Midterm2ReviewCanvasVersion2

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St. John's University **We aren't endorsed by this school
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MATH 112
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Mathematics
Date
Nov 14, 2023
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Math 112 Midterm 2 Summary Fall 2023 November Table of Contents Clicking on each section will refer you to corresponding page in the review packet. All solutions to the example problems corresponding to each section can be found in this pdf file. 1. Functions Linear Functions and Rate of change - Example 0.1 Transformations of Functions - Example 0.2: Part(a), Part(b) Combination and Compositions of Functions - Example 0.3 - Example 0.4 Inverse Functions - Example 0.5 - Example 0.6 2. Polynomial Functions Quadratic equations and optimization - Example 0.7 - Example 0.8 Graphing polynomial functions - Example 0.9: Part(a), Part(b), Part(c) Polynomial Division - Example 0.10: Part(a), Part(b), Part(c) Rational Root Theorem - Example 0.11: Part(a), Part(b), Part(c) - Example 0.12: Part(a), Part(b), Part(c) Fundamental Theorem of Algebra - Example 0.13: Part(a), Part(b) 3. Exponential Functions Exponential Functions - Example 0.14: Part(a), Part(b) - Example 0.15: Part(a), Part(b)
1. Functions Linear Functions and Rate of change (Section 2.5) - Linear Model : We can express any linear relation between the input x and output y as y = mx + b where m , the slope, is the rate of change . Example 0.1. Suppose it costs $ 30 to make a quilt, and it costs on average $ 40 per week to run my business. If x is the number of quilts I make in a week, and y is my weekly costs, find a linear relationship between x and y . What is the rate of change of the linear relation, and what are its units? Solution : Let's try to find the linear relationship y = mx + b . The first sentence "costs $ 30 to make a quilt" indicates that the slope m = 30. The second sentence "costs on average $ 40 per week to run my business" indicates that even when we don't make any quilts, there is still going to be an average $ 40 cost perweek. Hence, the linear relationship satisfies b = 40. (Plug in x = 0) We hence have y = 30 x + 40 as the linear relation. The rate of change is 30 $ /week.
Transformations of Functions (Section 2.6) - Domains and Ranges : Here is a list of domains and ranges of some functions. y = x : Domain : ( −∞ , ) , Range : ( −∞ , ) y = | x | : Domain : ( −∞ , ) , Range : [0 , ) y = x 2 : Domain : ( −∞ , ) , Range : [0 , ) y = x 3 : Domain : ( −∞ , ) , Range : ( −∞ , ) y = x : Domain : [0 , ) , Range : [0 , ) y = 3 x : Domain : ( −∞ , ) , Range : ( −∞ , ) y = 1 x : Domain : ( −∞ , 0) (0 , ) , Range : ( −∞ , 0) (0 , ) - Transformations : Here is a list of transformations of a function y = f ( x ). Tip : These transforma- tions also change the domains and ranges of the functions respectively. y = f ( x a ) : ( Assume a > 0) Horizontal shift to right by a y = f ( x + a ) : ( Assume a > 0) Horizontal shift to left by a y = f ( x ) + a : ( Assume a > 0) Vertical shift upwards by a y = f ( x ) a : ( Assume a > 0) Vertical shift downwards by a y = f ( c · x ) : ( Assume c > 0) Horizontal stretching of the function by a factor of 1 c y = c · f ( x ) : ( Assume c > 0) Vertical stretching of the function by a factor of c y = f ( x ) : Reflection along the y-axis y = f ( x ) : Reflection along the x-axis - Graphs : When graphing transformations of functions, there are two strategies: Strategy 1: Apply transformations mentioned above one step at a time. Strategy 2: Pick reference points on the graph, and identify where the reference points move to. Connect the reference points to draw a new graph. For example, suppose we want to draw the following function given y = f ( x ) y = 2 · f ( 3 · x ) + 5 Strategy 1: There are 3 operations to take. · Red : Horizontally stretch the function by a factor of 1 3 · Green : Vertically shift the function upwards by 5. · Blue : Vertically stretch the function by a factor of 2. Stategy 2: Suppose ( x 0 , y 0 ) is a point on y = f ( x ). The point ( x 0 , y 0 ) moves to a new point on y = c · f ( a · x ) + b via the following strategy: · x-coordinate: multiply x 0 by 1 3 ( Red ) · y-coordinate: multiply y 0 by 2 ( Blue ), then add by 5. ( Green ) In particular, ( 1 3 x 0 , 2 y 0 + 5) is the new point on the transformed function. Likewise, suppose we want to draw the following function given y = f ( x ) y = 2 · f ( x + 3 ) + 5 Strategy 1: There are 3 operations to take. · Red : Horizontally shift the function to the left by 3 · Green : Vertically shift the function upwards by 5 · Blue : Vertically stretch the function by a factor of 2. Stategy 2: Suppose ( x 0 , y 0 ) is a point on y = f ( x ). The point ( x 0 , y 0 ) moves to a new point on y = c · f ( a · x ) + b via the following strategy: · x-coordinate: subtract x 0 by 3 ( Red ) · y-coordinate: multiply y 0 by 2 ( Blue ), then add by 5. ( Green ) In particular, ( x 0 3 , 2 y 0 + 5) is the new point on the transformed function.
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