# 1 Math Review APEC 3001 Summer 2007. 2 Objectives Review basic algebraic skills required to successfully solve homework, quiz, and exam problems in this.

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1 Math Review APEC 3001 Summer 2007

2 Objectives Review basic algebraic skills required to successfully solve homework, quiz, and exam problems in this class. Review different ways of describe relationships, including –Functions –Tables –Surface Plots –Contour Plots Review the special class of linear functions. –What is the slope? –What is the intercept? –Solutions to systems of linear equations. Review the basic differential calculus skills required to successfully solve homework, quiz, and exam problems.

3 Types of Numbers Integers –…,-3, -2, -1, 0, 1, 2, 3,… Rational Numbers –An Integer Divided By An Integer (e.g. ½ = 0.5 & -4/3 = -1.333…) –All Integers Are Rational Numbers Irrational Numbers –Not Rational (e.g. = 1.4142… &  = 3.1415…) –Decimals Are Non-Repeating & Non-Terminating –Any Irrational Number Lies Between Two Rational Numbers Real –Includes All Rational & Irrational Numbers

4 Properties of Real Numbers ( a, b, & c are real numbers) Closure: a + b & a  b Are Real Numbers –1 + 2 = 3 & 1  2 = 2 Are Real Commutative: a + b = b + a & a  b = b  a –1 + 2 = 2 + 1 = 3 & 1  2 = 2  1 = 2 Associative: (a + b) + c = a + (b + c) & (a  b)  c = a  (b  c) –(1 + 2) + 3 = 1 + (2 + 3) = 6 & (1  2)  3 = 1  (2  3) = 6 Distributive: a  (b + c) = a  b + a  c & (a + b)  c = a  c + b  c –1  (2 + 3) = 1  2 + 1  3 = 5 & (1 + 2)  3 = 1  3 + 2  3 = 9 Identity: a + 0 = a & a  1 = a –1 + 0 = 1 & 1  1 = 1

5 Properties of Real Numbers Continued ( a, b, & c are real numbers) Inverse: b + (-b) = 0 & b  1/b = 1 –2 + (-2) = 0 & 2  1/2 = 1 Cancellation: –If a + b = a + c, then b = c –If a  b = a  c, then b = c (provided a is not 0) Zero Factor: –a  0 = 0  a = 0 –If a  b = 0, the a = 0, b = 0, or a = 0 & b = 0 Negation: –-1  (-a) = -(-a) = a –(-a)  (-b) = a  b –a  (-b) = (-a)  b = -(a  b)

6 Exponents (a & b are real numbers, m & n are integers) Exponent: –a n = a  a  …  a where a is multiplied by itself n times –2 4 = 2  2  2  2 = 16 Properties of Exponents: –a n  a m = a n+m (e.g. 4 2  4 5 = 4 2+5 = 4 7 = 16,384) –(a n ) m = a n  m (e.g. (4 2 ) 5 = 4 2  5 = 4 10 = 1,048,576) –a n  b n = (a  b) n (e.g. 2 2  3 2 = (2  3) 2 = 6 2 = 36) –a -1 = 1/a(e.g. 2 -1 = ½ = 0.5) Note that these properties will also hold if m & n are rational numbers, but we can then end up with complex numbers. In this class, we will choose rational number for m & n, but we will never use complex numbers.

7 Functions What is a function? –A function describes a definite relationship. –For example, the relationship between the area and radius of a circle is y = f(x) where y is the area x is the radius, & f(x) =  x 2 Causal Relationship –The way a function is written is sometimes meant to imply a causal relationship. –For example, y = f(x) means x determines y or x is the independent variable, while y is the dependent variable. –In this class, the way we write a function will not always imply causality.

8 Inverse Functions Many of the functions we will deal with in this class will be invertible. That is, we can write y = f(x) or x = f -1 (y). This also implies, we can write y = f(f -1 (y)). For example, if y = f(x) =  x 2, then

9 Other Descriptions of Definite Relationships Functions like y = f(x) =  x 2 provide a general and concise way to describe definite relationships, but they are not the only way. For example, we could have described the relationship between the area and radius of a circle using a Table:

10 Other Descriptions of Definite Relationships We could have also described the relationship between the area and radius of a circle using a Figure. x-axis: Often Used for Independent Variable y-axis: Often Used for Dependent Variable IMPORTANT NOTE: WE WILL NOT ALWAYS STICK TO THIS CONVENTION

11 Could Just As Well Drawn As

12 More Complex Relationships What is a function that describes the number of words in a book? –T = f(p, n) where T is the total number of words in a book p is the number of pages in the book n is the average number of words per page f(p, n) = pn Many of the functions we will look at in this class characterize the relationship between one variable and many others.

13 Again, We Can Use Tables

14 But, What About Figures? Surface Plot Not Very Easy to Draw

15 There Has To Be An Easier Way Contour Plot n = 100 n = 200 n = 300 n = 400 n = 500 Each contour shows the relationship between the total words and the number of pages holding the words per page constant.

16 Alternatively Contour Plot p = 50 p = 100 p = 150 p = 200 Each contour shows the relationship between the total words and the words per page holding the number of pages constant.

17 Surface Versus Contour Surface plots become virtually impossible to draw once we are describing a relationship with more than three variables. Contour plots are easy to draw regardless of how many variables you have. –Plot the relationship between two variables holding all the other variables constant. –Change one of the constant variables and repeat. –Repeat again as long as you need to in order to understand the relationship. IMPORTANT NOTE: AN UNDERSTANDING OF CONTOUR PLOTS IS ESSENTIAL IN THIS CLASS

18 Linear Functions Looks Like y = b + mx –b is referred to as the intercept The value of y when x = 0 –m is referred to as the slope How much y changes as x increases by 1 m =  y/  x where  y is the change in y (the rise) &  x is the change in x (the run) m > 0 implies a positive relationship m < 0 implies a negative relationship m = 0 implies no relationship –Inverse is x = - (b/m) + (1/m)y - (b/m) is the intercept for the inverse function 1/m is the slope for the inverse function

19 Graphical Example y = b + mx = 10 + 2.5x  y = 10  x = 4 m =  y/  x = 10/4 = 2.5

20 Some More Examples y = 10 + 2.5x y = 20 + 0 x y = 30 - 1.5x

21 Solving Two Linear Functions Suppose we have two linear equations: –y = b 1 + m 1 x –y = b 2 + m 2 x Question: –Is there a value for x that makes y the same for both functions? What if we take two functions from the previous slide: –y = 10 + 2.5x –y = 30 - 1.5x What is the answer to our question for these two functions?

22 If y is equal for both equations when x = x*, then we know 10 + 2.5x* = 30 - 1.5x* 10 - 10 + 2.5x* = 30 - 10 - 1.5x* 2.5x* = 20 - 1.5x* 2.5x* + 1.5x* = 20 - 1.5x* + 1.5x* 4x* = 20 4x* / 4 = 20 / 4 x* = 5 Is this right? y* = 10 + 2.5  5 = 22.5 y* = 30 - 1.5  5 = 22.5 So, it would seem!

23 Does this work in general? b 1 + m 1 x* = b 2 + m 2 x* b 1 – b 1 + m 1 x* = b 2 – b 1 + m 2 x* m 1 x* = b 2 – b 1 + m 2 x* m 1 x* - m 2 x* = b 2 – b 1 + m 2 x* - + m 2 x* (m 1 - m 2 )x* = b 2 – b 1 (m 1 - m 2 )x* / (m 1 - m 2 ) = (b 2 – b 1 ) / (m 1 - m 2 ) If (m 1 - m 2 )  0, then x* = (b 2 – b 1 ) / (m 1 - m 2 ) Is this right? So, it would seem, at least if (m 1 - m 2 )  0!

24 What is this thing we have found? What we have found is the point of intersection between two lines! Later it will become obvious why this point of intersection is of interest to us!

25 Example y = 10 + 2.5x y = 30 - 1.5x 5 22.5

26 Why doesn’t it work with m 1 = m 2 ? If m 1 = m 2 and b 1 = b 2, then the two lines are identical and any value of x will give you the same value of y. If m 1 = m 2 and b 1  b 2, then the two lines are different, but have the same slope. They are parallel, which means there is no value x* that will give you the same y*. There is no point of intersection!

27 Elementary Calculus Calculus Has Many Uses –Compute Areas & Volumes –Compute Velocities –Describe Functional Relationships –Find Extremes In this class, we will typically be interested in –describing functional relationships & –finding extremes. Both these uses can be accomplished using what are called derivatives or partial derivatives.

28 Derivative of a Function Consider the function y = f(x) We will denote the derivative of a function with only one variable using ’: f ’(x). The definition of a derivative is What does this definition tell us? –It tells us how the value of y is changing as x changes.

29 Graphical Interpretation y = f(x) = 10+15x-2x 2 h = 4 – 2 = 2 f(x + h) - f(x) = 38 – 32 = 6 f’(x) = slope as h goes to 0 or slope of line tangent to f(x) at x. Slope = (f(x + h) - f(x)) / h = 6 / 2 = 3

30 So why is this useful? What can we say about f(x) when f ’(x) > 0? What can we say about f(x) when f ’(x) < 0? What can we say about f(x) when f ’(x) = 0?

31 When f ’(x) > 0, increasing x increases f(x) or y!

32 When f ’(x) < 0, increasing x decreases f(x) or y!

33 When f ’(x) = 0, f(x) or y is at a maximum!

34 When f ’(x) = 0, f(x) or y is at a minimum!

35 To Summarize f ’(x) > 0 tells us there is a positive relationship between x and y given x. f ’(x) < 0 tells us there is a negative relationship between x and y given x. f ’(x) = 0 tells us y is at a maximum or minimum given x. Note: There is a huge difference between being at a maximum vs. being at a minimum! It would be nice to able to say which is the case. –What is happening to f ’(x) when x is increasing for a maximum? –What is happening to f ’(x) when x is increasing for a minimum?

36 For a maximum, f ’(x) is decreasing as x increases!

37 For a minimum, f ’(x) is increasing as x increases!

38 But how do we tell what is happening to f ’(x) as x increase? This is precisely what a derivative tells us! Can we take the derivative of a derivative? –Yes! –It is called the second derivative, which we will denote with ’’: f ’’(x). –The second derivative is defined as –It is just the derivative of the derivative! –If f ’’(x) < 0, it tells us the derivative or slope of the tangent is decreasing as x increases or that the x that solves f ’(x) = 0 is a maximum. –If f ’’(x) > 0, it tells us the derivative or slope of the tangent is decreasing as x increases or that the x that solves f ’(x) = 0 is a minimum.

39 Useful Properties of Derivatives You Need to Know Suppose we have three functions: f(x), g(x), and h(x) –If f(x) = g(x) + h(x), f ’(x) = g’(x) + h’(x). –If f(x) = g(x)h(x), f ’(x) = g’(x)h(x) + g(x)h’(x). –If f(x) = h(g(x)), f ’(x) = h ’(g(x)) g’(x). –If f(x) = h(x)/g(x), f ’(x) = (h’(x) g(x)- h(x)g’(x))/g(x) 2. Suppose a and n are real numbers: –f(x) = a, f ’(x) = 0. –f(x) = ax n, f ’(x) = nax n - 1

40 An Example Suppose f(x) = 10+15x-2x 2 f(x) = f 1 (x) + f 2 (x) + f 3 (x) where f 1 (x) = 10, f 2 (x) = 15x, and f 3 (x) = -2x 2. f ’(x) = f 1 ’(x) + f 2 ’(x) + f 3 ’(x) = 0 + 15 – 4x = 15 – 4x because –for f 1 (x) = 10, f 1 ’(x) = 0 –for f 2 (x) = 15x, f 2 ’(x) = 1  15  x 1 - 1 = 1  15  x 0 = 15 and –for f 3 (x) = -2x 2, f 3 ’(x) = 2  (-2)  x 2 - 1 = -4 x 1 = -4x

41 Another Example Suppose f(x) = x 2 x 3 f(x) = f 1 (x)f 2 (x) where f 1 (x) = x 2 and f 2 (x) = x 3. f’(x) = f 1 ’(x) f 2 (x) + f 1 (x) f 2 ’ (x) = 2xx 3 + x 2 3x 2 = 2x 4 + 3x 4 = 5x 4 because –for f 1 (x) = x 2, f 1 ’(x) = 2x 2 - 1 = 2x and –for f 2 (x) = x 3, f 2 ’(x) = 3x 3 - 1 = 3x 2 Notice that f(x) = x 2 x 3 = x 5, such that f ’(x) = 5x 5 - 1 = 5x 4, but things may not always simplify so nicely.

42 Yet Another Example Suppose f(x) = (x 2 - 5) 3 f(x) = f 1 (f 2 (x)) where f 1 (x) = x 3 and f 2 (x) = x 2 - 5. f ’(x) = f 1 ’(f 2 (x)) f 2 ’ (x) = 2(x 2 - 5) 2 2x = 4x(x 2 - 5) 2 because –for f 1 (x) = x 3, f 1 ’(x) = 3x 3 - 1 = 3x 2 and –for f 2 (x) = x 2 - 5, f 2 ’(x) = 2x 2 - 1 - 0 = 2x

43 One Final Example Before Moving On Suppose f(x) = (x 3 - 5) / x 2 f(x) = f 1 (x) / f 2 (x) where f 1 (x) = x 3 - 5 and f 2 (x) = x 2 f ’(x) = (f 1 ’(x) f 2 (x) – f 1 (x) f 2 ’ (x)) / f 2 (x) 2 = (3x 2 x 2 - (x 3 - 5) 2x) / x 4 = (3x 4 - 2x 4 + 10x) / x 4 because –for f 1 (x) = x 3 - 5, f 1 ’(x) = 3x 3 - 1 - 0 = 3x 2 and –for f 2 (x) = x 2, f 2 ’(x) = 2x 2 - 1 = 2x

44 What about functions with more than one variable? Suppose we have a function y = f(x, z) The derivative of f(x, z) with respect to x is defined as The derivative of f(x, z) with respect to z is defined as These are referred to as partial derivatives because we are taking the derivative of the function with respect to a single variable, while holding the other variable constant.

45 What do these partial derivatives mean? f x (x,z) tells us the relationship between y and x on a contour holding z constant. f z (x,z) tells us the relationship between y and z on a contour holding x constant. Solving f x (x,z) = 0 and f z (x,z) = 0, we can find the x and z that maximizes or minimizes y (there is also something called a saddle point that we will not deal with in this class). –Again, we can compute second partial derivatives to determine whether we are at a maximum or minimum, but explaining how is more than what we need to get into at this point.

46 An Example Suppose y = f(x,z) = x 0.4 z 0.6 –f x (x,z) = 0.4x 0.4-1 z 0.6 = 0.4x -0.6 z 0.6 –f z (x,z) = x 0.4 0.6z 0.6-1 = 0.6x 0.4 z -0.4

47 Surface Plot for y = f(x,z) = x 0.4 z 0.6

48 Contour Plot for y = f(x,z) = x 0.4 z 0.6 holding z constant. z = 5 z = 10 z = 15 z = 20 z = 25  f x (x,z) gives us slopes of tangents to contours holding z constant.

49 Contour Plot for y = f(x,z) = x 0.4 z 0.6 holding x constant. x = 10 x = 20 x = 30 x = 40 x = 50  f z (x,z) gives us slopes of tangents to contours holding x constant.

50 Skills for Success Manipulate Algebraic Expressions Represent Functions Algebraically and Graphically Find Solutions to Systems of Equations (Usually Two Equations) Take Derivatives and Partial Derivative of Functions with Exponents

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