Suppose we have the set of all real numbers, R, and two operations, +, and *. Then the following are assumed to be true.

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1 Algebra Review In this appendix, a review of algebra skills will be provided. Students sometimes think that there are tricks needed to do algebra. Rather, algebra is a set of rules about what one may and may not do to an equation. The general idea is to move toward a solution to an equation, although that need not be the only goal. A primary factor in doing algebra is actually doing it. One cannot be successful in doing algebra if one does not grab one s pencil and paper and use the pencil and paper to write out the steps. Only the very best students can do algebra in their head. A second fact is that student often do not want to write out each step; they want to combine steps and to two or more steps at the same time. This is bad practice. Do the algebra one step at a time. Otherwise, you may find yourself going back and doing the problem over and waste the time you thought you saved by doing more than one step at a time. The final piece of advice I would offer is that persistence pays off. If at first you don t get the problem solved, come back later and try again. If you make a mistake, go back and try again. So how, then, do we begin? Start with the assumptions that we make for the real numbers. These are sometimes called field axioms. Algebra starts with a set of elements (real numbers in our case) and two operations on those number. The axioms tell us the allowed actions that we can take with these operations and the outcome of these actions. The two operations are addition and multiplication. AXIOMS Suppose we have the set of all real numbers, R, and two operations, +, and *. Then the following are assumed to be true. Assumptions for + (1) There exists an element 0, 0 ε R, so that for any a ε R, a + 0 = a (2) For each a ε R, there exists another element of R, -a, so that a + (-a) = 0 (3) For any two a, b ε R, a + b = b + a (4) For any then a, b, and c ε R, a + (b + c) = (a + b) + c Assumptions for * (5) There exists an element 1, 1 ε R, so that for any a ε R, a * 1 = a (6) For each a ε R (a 0) there exists a unique a -1 so that a * a -1 = 1 (7) For any two a, b εr, a * b = b * a (8) For any three a, b, c ε R, a * (b * c) = (a * b) * c

2 Distributive Principle (links +, *) (9) For any a, b, and c ε R, a * (b + c) = a * b + b * c Remark: First, note the symmetry. The axioms 1 4 are similar to axioms 5 8. Axiom 1 and 5 asserts the existence of an identity element. Zero is the identity for + and one is the identity for *. Axioms 2 and 6 assert the existence of an inverse element. For addition, the inverse element of a is minus a. This introduces the idea of subtraction which is addition with negative numbers. Similarly for multiplication, we have a multiplicative inverse which introduces division (except for the element zero). The next axioms (3 and 7) tell us that the order we add or multiply does not matter. Thus whether we add b to a or a to b yields the same solution. If we multiply a times b is the same as b times a. These are the commutative axioms. Both multiplication and addition are commutative. Axioms 4 and 8 are called the associative axioms. They say it does not matter how we combine elements (for a given operation). Thus a added to the quantity b plus c is the same as c added to the quantity a plus b. A similar outcome holds for multiplication. The distributive principle is a very powerful tool (we will see it below). It links the two operations and says that if we have two numbers b and c to add and then multiply that sum by the number a, we would get the same outcome if we multiplied a times b and added that to a time c. In the context of numbers, these operations are pretty much second nature. So we know that 6 + (7 + 8) = (6 + 7) + 8 or that (3*(-1))*4 = 3*((-1)*4) or that 7 + (-7) = 0 or that -5*(-1/5) = 1 But the same principles apply to symbols. That is where algebra comes in! Here are some examples. See if you can identify the axiom used. x + x = 2*x (x + 4) + (x + 3) = 2*x + 7 x + (-x) = 0 x * (1/x) = 0 (for x 0) These axioms are useful in solving equations. An equation is an expression saying that two tings are the same or at least the two things have the same value somewhere. Consider the following. X 7 = 3

3 In this expression we are told that x 7 and three are the same. But clearly they are not the same for all values of x. If x = 100, then x 7 is 93 and 93 is not equal to 3! So what we are being told is that for some x, x 7 and 3 are the same. What is the value of x, if any, so that they are the same? We are being asked to solve the equation. Our objective is to find the value of x so that the two sides are the same. How do we do this? The answer is that we use the rules given above with one other important fact. What we do to one side of the equation we must do to the other. So to get x all by itself on one side of the equation (solve for x), we need to get rid of the -7 on the left hand side. We do this by adding 7 to both sides. x 7 = = + 7 x = x = 10 Exercises: Solve the following expressions. a. x + 4 = 12 b. 3x 7 = 42 c. 5x = 3x + 12 d. (1/7)x = 2 e. x 1/8 = 7/8 FRACTIONS Fractions sometimes cause students problems. One way to think about fractions is as the multiplicative inverse of an element. So a -1 = 1/a. Now consider how we can add fractions. The one thing to remember is that to add fractions, we need a common denominator. Suppose we want to add 1/a to 1/b. What would be a good common denominator? We want to combine the denominators into one. In this case, a good common denominator would be ab. To use this denominator, we will multiply each fraction (that is the fractions 1/a and 1/b) by 1 to get the denominator we want. The question is exactly how do we multiply by 1 to get the denominator in the form we want? We will multiply the 1/a by b/b. Note that b/b is 1. Similarly we multiply 1/b by a/a. Again, a/a is 1. We obtain the following.

4 b 1 a 1 b + + = b a a b aba Multiplying fractions is a bit easier. In this case, we simply multiply the numerators to get the numerator of the product and then multiply the denominators to get the denominator of the product. 1 1 a b = 1 ab Exercises: a. b. c. d. e. f. g. h. i. j. k x y x y 4 3 x 3 y x y x 7 x 2 5 y 1 2 x 5 x + 7

5 EXPONENTS The following notation is frequently used; we should investigate what it means. The notation is x a. First note that if a = 0, then x a = 1. This will be true for any real value of x. Now consider x a where a is not zero. Then we have the following. x a = 1x 4x 2x 4L 4 3x a times The notation means that we multiply x times itself a times. Now we need to know how we multiply terms like x a times x b. x a x b = 1x 4x 2x 4L 4 3x 1x 4x 2x 4L 4 3x = 1x 4x 2x 4L 4 3x = x a+b. a times b times a+ b times In short, when we multiply x terms where the x s are raised to an exponent, we add the exponents. Suppose we have an expression as shown below. What do we do then? Assume that a > b for the moment. 1x 4243 x L x a x a times b = x 1x 42 x L 43 x b times = 1x 4243 x L xx x L x a b times 1x 4243 x L x b times b times = 1x 42 x L 43 x = x a-b a b times To divide x terms raised to an exponent, we subtract the exponent of the denominator term from the exponent of the numerator term. Consider now ( x b ) a. What is this expression equal to? ( x b ) a = ( x x L x)( x x L x) 4 L ( x x L x) 4 b times b times b times a times = ( x x L x) = x ab a b times Note that we can write x ab = x ba = (x a ) b = (x b ) a. Also observe that x -b = 1/(x b ) = (x b ) -1. We can see this because x 0 /x b = 1/x b (why is this true?). Now by the subtraction rule, x 0 /x b = x 0-b = x -b. Now we also know that we can write x -b = (x b ) -1 by what we just did.

6 Exercises: Simplify the following expressions. a. x a x b b. x a x b-a c x a x -b d. (x a ) a e. (x a ) -b FRACTIONAL EXPONENTS What do we mean by x 1/2? One way to think about this is that the ½ exponent undoes an exponent of 2. We know from the above that (x 2 ) 1/2 = x. Hence, we know that the exponent of ½ is the square root. By similar reasoning, we can find that an exponent of 1/3 rd would be the cube root. These fractional exponents can be used in just the same way as the integer exponents. Exercises: Simplify the following expressions. a. (x 2 ) 1/3 b. (x 1/3 ) 2 c. (x 3 ) 1/3 d. (x a ) 1/2 e. (x 1/4 ) 2 f. (x 1/2 )(x 1/3 ) g. (x 2 )(x 1/4 ) EXPONENTS WHEN THE BASE IS DIFFERENT What happens when we have x a y a? In this case, the exponent is the same, but the base (the element we are raising to the power) is different. What can we do now? Stop and look at what we have.

7 x a y a = 1x * 4243 x * L * x * y * y * L * y a times a times Because multiplication is associative, we could put one of the y s between each x to obtain the following. x a y a = ( x * y) * ( x * y) L* ( x * y) a times Thus as long as the exponent is the same, and we are multiplying, we can combine the x and y terms and use a single exponent. MULTIPLYING FACTORS (FOIL) AND FACTORING Suppose that we wish to compute (x + y)(x + y) {or what is the same thing, (x + y) 2. What do we get? Some students think we will get (x + y) 2 = x 2 + y 2. But this is not right. The diagram below helps us understand why. y xy y 2 x + y x x 2 xy x x + y y We have put x + y on the horizontal axis and x + y on the vertical axis. When we multiply these, we get the area of the rectangle. The product x 2 and y 2 are shown in the diagram in the upper right and lower left. Clearly these two terms are not equal to (x + y)(x + y). We have left out xy in the upper left plus xy in the lower right. The proper way to find the product of (x + y)(x + y) is to use the distributive principle. (x + y)(x + y) = (x + y) x + (x + y) y (why is this statement true? Look at the distributive principle.)

8 = x 2 + xy + xy + y 2 = x 2 + 2xy + y 2 More generally, (x + a)(x + b) = x 2 + (a + b)x + ab. This may be known to you as the FOIL method. By the same logic, we can see that (x + y) 1/2 x 1/2 + y 1/2. Exercises: Multiply the following terms. a. (x 5)(x + 3) b. (x + 3)(x -1) c. (x 4)(x 2) Remark: A really powerful tool, one used to solve equations, is to work backward. Suppose we have x 2 + 7x + 12 = 0. How can we solve this? There ar two ways to proceed. First, we could rely on brute force and use the quadratic formula. But a more elegant and potentially fruitful method is to factor the expression. Factoring is to use the distributive principle. Are there values of a and b so that we can write x 2 + 7x + 12 = (x + a)(x + b)? One way to proceed is to multiply out the right hand side and then match up coefficients. (x + a)(x + b) = x 2 + (a + b)x + ab We will match up the following two equations. x 2 + 7x + 12 x 2 + (a + b)x + ab For these two to be equal, we need (a + b) = 7 and ab = 12. Can we find two numbers, a and b, so that a + b = 7 and ab = 12? How about a = 4 and b = 3 (or a = 3 and b = 4). So we have: x 2 + 7x + 12 = (x + 4)(x + 3) Thus x 2 + 7x + 12 = 0 requires that (x + 4)(x + 3) = 0. Hence x = -4 or x = -3 must hold. Clearly being able to express x 2 + 7x + 12 in terms of its factors was a very useful thing. Practice with these problems. Exercises: Factor these expressions.

9 a. x 2 + 8x + 16 b. x 2 7x + 12 c. x 2 3x + 9 d. x 2 x 12 e. x 2 + x 12 SPECIAL CUBIC CASE The above procedures provide a way to solve a number of quadratic equations. There are two cubic equations that are easy to factor. (x 1) 3 = (x 1)(x 2 + x + 1) (x + 1) 3 = (x + 1)(x 2 x + 1) Expressions with exponents over 3 are difficult to factor. TWO FUNCTIONS COMMONLY OCCURING IN ECONOMICS Equations of the form y = ax 2 + bx + c show up with some frequency in applied problems where a, b, and c are constants. This is one function from the family of conic sections; this is the parabola. The parabola has a general form given above, y = ax 2 + bx + c (there is also an x = ay 2 + by + c form). How do we graph this? First, suppose we could re-arrange the equation to get (y k) = a(x h) 2 where a, h, and k are constants. In this form, it is clear that x = h and y = k is on the graph of this function (why?). Next, if a > 0, and x is not equal to h, then y k > 0 must hold (why?). Clearly the smallest value y can take is k (why?). Hence as x moves either direction from h (assume a > 0), y rises. Also observe that the graph is symmetric about the vertical line, x = h. If x is, say, 2 more or 2 less than h, the (x h) = 2 or (x h) = -2. And (-2) 2 = 2 2. Hence the values of y at these x s will be the same. Thus we can easily draw a rough approximation of this function. For larger values of a, the parabola will be steeper. If a < 0, then the parabola opens downward. Go back through the analysis given above with the assumption that a < 0 holds. What differences are there?

10 y k h x The question remains. How do we get y = ax 2 + bx + c to look (y k) = a(x h) 2? The process of making this transformation is called completing the square. We proceed as follows. y = ax 2 + bx + c Divide both sides by a. (1/a)y = x 2 + (b/a)x + (c/a) Subtract (c/a) from both sides. (1/a)y (c/a) = x 2 + (b/a)x We will now add a constant to both sides to make the right hand side a perfect square. From our earlier work we know that (x h) 2 = x 2 2hx + h 2 We are looking for h 2, and we know that -2hx = (b/a)x (how do we know this?). So h = (-b/2a). Therefore h 2 = (b 2 /4a 2 ). So we add (b 2 /4a 2 ) to both sides of the equation. (1/a)y (c/a) + (b 2 /4a 2 ) = x 2 + (b/a)x + (b 2 /4a 2 ) (1/a)y (c/a) + (b 2 /4a 2 ) = (x + (b/2a)) 2

11 (1/a){y c + (b 2 /4a)}= (x + (b/2a)) 2 Hence we have the following. y c + (b 2 /4a) = a {(x + (b/2a)) 2 } {y ((4ac - b 2 )/4a)} = a {(x + (b/2a)) 2 } And we have (y k) = a (x h) 2, where k = ((4ac - b 2 )/4a) and h = - (b/2a). Exercises: Use the above methods to graph the following parabolas. a. y = x 2 + 6x + 36 b. y = x 2 6x + 36 c. y = x 2 4x + 22 d. y = x 2 + 2x + 14 e. y = x 2 + 2x 3 f. y = x 2 4x 3 Remark: A second useful function is another conic section, the rectangular hyperbola. The general form is (x h)(y k) = r where h, k, and r are constants. How do we graph this equation? First, consider the case when r is zero. Then we have (x h)(y k) = 0. Either x = h or y = k or both must hold. Graph these outcomes. Now consider the case of r > 0. How do we graph (x h)(y k) = r? In the case that r 0, neither x = h or y = k could possibly hold (why?). Thus the graph could not cross x = h or y = k. Now a good way to proceed is to solve for y. In this case, we would have the following. (y k) = r/(x h) y = {r/(x h)} + k As x gets close to h from the right (so x h > 0), x h goes to zero and r/(x h) goes to infinity. As x get close to h from the left (x h) < 0, x h goes to zero and r/(x h) goes to minus infinity (why minus and why infinity?). Now as x gets larger and larger, y goes r/(x h) goes to zero

12 and y approaches k. A similar expression hold as y approaches negative infinity. Thus the values h and k become asymptotes for the function. y k h x If r rises, the curves move further from the origin. If r is less than zero, then the curve move to the southeast and northwest quadrants of the asymptotes. Exercises: Graph the following. a. xy = 100 b. (x 10)(y + 4) = 3 c. (4x 8)(y 2) = 12 d. y = {(-12/x) + 3} e. y = {(12/x) + 3} f. xy 3x + 4y -12 = 10 g. xy + 2x 3y = 10 h. xy + 5x + 3y = 50 i. xy + 5x + 3y = 60

13 Remark: One final functional form is built on the rectangular hyperbola. Consider x a y b = r. Now if we solve for y, we get the following expression. y = (r/x a ) 1/b = r 1/b / x a/b Compare this to xy = r which transforms to y = r/x. These two forms are quite similar. The only difference is that in our new expression, in the denominator we have x a/b rather than x. While this is a significant difference, the general shape of the curve will be very much like that of y = r/x. We will see x a y b = r again later. SOLVING A SYSTEM OF EQUATIONS Suppose that you have three linear equations in three unknowns. While there are elegant matrix methods to solve these, one should know how to do it by brute force. Here is how. Consider these equations. 6x + 3y 7z = 12-2x - 4y + z = -5 3x 5y + 2z = 7 What does it mean to say that we have solved these equations? It means that we have found a value for x, a value for y, and a value for z so that if we put those values back into each equation, each equation would hold as an equality. Note that at this point, we presume that there are such values. In fact, if there are not, we will discover this fact in the process of trying to find the solution. In the body of the notes, you will learn ways to solve a system of linear equations. We will do it here by brute force. How can we start? The overall strategy will be to solve one equation for one of the variables in terms of another variable. Then we will plug that expression into the other two equations. So suppose we solve the second equation for z. We would plug the expression we get for z into both the first and third equation. We would then have two equations with only two unknowns, x and y. Here we go. From equation 2, we have z = x + 4y Plug into equation 1 and 3. 6x + 3y -7(-5 + 2x + 4y) = 12 3x 5y + 2(-5 + 2x + 4y) = 7

14 We can simplify these expressions; we start by multiplying out the term in parentheses. 6x + 3y x 28y = 12 3x 5y x + 8y = 7 Now combine like terms. -8x - 25y = -23 7x + 3y = 17 We can repeat the process we used above. Solve one of the equations for one variable and plug back into the other equation. Suppose we solve the second equation for y; we would get y = 17/3 (7/3)x. Plug into the other equation. -8x 25(17/3 (7/3)x) = -23 We can now solve for x. {25*7/3 - (24/3)}x = 25*(17/3) - 23 (151/3)x = 356/3 x = 356/151 We can now find both y and z. y = 17/3 (7/3)x. (Please carry out the computation). Now that we have both x and y, we can find z. z = x + 4y (Please carry out the computation). Exercises: 1. -2x + y 7z = 1 -x - y + 8z = -2 4x 5y - 2z = 3 2. x - 2y + 5z = -22-2x - 4y + z = 15 3x 5y + 2z = x + 4y 8z = 6 2x - 5y + 3z = 3 x - 3y - z = 12

15 INEQUALITIES 4. 3x + 2y z = 10 2x 4y + 4z = -2 2x + 4y 3z = w + 3x 4y + z = -3 -w + 2x + 3y 2z = 5 3w x 2y + 3z = 4-5w + 7x + 8y + 5z = - 12 Sometimes one is faced with an inequality, for example, 6x + 3y > 17. What do we do then? The primary factor you need to remember is that an inequality basically put you on one side of a line or on the other side. Every point on one side of the line is the greater to or equal and every point on the other side of the line is less than or equal. What line? Well, the line that you would get with the equality. So first turn the inequality into an equality by replacing the inequality with an equal sign, solve the equality to get the equation, and then figure out which side of the line satisfies the inequality. Continue the example. 6x + 3y > 17 Replace the inequality with an equality. 6x + 3y = 17 Now solve the equation (in this case, the equation is linear and is easy to graph). y = (17/3) 2x Graph the equation. We know the y intercept, where the line crosses the y axis; that is 17/3. We can find the x intercept, where the line crosses the x axis by setting y to zero and finding x. In this case you should get x = 17/6. We now have two points and a straight line is determined (see below). The only question now is which side of the line is which? We need a test point. The easiest point to take is the origin. So convert the equation back to its original inequality form and plug in zero for both x and y. 6x + 3y > 17 6*0 + 3*0 > 17?

16 17/3 17/6 This inequality is not true! So we know that the points where the inequality are satisfied are above the line. Exercises: 1. 7x 4y < x + 3y > x > y > x x y < x 2 6x + 2 TWO COMMON MISTAKES There are some common mistakes students make. I. (x + y) n = x n + y n This statement is not true unless n = 1. For n = 2, we already know that (x + y) 2 = x 2 + 2xy + y 2. This mistake occurs because students do not carefully read the notation. (x + y) n is (x + y) times itself n times. (x + y) n = ( x + y)( x + y)...( x + y) n

17 You can clearly see that this will not be x n + y n 2 x + 3x 7 II. In, we cannot divide the x 2 2 x/ + 3x 7 to get. 2 2 x x/ Suppose you had had some numbers rather than symbols, say You know you cannot divide the 10. That is this ratio 10 is not equal to ; the ratio is equal to 2. So you cannot do the division as shown. One must divide the whole numerator by the denominator, not just one term in the numerator. What follows is not an algebra review in the strict sense, but it is a way to find growth rates when one is a bit math phobic. Connections to more powerful methods are noted in the discussion. Growth Rates and How to Compute Them I. WHAT DO WE MEAN BY THE GROWTH RATE? Δ y By the growth rate of a variable y, I mean the percentage change in y,, y where Δ stands for the change in. Suppose we let y stand for real GDP. Δ y Then the growth rate of y is. If we say the growth rate is constant, we mean y that each year, the percentage change in y is the same as it was the last year. Note that that means that y is changing and that the change in y, Δy, is changing too. How do we know that? EXERCISES: Write an expression for the growth rate of the following variables a. x b. y*z c. y/z ADVANCED: For those who know calculus, it is easy to see that the growth rate dy is. But this is just d(ln y) where ln stands for the natural log function. If we y assume that the growth rate is constant, then d ln y = k. If we integrate both sides, we get d ln y = kdx ln y = k( x xo ) k ( x xo ) kx o y = e = Ae where A= e kx This for the growth rate to be constant, y must by the exponential function with the growth rate as part of the exponent.

18 II. FINDING Δf In this section we will show that if f = h*g, then Δf = h*δg + g*δh. Suppose that we have a product of two functions, and we know the growth rate of the each function. What is the growth rate of their product? In particular, we have f = g*h. What is the growth of f given that g has a constant growth rate of α and h has a growth rate of β? Will the growth of f be α*β or α + β or will we not be Δf Δ able to find the growth rate of f? The growth rate of f is or ( x * y). To find f x * y the growth rate, we need to find Δf and then divide by f. In this step, we find Δf. Suppose that g and h each change. The change in g is Δg while the change in h is Δh. Thus, after the change, g becomes g + Δg and h becomes h + Δh. After the changes, the new value of f is (g + Δg)*( h + Δh). We can now find the value of Δf, the change if f. The change if f will be the new value of f minus the old value of f. Δf = (g + Δg)*( h + Δh) g*h = g*h + h*δg + g*δh + Δg*Δh g*h = h*δg + g*δh + Δg*Δh. We will neglect the value of Δg*Δh. because it is likely to be small. We can see in the following graph what is happening. The change in f, Δf, is the L shaped area within the dotted lines. This area is made up of three sections, h*δg, g*δh, and Δg*Δh. g* Δh h + Δh Δg*Δh h h*δg g g + Δg Thus we take the change in f to be the two areas g*δh and h*δg. In short, we have Δf = h*δg + g*δh. EXERCISES: Fill in the following table:

19 h g Δh Δg Δf case case case ADVANCED: If you know calculus, you will recognize that what we have done here is the product rule of differentiation. The derivative of the product is the sum of the derivatives. III. IF f = g*h, THEN THE GROWTH RATE OF f IS THE GROWTH RATE OF g PLUS THE GROWTH RATE OF h. Δ f We are looking for the growth rate of f which is. In part II, we found Δf. So f Δ f g * Δh + h * Δg g * Δh h * Δg Δh Δ we can easily find = = + = +. f g * h g * h g * h h gg Δ f Δh Δ In short, = + or the growth rate of f equals the growth rate of h plus f h gg the growth rate of g. EXERCISES: Fill in the following table. growth of h growth of g growth of f case case case ADVANCED: If you recall from part I that constant growth path equations are the exponential functions, then it is easy to see that the above is true. Suppose g = e αx and h = e βx. Then if we want to find the growth rate of g*h, we want the growth rate of e αx * e βx = e (α+β)x. Thus the growth rate of g*h is (α+β). IV. FINDING Δ(1/h). In this section, we will use a trick to find that Δ(1/h) = -(1/h) 2 *Δh. Here is the trick. h * (1/h) = 1. This statement is always true. It is an identity. Thus when we find Δ of one side, it will equal Δ of the other side. This will only be true for identities. Now note that we have a product, and we may use what we developed in part II. Δ[h* (1/h)]) = h*δ(1/h) + (1/h)*Δh. This is Δ of the left hand side. Now this term must be equal to Δ of the right hand side. But the right hand side is 1 and does not change. Thus Δ of the right hand side is zero. Thus we have the following.

20 h*δ(1/h) + (1/h)*Δh = 0 Solve for Δ(1/h). h*δ(1/h) = -(1/h)*Δh Δ(1/h) = -(1/h) 2 *Δh This is the term we claimed we were to get, so we are done. V. IN THIS SECTION WE WILL SHOW THAT Δ(g/h) = (1/h)*((-g/h)*Δh + Δg). Notice that h g can be written as g * (1/h). We can use the result of II to write Δ g 1 = g*δ h 1 + *Δg h h 1 Now from our result in IV, we have an expression for Δ. Use it now. h Δ g 1 = g* Δh h h * *Δg = g * Δh + Δg. h h h Thus, we have the desired result. Δ g 1 = h * g * Δh + Δg. h h Alternative: My colleague, Kevin Quinn, points out that there is an easy way to Δg Δf Δh get this result. If f = g/h, then g = f*h. We already know that = +. It g f h Δf Δg Δh is a matter of subtraction to find =. f g h VI. THE GROWTH RATE OF g/h IS THE GROWTH RATE OF g MINUS THE GROWTH RATE OF h. g Δ h The growth rate of g/h is g. In part V, we computed the numerator, Δ. g h h Now we divide that expression by g/h and see what we get.

21 g Δ h = g h g h 1 Δ * = * g h * Δh + Δg * h g h h g EXERCISES: 1 Δh Δg = * * Δh + Δg = +. g hg h g growth of h growth of g growth of f case case case