeso 1 cose GZ1 I cose 1 I cose Limeso 621 cosG Everywhere Limeso in domain Li mes 1 1 cos IE1 1 I Inverse Limxsw cos39jL Everywhere Trigonometric side note amp Limxgoo I in domain cos 1 I G quot3 DEFINING THEOREMS Squeeze Theorem aka sandwichpinching theorem If fx S gx 2 hx when x is near a exceptpossibly at a and lim x9e fx lim x9e hx L then Iim x9e gx L Example Evaluate Lim x90 x2 cos 10x using Squeeze Theorem 1 Determine the function s boundaries SPRING 2015 MIDTERM 1 STUDY GUIDE MATH 1151 3 FEBRUARY CALCULUS Cosine amp Sine oscillate between 1 and 1 so 1s cos 10x 2 1 which means that 1x2 S x2 cos 10x 2 1x2 2 Take the limits of both sides Lim x90 lel S Lim x90 le COS 10x 2 Lim x90 lel 0 S Lim x90 le COS 10x 2 0 So Lim x90 x2 cos 10x must equal 0 Intermediate Value Theorem Suppose that f is continuous on the interval a b and let N be any number between fa and fb where fa fb Then there exist a number c in a b such that fc N DERIVATIVES

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