Make graph of g by adding c to the y-values. on the graph of f by c. multiplying the y-values. even-degree polynomial. graph goes up on both sides

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1 Reference 1: Transformations of Graphs an En Behavior of Polynomial Graphs Transformations of graphs aitive constant constant on the outsie g(x) = + c Make graph of g by aing c to the y-values on the graph of f constant on the insie g(x) = f(x + c) Make graph of g by subtracting c from the x-values on the graph of f multiplicative constant g(x) = c Make graph of g by multiplying the y-values on the graph of f by c g(x) = f(cx) Make graph of g by iviing the x-values on the graph of f by c En behavior of polynomial graphs positive leaing coefficient even-egree polynomial graph goes up on both sies o-egree polynomial graph goes up on right, own on left negative leaing coefficient graph goes own on both sies graph goes own on right, up on left page 1 of 62

2 Reference 2: Facts About Limits from Section 21 If lim = L an lim g(x) = M exist, where L an M are real numbers, then Theorem 21: lim k = k Theorem 22: lim x = c Theorem 23: lim [ + g(x)] + lim g(x) = L + M Theorem 24: lim [ g(x)] lim g(x) = L M Theorem 25: If k is a constant, then lim k = k lim = kl Theorem 26: lim ( g(x)) = (lim ) (lim g(x)) = L M Theorem 27: If lim g(x) = M 0, then lim Theorem 28: Theorem 3: Definition: Theorem 4: = lim g(x) lim g(x) = L M If n is a positive o integer, or if n is a positive even integer an lim = L > 0, n n then lim = lim If is a polynomial, then lim = f(c) If r(x) is a rational function an c is a real number that is in the omain of r(x), then lim r(x) = r(c) If lim = L = 0 an lim g(x) = M = 0, then then lim sai to have the ineterminate form 0 In this case, Theorem 27 0 cannot be use to etermine the limit g(x) is If lim = L 0 an lim g(x) = M = 0, then lim oes not g(x) exist accoring to the Definition of the Limit foun in Section 3-1 (Remark: In Section 22, the Definition of Limit gets expane, with the result that in some cases, we will say that the limit is infinity or negative infinity an will write lim = or lim = But g(x) g(x) this will not always be the case That is, sometimes the limit will not exist even with the expane efinition of limit) page 2 of 62

3 Reference 3: Contrasting Limit Methos from Sections 21 an 22 for Fining the Limit at x = c In Section 21, the limit at a particular x = c was efine in a way that the limit coul only exist if the limit was a number Otherwise, the limit i not exist In Section 22, the efinition of limit was expane The limit at a particular x = c coul still be a number (as in Section 21), but in some cases the limit coul turn out to be or (In these cases the limit woul not have existe using the more restrictive efinition of limit from Section 21) But even in Section 22, it is possible for the limit to not exist But when the limit i not exist using Section 22 methos, it was usually not for the same reason that the limit i not exist using Section 21 methos The three functions f, g, h presente here contrast the techniques an results of Sections 21 an 22 Function = function value at x = 5 Limit at x = 5 Using 21 Methos lim f(5) = (x 2)(x 5) (x 5) g(x) = x 2 x 5 (5 2)(5 5) (5 5) = 0 0 DNE (x 2)(x 5) (x 5) (x 2) ( ) = 5 2 ( ) = 3 ( ) You must justify these steps! g(5) = = 3 0 DNE Observe that lim (x 2) = 3 an that lim (5 2) = 0 Since lim (numerator) 0 an lim (enominator) = 0, Section 21 Theorem 4 tells us that lim g(x) = DNE h(x) = x 2 (x 5) 2 h(5) = 5 2 (5 5) 2 = 3 0 DNE Observe that lim (x 2) = 3 an that lim (5 2) = 0 Since lim (numerator) 0 an lim (enominator) = 0, Section 21 Theorem 4 tells us that lim h(x) = DNE Left Limit at x = 5 using 22 Methos Same metho an same result as in Section 21 That is, lim = 3 When x is close to 5 but slightly less than 5, the y-value will be y = number close to 3 neg num very close to 0 = very large neg number Conclue lim g(x) = When x is close to 5 but slightly less than 5, the y-value will be y = number close to 3 pos num very close to 0 = very large pos number Conclue lim h(x) = When x is close to 5 but slightly When x is close to 5 but slightly Right Limit at x = 5 using 22 Methos Same metho an same result as in Section 21 That is, lim + = 3 greater than 5, the y-value will be y = number close to 3 pos num very close to 0 = very large pos number greater than 5, the y-value will be y = number close to 3 pos num very close to 0 = very large pos number Conclue lim + g(x) = Conclue lim + g(x) = Limit at x = 5 using 22 Methos Same metho an same result as in Section 21 That is, lim = 3 Because the one-sie limits on t match, we conclue that g(x) DNE lim Because the one-sie limits match, we conclue that lim g(x) = page 3 of 62

4 Reference 4: Limits as x an What They Tell Us About Graph Behavior In the previous Reference 3, we foun the limits as x c where c is some real number Observe that in all three of the examples, we use the factore form of the functions f, g, h When investigating substituting a particular real number into a function, the factore form of the function is the best form to use, because it makes the arithmetic easier But when fining the limit as x, it is much better to use the stanar form of the function, because that form makes it easier to ientify the ominant terms In the three examples belwo, we will fin the limits as x of three rational functions f, g, h Notice that we use the stanar forms of the functions when fining the limits as x The results of these limits allow us to make conclusions about corresponing graph behavior Function factore form Function stanar form Limit as x Graph Behavior = 7(x 1)(x 5) 7(x 1)(x 5) 2(x 3)(x 5) g(x) = 2x(x 3)(x 5) = 7x2 42x x 2 16x + 30 g(x) = 7x2 42x x 3 16x x 7x 2 42x + 35 lim x x 2x 2 16x x 2 x 2x 2 7 x 2 = 7 2 Conclue that the graph of has a horizontal asymptote on the right with line equation y = 7 2 Since is a rational function, we know that it will also have a horizontal asymptote on the left with the same line equation 7x 2 42x + 35 lim g(x) x x 2x 3 16x x 7x 2 x 2x 3 7 x 2x Notice that numberator of the fraction is fixe at 7, while the enominator is getting huge an postive So as x gets bigger & bigger, the fraction will be getting closer an closer to zero Therefore, lim g(x) x x 7 2x = 0 Conclue that the graph of g(x) has a horizontal asymptote on the right with line equation y = 0 Since g(x) is a rational function, we know that it will also have a horizontal asymptote on the left with the same line equation h(x) = 7x(x 1)(x 5) 2(x 3)(x 5) h(x) = 7x3 42x x 2x 2 16x x 3 42x x lim h(x) x x 2x 2 16x x 3 x 2x 2 7x x 2 Notice that numberator of the fraction is getting huge an postive, while the enominator is fixe at 2 So as x gets bigger & bigger, the fraction will be getting more an more positive, without boun Therefore, 7x lim h(x) x x 2 = Conclue that the right en of the graph of h(x) goes up There is no horizontal asymptote on the right Since h(x) is a rational function, we know that it will also not have a horizontal asymptote on the left (A calculation similar to the one above woul fin that lim h(x) 7x x x 2 = That tells us that the left en of the graph of h(x) goes own page 4 of 62

5 Reference 5: Definitions of Rates of Change Definition of Average Rate of Change wors: the average rate of change of f as the input changes from a to b usage: f is a function that is continuous on the interval [a, b] meaning: the number m = f(b) f(a) b a graphical interpretation: The number m is the slope of the secant line that touches the graph of f at the points (a, f(a)) an (b, f(b)) remark: The average rate of change m is a number Definition of Instantaneous Rate of Change wors: the instantaneous rate of change of f at a alternate wors: the erivative of f at a symbol: f (a) f(a+h) f(a) meaning: the number m h0 h graphical interpretation: The number m is the slope of the line tangent to the graph of f at the point (x, y) = (a, f(a)) remark: The instantaneous rate of change f (a) is a number Definition of the Derivative wors: the erivative of f symbol: f meaning: f is a function To escribe a function, one must show how it prouces output for a given input For an input x, the output is the number f (x) f(x+h) h0 h graphical interpretation: For an input x, the output f (x) is the number that is the slope of the line tangent to the graph of f at the point (x, y) = (x, ) remark: The erivative f is a function Terminology of Position an Velocity Time: When our book uses mathematical functions to escribe the motion of objects, x is a variable that represents the elapse time Position: To say an object is moving in 1 imension means that it can go forwar or backwar in one irection but cannot turn In such situations, a single coorinate can be use to keep track of the position of the object A function calle the position function gives the value of the coorinate at a given time In our book, the position function is calle f That is, at time x, the coorinate of the object is the number average velocity: The wors the average velocity from time x = a to time x = b mean the same thing as the average rate of change of position from time x = a to time x = b That is, the number m = f(b) f(a) b a instantaneous velocity: The wors instantaneous velocity at time x = a mean the same thing as instantaneous rate of change of position at time x = a That is, the number m = f (a) = lim h0 f(a+h) f(a) h velocity: The wor velocity means the same thing as the wors erivative of the position function That is, the velocity is the function f page 5 of 62

6 Power Rule for Derivative Sum an Constant Multiple Rule Exponential Function Rule #1 Exponential Function Rule #2 Exponential Function Rule #3 Logarithmic Function Rule #1 Log Rule #2 (larger omain) (not use in our course) Logarithmic Function Rule #3 Reference 6: Derivative an Inefinite Integral Rules Prouct Rule Quotient Rule Chain Rule x xn = nx n 1 (a + bg(x)) = af (x) + bg (x) x x e(x) = e (x) x e(kx) = ke (kx) x b(x) = b (x) ln(b) x ln(x) = 1 x x ln x = 1 x x log b(x) = 1 x ln(b) (omain x > 0) (omain x 0) (omain x > 0) x ( g(x)) = f (x) g(x) + g (x) top(x) x bottom(x) = top (x) bottom(x) bottom (x) top(x) (bottom(x)) 2 outer(inner(x)) = outer (inner(x)) inner (x) x Power Rule for Inefinite Integral x n x = xn+1 + C when x 1 n + 1 Sum an Constant Multiple Rule a + bg(x)x = a ( x) + b ( g(x)x) Exponential Function Rule #1 e (x) x = e (x) + C 1 Exponential Function Rule #2 e (kx) x = e(kx) k + C Rule (not use in our course) 1 x = ln(x) + C (omain x > 0) x x 1 Rule (larger omain) 1 x = ln x + C (omain x 0) x x Logarithm Function Rule ln(x) x = x ln(x) x + C (omain x > 0) page 6 of 62

7 Reference 7: Business Terminology In our course, we will stuy hypothetical business examples in which a company makes an sells some item The simplifying assumptions are The items are manufacture in batches All of the items manufacture are sol, an they are all sol for the same price per item Here is the Business Terminology that we will be using Deman, x (small letter), is a variable that represents the number of items mae This souns simple enough, but there can be complications For example, in some problems, x represents the number of thousans of items mae Price, p (small letter), is a variable that represents the selling price per item The Price Deman Equation is just what it says: an equation that relates the Price p an the Deman x For example 2x + 3p = 10 coul be a Price Deman Equation In some situations, the Price Deman Equation can be solve for one variable in terms of the other For example, the equation above can be solve for p in terms of x It woul rea p = 2 10 x + When this 3 3 is one, notice that the equation escribes Price p as a function of Deman x We coul use function notation to inicate this, writing p(x) = 2 10 x Revenue, R (capital letter), is the amount of money that comes in from the sale of the x items that are mae Because of our simplifying assumptions liste above, we can say that Revenue = (number of items sol) (selling price per item) Revenue = Deman Price R(x) = x p(x) Cost, C(x) (capital letter C), is a function that gives the cost of making the batch of x items We say that the company Breaks Even when Revenue = Cost That is, when R(x) = C(x) Profit, P(x) (capital letter P), is a function efine as follows Profit = Revenue Cost P(x) = R(x) C(x) The expression Average Quantity, enote by the symbol Quantity, means Quantity That is, Average Revenue is R (x) = R(x), Average Cost is C (x) = C(x), an Average Profit is P (x) = P(x) x x The expression Marginal Quantity means The Derivative of Quantity That is, Marginal Revenue is R (x), an Marginal Cost is C (x), an Marginal Profit is P (x) The wor Marginal can also be put in front of the Average Quantities That is Marginal Average Revenue is R (x), an Marginal Average Cost is C (x), an Marginal Average Profit is P (x) x x page 7 of 62

8 Reference 8: Derivative Relationships f is positive on (a, b) f is negative on (a, b) f is zero on (a, b) f is positive at x = c f is negative at x = c f is zero at x = c Derivative Relationships on an interval (a, b) f is positive on (a, b) f is negative on (a, b) f is zero on (a, b) f is increasing on (a, b) f is ecreasing on (a, b) f is constant on (a, b) f is increasing on (a, b) f is ecreasing on (a, b) f is constant on (a, b) f is concave up on (a, b) f is concave own on (a, b) f is a straight line on (a, b) Derivative Relationships at a particular x = c f is positive at x = c f is negative at x = c f is zero at x = c of f at x = c slopes upwar of f at x = c slopes ownwar of f at x = c is horizontal of f at x = c slopes upwar of f at x = c slopes ownwar of f at x = c is horizontal f is concave up at x = c f is concave own at x = c page 8 of 62

9 Reference 9: The Substitution Metho for Inefinite Integrals Given an inefinite integral F(x) = x with an integran that involves a neste function (1) Ientify the inner function It will be a function of the variable x Write the equation inner(x) = u to introuce the single letter u to represent the inner function Circle the equation inner(x) = u (2) Fin u u u Your result shoul be an equation of the form = inner (x) Then use x x x to buil a new equation that expresses x in terms of u To o this, write x = 1 u ( u x ) 1 That is, x = u Circle this new equation x = 1 u inner (x) inner (x) (3) Notice that in steps (1) an (2) you have two circle equations Substitute these into the integran of your inefinite integral Cancel as much stuff as possible an simplify by moving multiplicative constants outsie the integral The result shoul be a new inefinite integral with an integran that is a function involving the variable u Important things to note at this point: There shoul be no x in the new inefinite integral If there is an x in the integral, then either you have mae a mistake or the original integral was not one that can be solve using the metho of substitution The new inefinite integral oes not involve a neste function (4) Solve the new inefinite integral by using the antierivative rules The result shoul be a new function involving u (5) Substitute u = inner(x) into your function from step (4) The result shoul be a new function of just the variable x This is F(x) (6) Check by fining F (x) If F (x) =, then your work was correct page 9 of 62