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## Exam 3 Study Guide

by: Rachel McCord

599

1

3

# Exam 3 Study Guide M119

Marketplace > Indiana University > Mathematics (M) > M119 > Exam 3 Study Guide
Rachel McCord
IU
GPA 3.8
Brief Survey of Calculus
William Orrick

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This study guide contains all the notes and important definitions and equations that will be on Exam 3 (from Chapter 4 and 5). It covers global and local minima and maxima, as well as marginal an...
COURSE
Brief Survey of Calculus
PROF.
William Orrick
TYPE
Study Guide
PAGES
3
WORDS
KARMA
50 ?

## Popular in Mathematics (M)

This 3 page Study Guide was uploaded by Rachel McCord on Wednesday April 8, 2015. The Study Guide belongs to M119 at Indiana University taught by William Orrick in Spring2015. Since its upload, it has received 599 views. For similar materials see Brief Survey of Calculus in Mathematics (M) at Indiana University.

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Date Created: 04/08/15
M119 Exam 3 Study Guide 1 10 11 12 A function has a global maximum at xp if fpgtfx for all x in the domain of fx the highest point of the function A function has a global minimum at xp if fpltF9x for all x in the domain of fx lowest point of function A function has a local minimum at xp if Fpltfx for all x in the some interval containing p Same rules apply for local maximum but reversed Local and Global maxmin can be turning points A critical point of a function is a point p in the domain where the derivative of the function is either zero0 or undefined The yvalue fp is called the critical value A function x is increasing on an interval when its values increase as x increases as function decreases in a similar manner Local max occurs where the function changes form increasing to decreasing Local min occurs where function changes from decreasing to increasing Another way to think about it is the CHANGE OF SIGNS First Derivative Test fx changed from increasing to decreasing means local max fx changes from decreasing to increasing means local min Steps are a Compute f x b Locate critical points by setting f x0 c Determine where f x changes sign i Graph ii Small intervals Second Derivative Test Function has horizontal tangent line at p and is concave up is a local min Function horizontal tangent and concave down is local max Steps are Compute f x and f x Find critical points Plug critical points into f x Negative numbers downward concavity local max e Positive numbers upward concavity local min Remember equotax NEVER touches zero never touches xaxis Inflection Points fx has an inflection point at xp if f changes from concave up to concave down or from concave down to concave up at p a The point where the slope peaks Fx is concave up on interval if f x is positive F9XO is concave down on interval is f x is negative Inflection point is where concavity changes Points are potential inflection points until they are tested to make sure they are inflection points The Surge Function Exercises such as fxxequotx For xgt0 the function rises rapidly to a peak and then decays essentially exponentially Can be used to describe concentration of a drug in the bloodstream as a function of time since the drug was administered A closed interval is one such as 25 An open interval is one such as 25 Finding global min and max is easiest on a closed interval a An open interval uses circles to indicate that it does not include endpoints Bound is unattainable goo9 0903 13 14 15 16 17 18 19 20 21 22 b A continuous function on a closed interval MUST have a global max and min The global maxmin can only be at a critical point or an endpoint All we have to do is plug the x coordinates of critical points and endpoints and see which give the greatest and least values c In an open interval there can be no global max because it is bounded above xp Profit equals revenue minus cost Represented by 7tq The marginal profit is 7t q R q C q or MCMR The Logistic Function Exponential Finite carrying capacity there is a limit As it reaches the carrying capacity growth will begin to slow and then almost come to a halt a Pt where LC and K are constants The variable t is usually time b If you change C only the shape of the graph doesn t change it will only shift the graph horizontally c The inflection point occurs when Pt05L ONLY FOR LOGISTIC FUNTIONS Definite Integral way to compute the area under a curve within a certain range LeftHand Riemann Sum the height of each rectangle is taken to be the value of the function at the left endpoint of the corresponding subinterval RightHand Riemann Sum the height of each rectangle is taken to be the value of the function at the right endpoint of the corresponding subinterval a Left and RightHand Sums must have rectangle of equal width equal intervals b Find division points between ab by taking banumber of rectangles you want Notation for Riemann Sums a Fx between xa and xb b Calculate the width of the subintervals change of Xban c Calculate the division points XOa x1a change of x x2a 2Change of X xnb d Compute fx0 fx1fxn Note fxn will not be used in lefthand sum Fx0 will not be used in righthand sum Definite Integral Notation f f xdx the area under the graph f x between x a and x b If fx dips below the xaxis we can t simply regard the definite integral as the area under the curve The definite integral is the limit of a Riemann sum as the number of subintervals approaches infinity Count areas below the axis as negative Fundamental Theory of Calculus If we know the rate of change or derivative of some quantity as a function of time graph and we know the quantity itself at some initial time then we can compute the quantity of a later time Area between two curves Compute the area of the two curvesgraphs Find intersection points Determine which graph is the top one Then subtract a ffxdx fgxdx 0R ffx gxdx b Similar for addition c f Cfxdx Cffxdx The Fundamental Theorem of Calculus The total change in a quantity over an interval is the area under the curve representing the rate of change of that quantity with areas below the x axis as negative The signed area under a curve represents a function is given by a definite integral of that function The rate of change of a quantity is its derivative a Fb Fa ff F tdt ffxdx b The average value of fx over the interval altxltb is ba

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