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by: Fizza Aslam

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Worksheet lesson 8 Math 124

Marketplace > University of Washington > Math > Math 124 > Worksheet lesson 8
Fizza Aslam
UW
Math 124
David H. Collingwood

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Detailed description and solutions of worksheet for lesson 8.
COURSE
Math 124
PROF.
David H. Collingwood
TYPE
Bundle
PAGES
4
WORDS
KARMA
75 ?

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This 4 page Bundle was uploaded by Fizza Aslam on Wednesday October 21, 2015. The Bundle belongs to Math 124 at University of Washington taught by David H. Collingwood in Fall 2015. Since its upload, it has received 63 views. For similar materials see Math 124 in Math at University of Washington.

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Date Created: 10/21/15
Worksheet Lesson 8 Question 1 If fx is a function then remember that we define x 2 fx h fx If this limit exists then f39x is the slope of the tangent line to the graph offat the point xfx Consider the graph offx below a b c d a For which values ofx does f39x not exist Since we know that the derivatives do not exist at the sharp corners look at the sharp corners on the graph which are marked in red in graph above So the answer here will be smallest value 2 and largest value 2 For which value x is f39x 0 To find derivative of graph 0 look at the graph and try to find out at which point the slope of tangent line to the graph is 0 It is marked in green line above So the point here will be Oquot3911 Hence our answer is x 0 This particular function f has an interval on which its derivative f39x is constant What is this interval Note that a derivative or slope is constant for a straight line and in the above graph the line exists from points 42 and 22 So the interval here would be x1x2 ie 42 Estimate the slope of fx on that interval Calculate the slope using the formula yZ y1 2 2 x2 X1 4 2 Question 2 Below is a graph of a derivative g39x Assume this is the entire graph of g39x Use the graph to answer the following questions about the original function gx a b w On which intervals is the original function gx decreasing Because g x represents slope of the gx At any point in g x if g xgtO it means that gx in that part has positive slope or so called increasing Similarly when g xltO then gx has negative slope or decreasing Now look at the parts in the graph where g xlt0 Decreasing parts are highlighted in yellow in the graph Hence the answer is 43 for the left most interval and O532 for the right most interval Now suppose gO 0 Is the function gx ever positive For this one if you look closely in the domain of g x where xltO you39ll see that there39s some part that g xgtO which means that the part before x approaches zero gx has positive slope However the slope of g x or gquotx in that part is negative meaning that although in gx the slope is positive it39s declining and approaching to negative So imagine the possibility of this situation the slope of gx where xltO starts off with negative slope and then positive slope and then declining towards O with the slope that is decreasing Do you think this line of graph pass the area where gxgtO Yes it does Look at the highlighted part in blue Question 3 Six graphs of functions are below along with six graphs of derivatives Match the graph of each function with the graph of its derivative Graphs of original functions are on left side below and their derivatives are at right side ll This is a continuous line with a positive constant slope so the graph must be a straight line on positive yaxis ml quot all A Ill During 1st part of this graph the function is decreasing so the slope is negative then there is a sharp corner where derivative doesn t exist and then increasing line that means positive slope So the graph of derivative should have a line with constant negative value of y then the empty circles show that derivative doesn t exist followed by a line with constant positive value of y dl n A The two lines on graph shows positive slope but with a jump so the derivative must be a line in positive yaxis with a break at x0 which shows no existence of derivative For the first part of graph the tangent line is positive becoming less positive as it reaches to zero So the derivative should be decreasing in a positive y quadrants until it reaches zero For the second part the derivative is positive and getting steeper El For the first part we can see it s a line with positive slope so the derivative has a constant positive y value until it reaches the sharp corner at x0 where derivative doesn t exist For the second part the derivative is positive and getting steeper

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