This occurs only when dx x = In fact, you’ve found the minimum of this function It occurs when x = 0; something that is easy to see upon graphing 4x2 + d ( x + 8x )=8x+8 Setting this equal to zero gives 8x + = Subtracting from both dx sides and dividing by gives x = -1 Once again you’ve found a minimum that occurs when x = -1 You should recognize this as an equation for a line As all lines have a constant slope, one hopes the derivative of this function is a constant In fact, the derivative of this IS a constant d (12x + 6) = 12 This is similar to using derivatives to find the MPC dx Algebra: A number of students have had trouble in class before because they have forgotten some common algebra rules Most of these rules involve the use of exponents; this section is intended to remind you of the correct use of exponents Typical 307 problems that use exponents include multiplying and dividing numbers with exponents For instance: 23×24 or 34/33 The rules dealing with exponents are that exponents, when pertaining to the same base number, are either added or subtracted when depending on if the base numbers are multiplied or divided When the base numbers are multiplied, as in the case of 23×24, the exponents are added without a change of the base numbers 23×24 = 27 When the base numbers are divided, as in the case of 34/33, the exponents are subtracted 34/33 = 31 Another frequently forgotten procedure includes how to deal with negative exponents A negative exponent indicates nothing more than an inverse In other words, 3-1 is nothing more than (1/3) and 3-2 is equal to (1/32 = 1/9) Practice with a few exponent problems: x ×x 32×23 32×22 10 y y 11 12 12 Answers: Because (1/3) + (2/4) = 5/6, x × x = x This problem has no way of easily simplifying because the base numbers differ and the exponents differ The only way to simplify is to actually take 32 (9) and 23 (8) and multiply them together (72) 62 −1 10 = y y 11 = 12 −3 12 We will use these procedures frequently when attempting to solve algebraic formulae that contain exponents The most frequent type of problem we will encounter are of the form: x3 = 2x2 To solve this problem for x, one must follow the common rules of algebra, the first being to get all of the common unknowns (in this case x) to one side I this by dividing x2 from both x3 sides and getting: = x Since we have just seen a way to simplify the left-hand side of this equation, it is straightforward to find that x = Here are some more for you to practice: 1 3x = x 12 13 = x 2 x 14 x = 2x Answers: 1 x x 6x = x = = x Since we are solving for x (and not x to x the one-half), we need to “get rid of the one-half power exponent To this, we will square 12 3x = both sides of the equation: = x 62 = x 2 What is x 2 ? This is simply another multiplication problem as above: x 2 = x × x 2 =x +1 2 = x = x The answer to this problem therefore is x = 36 3 2 13 = x = = × = x x x x Like above, we are solving for x, and 1 2 x x not x to the three-halfs How we get rid of this exponent? Well, in the answer to 12, we squared both sides of the equation and found that we ended up with x1 Why did we square? Because we wanted to eliminate the one-half power and squaring did the job It turns out, that we always want to take the inverse of the power we want to eliminate In this case the power we want to eliminate is three-halfs so I will take the two-thirds power of both sides: Now it turns out = 3.301 but what is x 2 3 2 =x 3×2 2×3 = x = x In this problem, x = 3.301 x = 2x 6x = x 6x =1 x 2 ? It turns out that we multiply exponents when we have an exponent to an exponent In this case x 14 = x = x 2 = x= 36 ... most frequent type of problem we will encounter are of the form: x3 = 2x2 To solve this problem for x, one must follow the common rules of algebra, the first being to get all of the common unknowns... solving for x (and not x to x the one-half), we need to “get rid of the one-half power exponent To this, we will square 12 3x = both sides of the equation: = x 62 = x 2 What is x 2 ? This is simply... solving for x, and 1 2 x x not x to the three-halfs How we get rid of this exponent? Well, in the answer to 12, we squared both sides of the equation and found that we ended up with x1 Why did we