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Calculus and Analytic Geometry I

by: Theresia Hyatt DDS

Calculus and Analytic Geometry I MATH 151

Marketplace > Ohio State University > Mathematics (M) > MATH 151 > Calculus and Analytic Geometry I
Theresia Hyatt DDS
GPA 3.8


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This 6 page Class Notes was uploaded by Theresia Hyatt DDS on Monday September 21, 2015. The Class Notes belongs to MATH 151 at Ohio State University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/209943/math-151-ohio-state-university in Mathematics (M) at Ohio State University.


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Date Created: 09/21/15
The quotSqueezequot Theorem If f x S gx S hx for any x in an interval around the point a and if lim fx lim hx L then we may assume that limgx L sin x For example let fx l E and gx and hx l x sin x In this case f x S gx S hx Suppose you are asked to evaluate ling We are unable to H x sin x evaluate lim using any known means So we will nd liml E l and liml l xgt0 xgt0 xgt0 x smx l as well x According to squeeze theorem now we claim that lim Xgt0 Compute the following limits if they exist CAUTION AVOID 6 1 Equot 5 gt P lim Limits 0 AND EFORMS 00 lim 2x 11 Direct substitution Answer 15 xgt72 2 11m x2 4 Direct substitution results in the 00 form So factor the denominator rail x cancel x 2 substitute Answer 14 lim xx 2 Answer DNE why lim xx 2 Answer 0 rgt2 x 1 2 The denominator gets smaller and smaller in this case so the rational expression will get bigger and bigger Also visualize the graph Answer oo lim x 2 Answer 00 see 5 HP x 2 x 3 4 The denominator gets bigger in this case so the rational expression will get smaller and smaller Also visualize the graph Answer 0 0 to be speci c 3 11m 4 The denominator gets bigger in this case so the rational expression will xgtroo x get smaller and smaller Also visualize the graph Answer 0 0 to be speci c lim 3x4 Caution Eform Divide both the numerator and denominator by x 00 xgtoox You will get limi4 Think about it 1 7 x lim xgt7oo 2 3 5 2 11 Caution Eform Try dividing both the nr and the dr by x3 7 so 5 11 2 i 73 2 You will get lim x 7x approaches 6which approaches 00 xx3 5x2 11 Hi 2x3 6x2 7 denominator by x3 as in the previous problem Observe what happens Did it help Caution Eform Try dividing both the numerator and 00 Continue Continuity Find the values at which the following functions are discontinuous if any 2x 12 fx xx24 1 13 x g x211 14 hx3x 7 15 fxx2 9 2 16 7 f X Ti 16 Find the points at which the following function tx is discontinuous In each case explain the reason Also nd the points at which tx is left continuous and light continuous 17 The symbol 7 Leibniz Consider the functiony 3x4t l lxzts 2t2 In this case fix represents the derivative of the function 2 with respect to the variable g rate of change of y with respect to X Note that the notation diy does not represent division While nding wheny 3x4t l lxzt5 2t2 you must treat all letters except for x as dy constants Thus 67 3 4x3 t l l 2xt5 0 12x3t 22m5 x How about Tywheny 3x4t llxzts 2t2 1 You will be nding the derivative of the function 15 with respect to the variable t During this process x will be treated as a constant 3x4 55x2t4 4t dt d The function whose derivative y is being found dx The variable in the function Sections 41 amp 42 Maximum Minimum occurs at the critical points Critical points include endpoints of the closed interval indicated and points where f1 is zero or undefined lnterval s in which the graph is increasing Give closed interval except at poinU s where the function is undefined Intervals in which the graph is decreasing Give closed interval except at poinU s where the function is undefined Intervals of monotonicity Intervals where the graph is either increasing or decreasing No shift from increasing to decreasing or vice versa Concave up Give open intervals Concave down Give open interval s In ection point Points were the graph of the function shifts from concave up to concave down and vice versa At the possible in ection points f11 will be either 0 or undefined Relation between f and f1 when f in increasing in that interval the slope of the tangents will be positive and therefore f1 gt 0 When f is decreasing in that interval the slope of the tangents will be negative and therefore f1 lt 0 Relation between f and f11 when f is concave up f11gt0 when f is concave down f 11lt 0 Sketching the graph of f from the information regarding f1 and f11 Use f1 to identify intervals of increasingdecreasing behavior Use f11 to identify intervals of concave updown behavior Find in ection points Pay attention to x and y intercepts if any exists


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