December Exam Summary

December Exam Summary 1 Lines and Distances 1.1 List of Concepts Distance between two numbers on the real number line or on the Cartesian Plane. Increments. If A = (a 1, a 2 ) and B = (b 1, b 2 ), then d(a, B) = ( x) 2 + ( y) 2 = (b 1 a 1 ) 2 + (b 2 a 2 ) 2. The midpoint M between A = (a 1, a 2 ) and B = (b 1, b 2 ) is given by: ( a1 + b 1 M =, a ) 2 + b 2 2 2 The slope between two points (x 0, y 0 ) and (x 1, y 1 ) is: Equation of a line. m = y x = y 1 y 0 x 1 x 0 (y y 0 ) = m(x x 0 ) or y = mx + b. (1) Obtaining the line equation from a point and a slope or from two points: (y y 0 ) = m(x x 0 ) with m = y 1 y 0 x 1 x 0. (2) Parallel lines: Two lines y y 0 = m 0 (x x 0 ) and y y 1 = m 1 (x x 1 ) are parallel if and only if the slopes are equal: m 0 = m 1. Perpendicular lines: Two lines y y 0 = m 0 (x x 0 ) and y y 1 = m 1 (x x 1 ) are perpendicular if and only if the product of their slopes is 1: m 0 m 1 = 1. 1.2 Suggested Exercises from the Textbook Section 0.2: All practice problems except for Practice 3 and Problems 1 to 8, 10a, 10bc, 11, 15 to 18, 19, 20, 21, 27, 28, 29. 1

1.3 More Exercises Compute the cordinates of the midpoint M between the two points A = (x a, y a ) and B = (x b, y b ) and show it is equidistant from A and B, that is dist(a, M) = dist(m, B). Show how the Pythagorean theorem on right triangles (A 2 + B 2 = C 2 ) appears in the distance equation for two points using a figure to support your argument. 2 Functions 2.1 List of Concepts The definition of a function: A function f takes an element from a set of inputs X and gives out an output Y. For any given input x there is only one possible output f(x). You can check this on a graph with the vertical line test. Identifying the domain a function. The domain of a function f is the set of all possible inputs. That is all the values of x such that f(x) exist. In general they exclude inputs that would divide by zero or have a negative number exponent 1 n, for an even number n, but there could be predetermined restrictions on the domain, such as in some piecewise functions. If we have, for example, f(x) = then the domain of f is [0, 3). { 2x + 5 if x [0, 2] 8 if x (2, 3), The range of a function f are the set of all possible outputs f(x). Reading and building graphs of functions. Each point on the graph of a function f(x) corresponds to one of the (x, f(x)) pairs. The input is always on the horizontal coordinate and the output is on the vertical one. Sketching the graph of a function can help you finding where it is increasing or decreasing as well as where the limit exists or not. It can be a good idea to sketch the graph of a function. Not all graphs are graphs of a function. Remember the vertical line test: if one input has more than one output, then we are not looking at the graph of a function (a circle is not a function, for example). Combining functions through addition, multiplication, division (rational functions). (3) 2

Piecewise functions: A piecewise function is a function whose formula, or rule, changes depending on the value of the input. Example: The function given in (3) is a piecewise function. Example: Absolute value function { x if x [0, ) x = x if x (, 0), Whenever there is an absolute value in a function, you should have the reflex of writing it as a piecewise function. Example: 1 x + 1 = { { 1 x + 1 if 1 x + 1 > 0 1 1 x 1 if 1 x + 1 < 0 = x + 1 if x (, 1] (0, ) 1 x 1 if x ( 1, 0), Function composition: These are invoked when we use the output of a function f, f(x), into another function g. The composition of g(x) with f(x) is then noted g(f(x)). To compute the value of a function composition, you first compute the value of the inside function f(x) and input that value into g. To find the domain of a function composition g(f(x)) you need both domains of f and g. The domain of the composition is then the domain of f without all the inputs x such that f(x) is not in the domain of g. Example: Find the domain of g(f(x)) with g(x) = 1 x and f(x) = x2 1. The domain of f is all the real numbers and the domain of g is R/0. Thus the domain of the composition g(f(x)) = 1 x 2 1 are all R (domain of f) except for values for which f(x) = 0. We see that x 2 1 = (x 1)(x + 1) = 0 if x = 1, 1 so the domain of the composition g(f(x)) is (, 1) ( 1, 1) (1, ). 2.2 Suggested Exercises from the Textbook Section 0.3: Practice problems 3, 4, 5 and Problems 1 to 5, 10, 11, 20 Section 0.4: Practice problems 1 to 4 and Problems 1 to 9, 10, 11, 12, 13, 15 to 18 (except for the bracket function). 2.3 More Exercises Find the domain of the function x2 1 3 3x f(x) = 3 + 2x + x x 5 by finding the domain of each term in the sum, then combining them. 3

Write the formula and sketch the graph of a piecewise function going through the points ( 5, 12), ( 1, 4) and (3, 8). Write the formula of a piecewise function going through the points (x a, y a ), (x b, y b ) and (x c, y c ). Let f(x) = 1 x, g(x) = x and h(x) = 2x + 1. Find a formula for the function f(g(h(x))). What is its domain? 3 Position, Velocity and Rates of Change 3.1 List of Concepts Definition of average velocity on a time interval [t 1, t 2 ] from the ratio of increments. v avg = x(t) = x(t 2) x(t 1 ). t t 2 t 1 This is the slope of the secant line between (t 1, x(t 1 )) and (t 2, x(t 2 )) (straight line between the two points). Definition of instantaneous velocity: Instantaneous velocity v(t o ) is the slope of the tangent line to x(t) at t = t o. The slope of the tangent line is obtained by computing the slope of the secant line for smaller and smaller interval until we are essentially at one point. Reminder. Average is on an interval [t o, t 1 ] and Instantaneous is on a specific time t o. Definition of growth of some quantity over time. For example population growth is the rate of change in population over time. Growth and velocity are rates of change of some function (quantity, position and velocity respectively) over time. Other rates of change (Profits over cost, temperature over energy for example) Computation of average growths or velocities from a table of values. Just use the secant slope formula with the two points at the boundaries of your time interval. Estimation of the instantaneous velocity from a position table of values. Just compute the average velocity for the smallest interval containing t o. If there are two of them, take the average of the two or combine the two intervals and then compute the average velocity (both give the same result). Estimation of the instantaneous growth from a quantity table of values. 4

Units of rates of change. For example you can measure velocity in meters per second (m/s), acceleration in meters per second squared (m/s 2 ) and the growth of a stock value in dollars per hour ($/h). Difference quotients and their relation with the slope of the tangent line. A difference quotient is a quotient of the form: x(t o + h) x(t o ). h They are obtained by computing secant line slopes for (t o, x(t o )) and (t o + h, x(t o + h)). You get towards the slope of the tangent line if you compute difference quotients for smaller and smaller values of h (both negative and positive). The slope of the tangent line of a function gives the rate of change of that function. If the slope is positive the function is increasing and if the slope is negative the function will be decreasing. If the slope of the tangent line is zero the graph will be horizontal (flat) there. Being able to identify position and velocity given two graphs. 3.2 Suggested Exercises from the Textbook Section 1.0: All practice problems and problems at the end of the section. You might need a calculator to solve some of those problems. While they are worth doing, these computations won t be necessary in quizzes or exams. Be careful as sometimes a problem looks like it might need the use of a calculator, but in fact there is a trick to solving it without one. 3.3 More Exercises Using sketches, show how the slope of the secant line on a position graph gives the average velocity over a time interval [t 1, t 2 ] and the slope of the tangent line gives the instantaneous velocity at a time t = t o. Here is a table of profit over a week at a boutique. Monday is day 1 and Sunday is day 7. time (day) Profit ($) 1 78 2 62 3 65 4 72 5 91 6 135 7 150 5

On what day did we observe the greatest profit? Between which two days did we observe the greatest profit growth? What is that growth? What is the unit of profit growth? Let p(s) = 1 2+s. Compute the difference quotient p(s+h) p(s) 2 h value of h. Simplify your solution as much as possible. for some 4 Limits 4.1 List of Concepts Definition of a limit (in your own words) and notation. lim f(x) = L x a This is the limit of f(x) as x goes to a. This means that if our input gets closer and closer to a our output gets closer and closer to L. The limit is always a single value and x approaches a from both directions. The limit of f(x) as x approaches a has nothing to do with the value f(a). It is possible to have any combination of existing or non existing limit and f(a). Example: for f(x) = x x, f(0) is not defined but the limit exists: lim f(x) = 1. x 0 Computing limits graphically and numerically for polynomial functions, rational functions and piecewise functions of polynomials and rational functions. If we are provided the graph of a function, the limit lim x a is the value on the vertical axis that the curve leads to when your input approaches a, provided that value is the same from both directions. It does not depend the designated value of the function at a. In our previous example, we would have a circle at x = 0 but since the function leads us to 1 near x = 0 (but not exactly at 0) we have our limit. If we are provided with a formula we can input values closer and closer to a (but never exactly a) and see if we are approaching a single value from both directions. Sometimes that value is not clear (if it is a square root for example) and we can only estimate the limit in that case. One-sided limits. When we are only looking at values of x close to a from one side we can compute one-sided limits. If we need to find lim x c ± f(x) then if the sign is we only look to the left of c (c < 0) and if the sign is + we only look to the right of c (c > 0). 6

We say that the limit lim x a f(x) exists if and only if both one-sided limits exist and are equal. If and only if means that you cannot have one without the other. Knowing when a limit does not exist. A limit does not exist if the onesided limits do not match, if either one-sided limit does not exist or if the function is not defined around x = a. Example: lim t 1 t DNE because t is not defined anywhere near 1. The near is important here as we are not interested at exactly 1. { 1 if x 2 Example: Let f(x) = If we take the left limit at x approaching 2, where the function has the x formula, we x if x < 2. find lim f(x) = lim x = 2. x 2 x 2 If we take the right limit at x approaching 2, where the function has the 1 formula, we find lim f(x) = lim 1 = 1. x 2 + x 2 + The two limits are not equal so we say lim x 2 f(x) DNE, even if we have f(2) = 1. Example: Find the limit We know that if x 3 2x 2 6x lim t 3 x 3. (4) 2x 2 6x x 3 = (x 3)(2x) x 3 = 2x. Thus when we take the limit, where we are not looking at x = 3 the limit should be 2x 2 6x lim = lim 2x = 6. t 3 x 3 t 3 So we have a limit even though the function is not defined at x = 6. Limit theorem for two functions that are equal everywhere except at one specific input (No Name Theorem). If two functions f(x) and g(x) are equal everywhere except if x = a and lim x a f(x) = L then lim x a g(x) = L. Know that the limit of a sum is the sum of the limits if they exist, similar for subtraction and multiplication. Also true for division if we don t divide by two and roots if we avoid negative numbers for even roots. These are used to compute limits of more complicated functions. For example, for 7

the division rule we could still have a limit if L = 0, like in our previous example, but we could also have no limit, as 1 lim x 0 x = DNE. Direct Substitution theorem: if f(x) is either a polynomial or a rational function of polynomials such that a is in its domain, then we have lim f(x) = f(a). x a This is a direct result of our limit laws. Algebraic method to compute limits. We can solve limits algebraically if we simplify them to a form that is compatible with direct substitution or a limit law. 4.2 Suggested Exercises from the Textbook Section 1.1: All practice problems and problems 1 to 6, 8 to 11, 13, 16a), 17a), 18, 19, 20. You might need a calculator to solve some of those problems. While they are worth doing, these computations won t be necessary in quizzes or exams. Be careful as sometimes a problem looks like it might need the use of a calculator, but in fact there is a trick to solving it without one. Section 1.2: All practice problems except for practice 6 and problems 1 to 8, 9, 11, 16 to 18, 20 (try to use that list method for a challenge). 4.3 More Exercises Let the following piecewise function for some constant b { 2x + 4 if x 2 f(x) = b if x < 2, (5) Compute the one-sided limits as x goes to 3, 2 and some x o (any constant). For what value(s) of b does all the limits (not one-sided) exist? (Challenging) Let the following piecewise function { x 2 1 if x Q f(x) = 1 if x / Q. (6) Does the limit lim x 0 f(x) exist? If so, what is its value? What theorem did you use to show that the limit of f(x) does or does not exist? 8

5 Continuous Functions 5.1 List of Concepts The definition of continuity at a single point: If 1. f(a) is defined, 2. lim x a f(x) exists and 3. lim x a f(x) = f(a), we say that the function f is continuous at x = a. If any of the three conditions fail, we say that f is not continuous or discontinuous at x = a. Continuous function on an interval. A function f is continuous on an interval [a, b] if it is continuous at c for all possible values of c [a, b]. Polynomials like x 8 25x + x 3 + 3, 3x + 10 and 3 are all continuous over all R, or (, ). This is a result of the Direct Substitution Theorem. By the same logic, rational functions of polynomials are continuous on their domain. Even if sometimes the limit exists where the function is not defined (like in (4)), but in that case, condition 1 is not satisfied and thus the function is not continuous there. Informally, you can say that a function is continuous over an interval if you can draw its graph without lifting your pen. Discontinuities thus include jumps and gaps in the graph. A function is not continuous if (1) it is not defined at x = a, (2) its limit at x = a is not defined or if both exist, but are not equal. Similar to how you can add, multiply limits you can add, multiply continuous functions to get other continuous functions. Be wary of divisions by zero or even roots of negatives! Example: If both f and g are continuous at x = a, then f + g is also continuous at x = a, f/g is continuous at x = a if g(a) 0, (f(a)) 1/n is continuous if f(a) > 0,... (Extra information! Not (directly) on the midterm) Continuity is preserved by composition, so if both f and g are continuous at x = a, then lim f(g(x)) = f( lim g(x)) = f(g(a)) x a x a and thus, f(g(x)) is also continuous. You can usually work that out separately using limit laws. Continuity of piecewise functions. If you need to check continuity of a piecewise function, there are two cases to worry about. 9

1. If you check at the boundary between two cases, you need to use onesided limits first to determine if the limit exists (criterion 2). The function will have different forms on each one-sided limit because you check right at where the change happens. Once you have that limit, you can check the second condition and then the other two. 2. If you are not at a boundary, you only worry about one form and you can compute the limit as though the function only had that form. You have to use the right one, though. The intermediate value theorem (IVT). If f is continuous on the interval [a, b] for every value L between f(a) and f(b) there is some number c [a, b] such that f(c) = L. This means that any continuous function f takes every value between f(a) and f(b). Is is important to show that the function is continuous on the interval before using the result of the theorem. x x Example: is continuous for every real number except for x = 0, but it has no roots. This means f(c) 0 for any c even if f( 1) = 1 is negative and f(1) = 1 is positive. IVT fails because the function is not continuous over [ 1, 1]. IVT is often used to find roots of a function. If the function is positive for some input value and negative for some other input value, then there must be a root in between those two inputs. Example: Let f(x) = x 2 1 We know that f( 2) = 3 > 0, f(0) = 1 < 0 and f(2) = 3 > 0. We also know x 2 1 is continuous for all real numbers, so we know there is a root between 2 and 0 and there is another one between 0 and 2. We know the two roots are distinct because f(0) 0 and 0 is the only common value in bth intervals. 5.2 Suggested Exercises from the Textbook Section 1.3: All practice problems and problems 1, 2, 3 a), b), f) and h), 5, 6, 7 a), b), d) and e), 12, 13, 14, 17 (you do not have to use the bisection method), 19, 20, 21, 22, 23. 5.3 More Exercises Let the following piecewise function for some constant b { 1 f(x) = x if x 2 x+4 x 2 16 if x < 2, (7) 10

Is the function continuous at x = 4? At x = 2? At x = 0? Over what intervals is the function continuous? Show that the function W (v) = v 8 5v 5 3v 8 has two roots in the interval [ 2, 2]. 6 Velocity, Tangent lines and Differentiability 6.1 List of Concepts Finding the slope of the tangent line of a function f(x) at x = a using the limit of the difference quotient: f(x + h) f(x) m tan = lim. h 0 h Finding a velocity at a time t = t 0, being given a position function x(t): Velocity units. x(t o + h) x(t o ) v(t o ) = lim. h 0 h Finding the equation of the tangent line to a function f(x) at x = a: Find m tan, then use the point-slope formula to find: y = m tan (x a) + f(a). If you are asked to find multiple values of the slope of the tangent line to a single function f, it might be worthwhile to first compute the derivative: f f(x + h) f(x) (x) = lim, h 0 h then use the function f (x) to compute your tangent slopes directly. That way you only need to compute a limit once. We say a function is differentiable at x = a if its derivative is defined at x = a (in other words if f (a) exists). A function needs to be continuous in order to be differentiable (remember that a jump discontinuity has no slope). Moreover a function needs to be defined where it is differentiable (f(x) is in the limit). A function can be continuous at a point but not differentiable there (corners). This also means that if a function is differentiable, then its derivative is continuous where it is differentiable (again think of corners). Figure 1 shows examples of functions that are continuous, but not differentiable. 11

y a b c x Figure 1: Functions that are continuous, but not differentiable: at x = a, the function has a vertical tangent line. This is called a cusp. At x = b we have another vertical tangent. This is a similar behaviour as the cube root function at x = 0. At x = c we have a corner, even though the function is decreasing on either side of it. 12

Relation between the sign of the derivative and increasing or decreasing functions: A function is increasing where its derivative is positive and it is decreasing where its derivative is negative. Example Compute the slope of the tangent line to f(x) = x 2 at x = 0, x = 1 and x = 5. Compute the equation of the tangent line to f(x) at x = 3 We could compute the slope at x = 1 first: f(0 + h) f(0) (0 + h) 2 0 2 m tan = lim = lim h 0 h h 0 h h 2 = lim h 0 h = lim h h 0 = 0. For the other two points we will compute the derivative first: f f(x + h) f(x) (x + h) 2 x 2 (x) = lim = lim h 0 h h 0 h x 2 + 2xh + h 2 x 2 = lim h 0 h = lim h 0 2x + h = 2x. = lim h 0 2xh + h 2 h Thus, the slope at x = 0 is f (0) = 0 and the slope at x = 5 if f (5) = 10. To compute the equation of the tangent line at x = 3, we first need a slope: m = f ( 3) = 6. We also know the point ( 3, f( 3)) = ( 3, 9) is on the line (tangent line intersects the graph at x=a), so we can use the point-slope method to find the line equation: y 9 = 6(x ( 3)) y = 6x 9. 6.2 Suggested Exercises from the Textbook Section 2.1 Practice problems 1, 2, 3 and problems 1 to 16, 18 to 20, 22 to 28, 32 to 34, 37. 6.3 More Exercises using the limit definition of the derivative, compute the derivative of the function f(x) = 1 x. Let U(t) := 2t 2 + 5t + 13, find all the values of a where the tangent line to U(t) has horizontal slope at t = a. 13

Let g(x) = x + 1. Knowing that g (3) = 1/4, compute the equation of the tangent line to g(x) at x = 3. Show, using an example with a diagram, why a function cannot be differentiable where it is not continuous. Let the following piecewise function for some constants a and b: { mu + b if u 2 T (u) = 2 u if u < 2, (8) Find the values of a and b that make T (u) differentiable on (, ). x Differentiate f(x) = x 3 +3 using the limit definition of the derivative. To differentiate means to find the derivative. Find the intervals where the function T (s) = 1 s + (x 3)2 is increasing and where it is decreasing. 7 Derivative Tools 7.1 List of Concepts Power rule Derivative of a constant k R d dx (xr ) = rx r 1, for x 0. (k) = 0 A function is differentiable at x = a if the derivative exists at x = a (if f (a) exists). The relation between the sign of the derivative and whether the function is increasing or decreasing. Sum and multiplication by a constant rules: (f(x) + g(x)) = f (x) + g (x) (k f(x)) = k f (x), for k R and provided f and g are differentiable where we are interested. Exponential functions f(x) = a x f (x) = ln(a) a x d dx (ex ) = e x 14

7.2 Suggested Exercises from the Textbook Here is an exhaustive list of problems from the textbook that were relevant last week. Problems in bold are more challenging, but worth a try. Section 2.2: Practice problems 1, 2, 4, 5, 11 and problems 1, 2, 13 to 16, 19, 20, 22 to 25, 28, 29, 34, 35, 38 to 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54. 7.3 More Exercises Find the equation of the line tangent to f(x) = e x and goes through the point ( 2, 1). Hint: (z + 1)e z = 1 if and only if z = 0 8 Basic Differentiation Rules: Sum, Products and Quotients 8.1 List of Concepts Sum and multiplication by a constant rules: (f(x) + g(x)) = f (x) + g (x) (k f(x)) = k f (x), for k R and provided f and g are differentiable where we are interested. Product Rule: (f(x) g(x)) = f (x)g(x) + f(x)g (x), provided f and g are differentiable where we are interested. Quotient Rule: ( ) f(x) = f (x)g(x) f(x)g (x) g(x) (g(x)) 2, provided f and g are differentiable where we are interested. You can always do the bookkeeping of your progress in these examples, as you did with the chain rule. Example: Find the derivative of R(x) = x3 +2x 2 x. Solution: We can write R(x) = f(x) g(x) with: f(x) = x 3 + 2x 2, f (x) = 3x 2 + 4x g(x) = x, g (x) = 1 2 x. 15

Then we just have to put everything in its right place: R (x) = f (x)g(x) f(x)g (x) (g(x)) 2 = (3x2 + 4x) x x 3 +2x 2 2 x x 8.2 Suggested Exercises from the Textbook Here is an exhaustive list of problems from the textbook that were relevant last week. Problems in bold are more challenging, but worth a try. Section 2.2: Practice problems 1, 2, 4, 5, 6 (not the part with the cosine function), 11 and problems 1 to 7, 13 to 20, 22 to 26, 28, 29, 31, 33 to 35, 38 to 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55. 8.3 More Exercises Use ( the ) result from the last section to find the derivative of V (x) = x 2 +. 1 3 x x 3 +3 Let U(t) := 2x2 +5x+13 x+4, find all the values of a where the tangent line to U(t) has horizontal slope at t = a. Let the following piecewise function for some constants a and b: { mu + b if u < 2 T (u) = ( u)(u 3 5 + 24u 2 + 5) if u 2, (9) Find the values of a and b that make T (u) differentiable on (, )? Using the product rule, find a formula to find the derivative of a product of three differentiable functions f(x), g(x), h(x). 9 Trigonometric Functions 9.1 List of Concepts Always use radians! Sine and Cosine functions: Coordinates of a point on the unit circle. Let (x, y) be a point on the circle and θ is its angle with the positive x-axis, then: x = cos(θ) and y = sin(θ) Relation of Sine, Cosine and Tangent functions to ratios of sides lengths in a right triangle. Mnemonic: SOH CAH T OA. 16

The tangent function tan(x) = sin(x) cos(x). Periodic functions: A function f(x) is periodic if there is some value P such that f(x + P ) = f(x) for all x in the domain of f. All integer multiples of 2π are periods of sin(x) and cos(x). We call P = 2π the minimal period of sin(x) and cos(x) because it is the smallest positive period. Table of values for trigonometric functions: x \f(x) sin(x) cos(x) tan(x) 0 0 1 0 π/6 1/2 3/2 3/3 π/4 2/2 2/2 1 π/3 3/2 1/2 3 π/2 1 0 Undefined Table of identities for trigonometric functions: a \f(a) sin(a) cos(a) tan(a) π x sin(x) cos(x) tan(x) π + x sin(x) cos(x) tan(x) x sin(x) cos(x) tan(x) x + 2π sin(x) cos(x) tan(x) Pythagorean Theorem identity: sin 2 (x) + cos 2 (x) = 1. Derivatives of trigonometric functions: d sin(x) = cos(x) dx d cos(x) = sin(x). dx You may remember that d dt tan(t) = sec2 (t), but you have to know what that secant function means: sec(t) = 1 cos(t), sec2 (t) = 1 cos 2 t. 17

9.2 Suggested Exercises from the Textbook Here is an exhaustive list of problems from the textbook that were relevant last week. Problems in bold are more challenging, but worth a try. Section 2.1: Practice problem 6 and problems 17, 35, 36. Section 2.2: Practice problems 3, 7, 8, 9 and 10 and problems 8 to 12, 21, 13 to 20, 30, 32, 36, 37, 52. Section 2.3: Practice problems 3 to 8 and problems 11, 12, 14, 15 to 17, 19, 22, 27 to 37, 38, 39, 42, 45 to 48. 9.3 More Exercises Find three solutions to 2 sin(x) = 1 in the [0, 4π) interval. Let the following piecewise function for some constants a and b: { sin(v) + a cos(v) if v < π T (v) = 2 bv 2 + 2 if v π 2. (10) Find the values of a and b that make T (u) differentiable on (, )? What is the smallest positive period of tan(x)? T (y) = tan(y) differentiable? Where is the function Find the equation of the tangent line to f(x) = ex +1 sin(x) at x = 3π 2. 18

10 Exponential and Logarithm Functions 10.1 List of Concepts Exponential functions vs polynomials: Natural exponential e x. 2 x or x 2 Derivative of the exponential function: d dx ex = e x. Logarithm functions: The logarithm function gets the exponent out of an exponential expression. If c = b n, for some b > 0, b 1 and n R, then we have: log b (c) = log b (b n ) = n. The natural logarithm, noted ln is the logarithm in base e, that is, for any x we have ln(e x ) = x. If we wish to change the base of a logarithm, we need to do the following computation: log b1 (c) = log b 2 (c) log b2 (b 1 ), for example, if we had to compute log 3 (e 2 ) we would do: log 3 (e 2 ) = log e(e 2 ) log e (3) = ln(e2 ) ln(3) 2 1.0986 1.8205. Main identity for logarithms: 1. log b b x = x for all b > 0, b 1 and x real. 2. b log b (x) = x for all b > 0, b 1 and x strictly positive. The derivative of the natural logarithm: d dx ln(x) = 1, for x > 0. x We showed this using the chain rule, but you only need to know the above result. It is important to specify that x > 0 because ln(x) is not defined for x 0 and thus cannot have a tangent line. 19

10.2 Suggested Exercises from the Textbook Here is an exhaustive list of problems from the textbook that were relevant last week. Problems in bold are more challenging, but worth a try. Section 2.3: Same as in Trigonometry section, the exercises are scrambled between the two categories. Section 2.5: Practice problems 1 to 4 and problems 1 to 29, 30 to 42. 10.3 More Exercises Given that ln(5) 1.6094, compute log 5 (e 30 ). Solve for t in the following expressions: 1. e 2t + 30 = 24 2. 8e 1 1 t = 5. 11 Chain Rule: Derivative of a Function Composition 11.1 List of Concepts Chain rule for the derivative of function compositions: if f and g are differentiable function, and h(x) = f(g(x)), then h (x) = (f(g(x))) = f (g(x)) g (x). To find these derivatives we have to identify the interior (g(x) previously) and exterior (f(x)) functions and their derivatives and proceed methodically. Derivative of the natural logarithm function. Let f(x) = ln(x), then f (x) = 1 x. Derivative of the exponential function for any base. Let a > 0 and a 1, then d dx ax = a x ln(a). example: Find the derivative of W (v) = ln( v 2 ). Solution: We first do the decomposition of the absolute value function: { ln(v 2) if v > 2 W (v) = ln(2 v) if v < 2, 20

remember that there is no output at v = 2 because ln 0 does not exist and that there is no possible negative number inside the logarithm because of the absolute value. If you have trouble with the decomposition, go back and look at a few examples from the class notes. Now we have to take the derivative of each case: { 1 W (v) = v 2 if v > 2 1 2 v if v < 2, = 1 for v 2. v 2 Normally you would also need to check at v = 2 to see if the derivatives connect, but in this case the function is not defined there, so its derivative cannot exist either. Note also that the x > 0 condition attached to the derivative of the logarithm also vanishes because of the absolute value. 11.2 Suggested Exercises from the Textbook Here is an exhaustive list of problems from the textbook that were relevant last week. Problems in bold are more challenging, but worth a try. Section 2.4: All practice problems and problems 1 to 37, 38 to 47, 48 to 71, 72 to 83. 11.3 More Exercises Let T (u) = { au 2 + b if u 1 u2 + 3 if u < 1 Find the values of a and b that make T (u) differentiable on (, )? Find the derivative of g(t) = ln(tan(t)) and find the equation of its tangent line at t = π 4. Find the derivative of g(t) = e 2 t at t = 2. and find the equation of its tangent line 12 Exponential Growth 12.1 List of Concepts When the rate of change of some quantity is proportional to that quantity (in other words, the more of it we have, the quicker it grows), we say that quantity follows exponential growth (if increasing) or exponential decay (if decreasing). 21

Such functions have the following form: y(t) = y(0)e ( kt), where y(t) is our quantity at time t, y(0) is out quantity at time 0, also known as initial quantity and k is called the relative rate of change. If we compute the derivative of y(t), we find: y (t) = k y(t), and this is why k is relative, the rate of change depends on the quantity y(t) multiplied by the relative rate k. It is important to differentiate between relative rate of change over a period of time and instantaneous relative rate of change. For example, if a problem says: The population triples each year or The annual percentage yield is of 5%, you have to find the parameters y(0) and k by comparing a result at a given time and then at a later time. If we specifically refer to the instantaneous relative rate of change, this is the value of k. Quantities that are modeled using exponential models: Population (growth or decay), Investment (Continuous Coumpound Interest), Temperature (Newton s Law), Radioactivity, disease modeling (under some constraints). Example: A population of bacteria doubles every 10 minutes. If we initially had 100 bacteria in our colony initially, find a formula that models the total population with time using exponential growth. Then find the rate of change in the population at 20 minutes. Solution: From exponential growth, we know that y(t) = y(0)e kt, with y(0) = 100. We know that after 10 minutes the population doubled, so we have y(10) = 100e k 10 = 2 100 e k 10 = 200 100 k 10 = ln(2) k = ln(2) 10 0.0693. The formula is thus y(t) = 100e ln(2) 10 t. We could use the chain rule to find the rate of change, but we already know it will be k y(20), so y (20) = 100 ln(2) 10 e ln(2) 10 20 = 10 ln(2) e ln(2) 2 9.803 bacteria min. 22

Example: We deposit 15000$ in an account with continuous compound interest with a yearly percentage yield of 4%. After how many years do we double our investment? At that time what is the rate of change in our investment? Solution: Our investment s worth follows exponential growth: y(t) = 15000e kt. After one year we have 4% more in our account: y(1) = 1.04 15000 = 15000e k 1.04 = e k k = ln(1.04) 0.0392 The formula giving us the account s worth is then y(t) = 15000e ln(1.04) t. To find the time t d when our investment doubles, we need to find: y(t d ) = 2 15000 = 30000 = 15000e ln(1.04) t d 2 = e ln(1.04) t d At this time, the rate of change is simply: ln(2) = ln(1.04) t d t d = ln(2) 17.67 years. ln(1.04) y (t d ) = k y(t d ) = ln(1.04) 30000 1176 dollars per year. Note: We could solve this problem using the following form for y(t): y(t) = 15000(1.04) t, and we would have obtained the same result. Example: An initial investment of 2000$ has an instantaneous growth rate of 3% of its current amount, in dollars per year. After how many years has the investment doubled? Solution: What the problem statement means is that if y(t) is the value of the investment after t years, it satisfies y (t) = ky(t) = 0.03 y(t) at any time t. From exponential growth we know that the function satisfying this is: y(t) = y(0)e kt = 2000e 0.03t. 23

To find the doubling time, we set y(t) = 2 2000$ = 4000$ and solve for t: 4000 = 2000e 0.03t e 0.03t = 2 13 Second Derivative 13.1 List of Concepts t = ln(2) 0.03 23 years. The second derivative f (x) of a function is the derivative of the derivative. f (x) = (f (x)) = d2 f dx 2 It s also the rate of change of the rate of change. For example, the second derivative of the position with respect to time is the acceleration, or the rate of change of velocity with respect to time. Example: We drop a ball from a height of 10 meters. Its height, in meters, with respect to time, in seconds, follows the function h(t) = 4.9 t 2 + 10 (m) The ball s vertical velocity is the derivative of the height function v(t) = dh dt = 9.8 t (m/s). This means that the ball is going down (negative sign) and its speed (absolute value) keeps increasing. To get the acceleration we take another derivative a(t) = dv dt = d2 h dt 2 = 9.8 (m/s)2. That means the velocity decreaces steadily at a rate of 9.8 meters per second each second. Example: Find the second derivative of g(x) = e x cos(x) and all values of x where the second derivative is zero. We first use the product rule for the first derivative: g (x) = e x cos(x) e x sin(x) = e x [cos(x) sin(x)]. We use the product rule again to get the second derivative: g (x) = e x [cos(x) sin(x)] + e x [ sin(x) cos(x)] = 2e x sin(x). 24

To solve for the zeros of the second derivative we only need to find the roots of sin(x), as 2e x is never equal to zero and we get g (x) = 0 if x = n π, for any integer n. 25