Welcome (Back) to IB Math - High Level Year 2. Some things to know: 1. HW yup. You know you love it! Be prepared to present. 2.

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1 Welcome (Back) to IB Math - High Level Year 2 Some things to know: 1. HW yup. You know you love it! Be prepared to present. 2. Content: 3. Grading Ultimately, you need to pass the IB exam! 40% Tests 20% Quizzes 30% Presentations (HW!) 10% Classwork 4. Bring: Notebooks ($3!), pencil(s), and you! 5. Homework judgment. Practice to proficiency > Multi part problems. Do last row if confident. Do last column for more gradual progression. 6. Lots of info at Welcome 1

2 HL Core Mind Map HL Calc Option Mind Map Calculus Overview Unit Plan is available on web page. Overview 2

3 1. Understand the idea of instantaneous vs. average rate of change 2. Understand why the instantaneous rate of change is the slope of the tangent to a curve at a point. 17D: #1*,2a (Rates of change) 17E: #1 3 all (The Derivative) distance (km) Jessie is driving away from school in her new car. The graph to the left represents how far away she is at any given time. At t = 0, how far away from school is she? t = 2?, t = 4? Describe her motion. What is her average speed in the first two minutes? What is her average speed in the next two minutes? The speed limit is 2 km/min. When will she "officially" be speeding? Hint: The equation for the curve is d(t) = ½t Try graphing it on your calculator. Your zoom feature might help. Geogebra Demo: Derivatives time (min) With a partner, complete the following: What have you concluded so far? The gradient of the tangent to y = f(x) at x = a is the instantaneous rate of change in f(x) with respect to x at that point. 8/31 17D:Rates of change 3

4 1. Understand the meaning of the derivative function. Given a non linear function, f(x), it would be nice to come up with a related function that describes the rate of change (or slope or gradient) of the original function at any point x. The Derivative Function The function that describes the gradient of y = f(x) is called its derivative function and is labeled f'(x) (read as "eff prime of x") The value of f'(a) is the gradient of the tangent to f(x) at the point where x= a We will discuss how to find this function next. But you can begin to work with certain questions already: 17D: #1*,2a (Rates of change) 17E: #1 3 all (The Derivative) On #1, use the method of Investigation 4 that we did in class. 8/31 17E:The Derivative Function 4

5 17D: #1*,2a (Rates of change) 17E: #1 3 all (The Derivative) 1. Understand limits as distinct from evaluating an expression. 2. Evaluate limits at infinity 17A: #1cf,2,4 (Limits) 17B.1: #1,2cde (Limits at infinity) 17B.2: #1def,2a,b.ii,4,5c (Limits at infinity) On your calculator, graph the function What is the value of the function at x = 2? What is the value of the function near x = 2? Hmmm. How can we describe this more formally? Limits exploring a function as the variable approaches a value that might not be well defined. > is not the same as evaluating the function f at a. > Notation and limits in the context of asymptotes. > Evaluating limits requires special approaches and rules Informal Definition of a Limit: There is a more precise definition known as the epsilon delta approach. I will describe it if we have time. Limits help us analyze functions in two ways: Understand a function as x gets very large or small Understand functions that are not continuous Definition of a continuous function A function is considered continuous at x = a if and only if: 1. f(a) is defined 2. exists 3. Indeterminant forms Expressions that evaluate to ratios involving 0 and/or are called indeterminant and require special treatment. You must rearrange expressions to see what value the expression will approach. A common trick is to divide the numerator and denominator by some power of the variable. Remember our work with rational functions! To find limits, simplify the expression, sometimes by factoring, to eliminate discontinuity /3 Some properties of limits make them easier to work with Geogebra epsilon delta proof 9/2 17A:Limits 5

6 1. Understand limits at asymptotes Limits arise when looking at functions with asymptotes (often these are rational functions or functions of the form p(x)/q(x) where p and q are both polynomials). Consider To explore asymptotes, factor as much as you can and cancel terms. You might see quadratics, even though it's really a ratio of linear expressions (with a twist). Note a problem at x = 7. Division by zero creates a vertical asymptote. To explore it we need to look at the sign of f very close to but on either side of x = 7. When x is just < 7, num is >0, denom is <0) so f is big and negative: When x is just > 7, num is >0, denom is >0) so f is big and positive: Horizontal asymptotes Let's also look at what happens as x gets very large. You may recall that: As x gets large, the highest power of x in a polynomial dominates the value of the function. In our case, the numerator and denominator are both quadratic Thus, as x gets large, they both approach infinity at the same rate. One way to see this is to divide the top and bottom by the highest power of x: The terms involving x in the denominator will approach 0 so we're left with But there is more to learn here. How does the function approach 2? One way is to use a sign diagram: Choose some test points: + + Big negative: get squared: positive 3 A bit above 3: + num, denom = 7 Big positive: get squared: positive + The function is large at 7, so has to come down to 2 on the right and go up to 2 on the left Note: This approach is only valid for certain functions like hyperbolas (a ratio of linear functions) Better to use long division (or synthetic division if the denominator is x ± c): As x > + this "remainder" is positive so the function is "above" 2, hence approaching from above and As x > this "remainder" is negative so the function is "below" 2, hence approaching from below and A final approach involves our favorite HL trick simply notice that 6 = Either way we end up with 2 plus some "remainder". As x gets large positive, is that remainder positive (the graph is above 2), or negative (the graph is below 2) Note that as x +, f(x) 2 from above since Likewise, as x, f(x) 2 from below since If all you care about is where the asymptote is, you can just imagine ignoring the lower order terms. What do you think this would give? These two limits (+ and ) define a horizontal asymptote of the function. Horizontal asymptotesin rational functions occur as x ± when the degree of the numerator and denominator are equal. The H.A. is at y = "the ratio of the two leading coefficients" Two other things to note while we're here. The zeros of a rational function are the values of x that make the numerator zero (assuming they do not cause an issue in the denominator). In our case, x = 3 is a zero. Secondly: Notice that x = 3 is not a zero. Even though it makes the numerator zero, there is also an x + 3 in the denominator. So there is a problem in the original function at x = 3 since it would result in 0/0. It's called a hole or point discontinuity at x = 3. Notice that after we cancelled the (x + 3), the problem is not apparent! f(x) is not defined at x = 3. But we can say that: It appears that we just plugged 3 into the reduced form which works in this case (but not always...) We'll finish by looking at the graph of the function: Oblique Asymptotes Consider The degree of the numerator is one larger than the degree of the denominator. So when you do the division, you'll end up with a "linear" expression, possibly with some remainder. If there's no remainder, you'll get a hole. But look at doing the division (use long division or synthetic division shall we digress?) From this we can see that as x approaches ±, f will approach the line x 1. This is called an oblique or slant asymptote. Summary of Behavior of Rational Functions 1. When the degree of the numerator is greater than the degree of the denominator by more than one, the function will diverge as x approaches ±. 2. When the degree of the numerator is one greater than the degree of the denominator, the function will approach an oblique (slant) asymptote as x approaches±. 3. When the degree of the numerator is equal to than the degree of the denominator, the function will approach a horizontal asymptote as x approaches ±. The y value of the asymptote is the ratio of the coefficients of the leading terms. 4. When the degree of the numerator is less than the degree of the denominator, the function will approach a horizontal asymptote at y = 0 as x approaches ±. Try a few: ln(x 3 9x) 17A: #1cf,2,4 (Limits) 17B.1: #1,2cde (Limits at infinity) 9/2 17B:Limits at infinity 6

7 17A: #1cf,2,4 (Limits) Present 4b, 5 verbal 17B.1: #1,2cde (Limits at infinity) Present 2de 17B.2: #1def,2a,b.ii,4,5c (Limits at infinity) Present 1ef,2,5c 1. Understand limits of expressions involving trig functions With a partner, do Investigation 2 on p C: #1acef,2,3 (Limits with trig) 17F: #1,2b,3c,4b,5d,6,7a,8d (First principles) QB: #5,45 (First principles) We will have a short quiz next time on Chapter 17 ideas Certain trigonometric limits can be evaluated by making use of an interesting feature of the behavior of the function This can be proven in various ways. The book gives a geometric proof (p. 516). We will see it again when we look at series and it can also be interpreted as the slope of the sin function at the origin (if done in radians). From the above result, other limits involving trig functions can be calculated: since the limit of a quotient is the quotient of the limits if the denominator limit 0 Consider Note that as x, 1/x 0 so we can rewrite this using a substitution of variables let n = 1/x. That idea is a useful one when working with limits. 9/7 17C:Trig Limits 7

8 1. Understand the definition of the derivative 2. Find derivatives of certain functions from first principles OK, let's get more formal about this: How do we find the derivative function more rigorously? Geogebra Demo: Derivatives Summary of a Derivative is called the limit quotient. Finding the derivative of a function by evaluating the limit quotient is called using first principles. The f' notation is associated with Isaac Newton, and thus is more common in Western European texts (particularly those with a British connection...). Another notation was developed at roughly the same time by Gottfried Leibnitz which uses differentials. This notation is very important in further Calculus. Instead of using h, Leibnitz references the Δx and Δy using the lower case δx and δy. Then the approximate gradient is: The exact derivative function is then...pronounced "dee why, dee ex" Either definition can be used to find the derivative function at any point x. But you can also use first principles to find a derivative at a single point, say, a. Try one Be prepared to dust off some trig identities. We will have a short quiz next time on Chapter 17 ideas 17C: #1acef,2,3 (Limits with trig) 17F: #1,2b,3c,4b,5d,6,7a,8d (First principles) QB: #5,45 (First principles) 9/7 17F:First Principles 8

9 17C: #1acef,2,3 (Limits with trig) Present 1ef,2,3 17F: #1,2b,3c,4b,5d,6,7a,8d (First principles) Present 1,4b,6,7a,8d QB: #5,45 (First principles) Present 5,45 Quiz 18A: #1 6 last col,7 (Rules of differentiation) QB: #1,14 (IB Practice) 1. Use rules of differentiation to find derivatives of functions involving powers of x. The result of 17F #1is fundamental and is very useful. Derivative of a power function We will explore many of the properties of differentiation using a power function to practice. Here are the first set of rules. Can you prove them from first principles? The following instructions are all different versions of saying the same thing: Find f'(x) Find Differentiate with respect to x Find the gradient function of f(x) Find the gradient of the tangent to f(x)...usually at a given point, rather than generally How about: find the slope of the line tangent to the curve at x = 2 18A: #1 6 last col,7 (Rules of differentiation) QB: #1,14 (IB Practice) 9/9 18A:Simple Differentiation 9

10 18A: #1 6 last col,7 (Rules of differentiation) QB: #1,14 (IB Practice) Return Quizzes 1. Recognize function compositions 2. Understand and apply the chain rule 18B.1: #1bdf,2cd (Composition review) 18B.2: #1 3 last col,5,6 (Chain rule) We begin this section by reviewing composite functions g x f f(x) g(f(x)) = (g f)(x) x f(x) = 3x + 7 Square & (g f)(x) = (3x + 7) Times add 3 The composite of two functions is created by using the output of one function as the input to the other function. Some properties: (f g)(x) is not the same as (g f)(x) in general The range of the first function in a composition is the domain of the second. Try: Given f(x) = x and g(x) = 2x + 4 find (f g)(x) and (g f)(x) (f g)(x) = (2x + 4) (g f)(x) = 2(x 2 + 7) + 4 You'll do a little more practice with composites in 18B.1 Consider the function x 2 whose derivative is 2x. What is the derivative of (2x) 2? Now try differentiating (2x + 3) 2. Do it with and without expanding first. Finally, try differentiating (x 2 + 3x + 4) 2 by guesswork and then by expanding it out. Try finding the derivative of (2x 2 ) 4 two different ways From these short examples, but with no formal proof, we can see the chain rule at work. The Chain Rule If f(x) = g(h(x)) = (g h)(x) then f'(x) = g'(h(x)) h'(x) Not recognizing the need to use the chain rule is probably the single most common source of errors in differentiation! You need to be able to recognize when a function is a composite of other functions. It takes practice. Although the concept is simple, it can get complex. It can help to introduce another variable to keep track of your work. What about Chains can have lots of links! Do 18B.2 thoroughly. This is like learning your times tables. The idea is to master it, not just to know how to do one. PRACTICE! 18B.1: #1bdf,2cd (Composition review) 18B.2: #1 3 last col,5,6 (Chain rule) An informal proof: 9/12 18B:Chain Rule 10

11 18B.1: #1bdf,2cd (Composition review) 18B.2: #1 3 last col,5,6 (Chain rule) Try QB #1 & 14 now Present 1f,2d Present 2cfi,3ef,5,6 let's prove it! 1. Recognize products of functions 2. Understand and apply the product rule 18C: #1 3,last 2,4 (Product rule) 18D: #1def,2bd,4 (Quotient rule) QB: #41,54 (IB Practice) What happens when you take the derivative of a product of two functions? Let's look at this from first principles: Add zero! The function f must be continuous in the vicinity of x. There are lots of notations for this all boil down to the same thing! Another proof using Leibnitz notation is given in the book. The Product Rule Try a couple: (Don't forget the Chain Rule) 9/14 18C:Product Rule 11

12 1. Recognize quotients of functions 2. Understand and apply the quotient rule in appropriate places! Consider a function Making use of the product rule, derive a formula for f'(x) in terms of u, u', v, & v' The Quotient Rule Try a couple: Please stack chairs! Again, these skill need to be second nature! No pain, no gain......train your brain! 18C: #1 3,last 2,4 (Product rule) 18D: #1def,2bd,4 (Quotient rule) QB: #41,54 (IB Practice) 9/14 18D:Quotient Rule 12

13 18C: #1 3,last 2,4 (Product rule) Present 2f,3d,4 18D: #1def,2bd,4 (Quotient rule) Present 2d,4 QB: #41,54 (IB Practice) Present all 1. Understand and use implicit differentiation 18E: #1&2 last row, 3 (Implicit differentiation) 18F: #1&2 last row,3c,5,7,9 (Exponentials) QB: #28,55 (IB Practice) Sometimes, it's hard to rewrite a relation in the form of a function of y. For example consider the relation x 2 + y 2 = 9 Solving for y is messy. But we can consider y to be implicitly related to x with a derivative of. We can differentiate the above recognizing this and incorporating the chain rule when we differentiate y: We can now solve for the derivative in terms of x and y to yield: r Notice that this is correct as the slope of the radius is y/x and the slope of the tangent is its opposite reciprocal. We can use this approach in more complex situations: If 2x 2 3y 2 = 2, find the two values of when x = 5. Notice that, in general, the chain rule tells us that if y is a function of x, then and even more generally 9/16 18E:Implicit Differentiation 13

14 1. Find and use derivatives of exponential functions We now turn our attention to derivatives of other functions. Like power functions, we will develop shortcuts from the definition. We begin with exponentials. Recall the graphs of exponential functions of the form y = a x For base > 1 the function increases, for 0 < base < 1 the function decreases. Let's find the derivative of this function. Since we don't have a formula yet, we need to start from first principles. Let f(x) = a x. Then: Now we notice something interesting: So we can rewrite our result as: Derivative of an exponential If f(x) = a x then f'(x) = f'(0)a x Whoopdeedo what good is this if we don't know f'(0)? Let that sink in a moment! What is the derivative of 2 x? It's 2 x times the slope of 2 x at x = 0! Point taken. Can I find the slope of 2 x at x = 0? How about by using a calculator! Use MATH/[8] nderiv(2 x,x,x). It may take a while to calculate. f'(0) Are we going to do that every time? No! Do you recognize that number? Let's find the value of the base that gives us a slope of 1 at x = 0. Experiment with different bases (remember, larger bases create steeper curves) You may have found a guess, but let's look at this algebraically. The question, again, is "for what base will we get f'(0)= 1"? That is what value of a makes: or For this to be true, the numerator has to approach h in the limit. So: Now substitute and notice that as So we can rewrite: Raising both sides to the n th power gives our answer! So, as it turns out, the function e x has a slope of one at x = 0. It has the very special property that: It is a function whose derivative is itself! That is, the value of the function at any point x is also the slope of the function at that point! Pretty natural, eh? We will come back to our original question about the derivative of a x but for now let's work a little with the base of e (where we don't have to worry about that pesky f'(0). All the differentiation tools in your belt work for this function even in combinations! Don't forget about the chain rule! The derivative of a function of the form e f(x) requires it! Sub HW i = 0 Do while (i < 100) Recite ("Repetition is my friend",volume = i) i = i + 1 End Do End Sub 9/16 18F:Exponentials 18E: #1&2 last row, 3 (Implicit differentiation) 18F: #1&2 last row,3c,5,7,9 (Exponentials) QB: #28,55 (IB Practice) 14

15 Chapter Test next Monday, 9/26 18E: #1&2 last row, 3 (Implicit differentiation) Discuss as needed 18F: #1&2 last row,3c,5,7,9 (Exponentials) Present 9 QB: #28,55 (IB Practice) Present (quickly) 1. Find and use derivatives of logarithmic functions A quick review: log b x is the power to which b is raised to get x b x = a x = log b a The log base e is called the natural logarithm or ln e x = a x = ln a e ln a = a and ln e a = a ln x is the inverse of e x Domain of e x is x R, Range of e x is Domain of ln x is x > 0, Range of ln x is y R 18G: #1 3 last row,5,7 (Logarithms) QB: #12 (IB Practice) y = e x Rules of Logarithms (any base) log b b x = x What is the derivative of y = lnx? We use the chain rule here: Extending this to arguments that are functions of x and using the chain rule, we get the more general case: Derivative of natural logarithms Can you generalize to logs of any base? Of course, this is usually going happen in a larger context. Use laws of logarithms! 18G: #1 3 last row,5,7 (Logarithms) QB: #12 (IB Practice) Practice doesn't make perfect. Practice reduces the imperfection. Toba Beta, "Master of Stupidity" 9/19 18G:Logarithms 15

16 18G: #1 3 last row,5,7 (Logarithms) Present 5,7 QB: #12 (IB Practice) Present 1. Find and use derivatives of trigonometric functions 18H: #1 6 last col (Trig functions) 18I: #2,3def,4bc,5ac (Inverse trig functions) QB: #7,18,20,27,46 (IB Practice) QB: #2 (IB Practice) A quick review of trig functions The derivative of sinθ is the rate of change (slope) of sinθ as θ changes. We can see that the slope oscillates from positive to zero to negative to zero and back with the same period as sinθ (2π). This gives us a hint as to what we can expect... Find the derivative of sinx from first principles. A hint: Recall the identity: cos(2θ) = 1 2sin 2 θ so cos(2θ) 1 = 2sin 2 θ and cos(θ) 1 = 2sin 2 (θ/2) and There is an alternative derivation in the text using the identity: Find the derivative of cos x (Hint: cos x = sin (x + π/2)) Find the derivative of tan x Don't forget the chain rule (ever!) Summary: for x in radians: Some applications: A couple more for the road: Find the derivatives of secx, cscx, and cotx. 9/21 18H:Trig Func 16

17 1. Find and use derivatives of trigonometric functions First of all, notice that the trig functions are not one to one. (What do I mean by that?) They don't pass the horizontal line test. Their inverses are not functions! But we can define inverse trig functions over restricted domains! Let's take a look at the inverses graphically (reflections around y = x) Note the domains and ranges of each function and its inverse carefully. I am not one over sinx! Now let's look at the derivative of y = sin 1 (x). Hmmm...this means that: sin(y) = x which we can differentiate implicitly. Your turn... Going back to definitions: If y = sin 1 (x) then cos (y) = So Notice the domain constraint which is the range of sin. You will derive the other inverse derivatives in your homework. 18H: #1 6 last col (Trig functions) 18I: #2,3def,4bc,5ac (Inverse trig functions) QB: #7,18,20,27,46 (IB Practice) QB: #2 (IB Practice) 9/21 18I: Inverse trig functions 17

18 18H: #1 6 last col (Trig functions) 18I: #2,3def,4bc,5ac (Inverse trig functions) QB: #7,18,20,27,46 (IB Practice) QB: #2 (IB Practice) Chapter Test Monday, 9/26 Quick Quiz 1. Find and use derivatives of trigonometric functions 18J: #2def,4 18 even (Higher derivatives) QB: #37 (IB Practice) QB: #6,15,33,35,25,29,44,47,49,50a,51,57 (Review) Consider the height of a ball thrown up from the top of a 60 foot building at an initial speed of 12 feet per second. The ball's height can be modeled by: h(t) = 16t t + 60 The change of position over time is also known as velocity. Using derivatives, we can find instantaneous velocity by letting Time 0 Acceleration is the rate of change of velocity. That is: Abstractly, f''(x) or f (2) (x) is the second derivative of f with respect to x. It represents the the "slope of the slope" or the curvature of f. We will talk a lot more about the meaning of second derivatives in the next chapter. One can continue to take the derivative as many times as you want. Notation: if y = f(x) then: 1 st derivative = slope of f(x) 2 nd derivative = slope of f'(x) = slope of the slope of f(x) 3 rd derivative = slope of f''(x) = slope of the slope of the slope of f(x) n th derivative = slope of f (n 1) (x) In some physical applications, the derivatives have important meanings, and thus, special names:...or... Chapter Test Monday, 9/26 18J: #2def,4 18 even (Higher derivatives) QB: #37 (IB Practice) QB: #6,15,33,35,25,29,44,47,49,50a,51,57 (Review) By doing this HW, you are practicing your derivatives, and your algebra. It is not sufficient to understand that you can do this or how to do it. It is sufficient to do it. 9/23 & 26 18J: Higher Derivatives/Quiz 18

19 Return Ch 18 Tests Need to add QB Problems into HW for 19ABC. 19A: #1ef,2cd,3d,4c,5c,6 24 even last one (Tangents and Normals) 1. Use derivatives to find lines that are tangent and normal to curves. Recognizing that a derivative is the gradient of a curve at a point, one can use differential calculus to explore problems involving lines that are tangent and normal to curves. Consider the diagram: If we find the derivative at some point, A(a, f(a)), we have the slope at x = a and a point. Remember point slope form of a line? y y 1 = m(x x 1) where (x 1, y 1) is a point on a line with slope m. So we can find the equation of the line tangent to the curve: Equation of a Line Tangent to f at x = a y f(a) = f'(a)(x a) Similarly, the equation for a line normal to a curve at a point A(a, f(a)) is given by: Equation of a Line Normal to f at x = a The formulas are useless, but understanding the idea, along with other things we know about lines, can help us to answer many types of questions: (By "tangent" we mean "the line tangent to the curve") Remember that a line with slope a/b going through point (x 1, y 1) can be written directly as ax by = a(x 1) + b(y 1) So in this case, the line through (1, 2) with slope 2 is 2x y = 2(1) 2 or 2x y = 0 Some more thoughtful applications... A line "normal" to something is perpendicular to it. What does it mean for a tangent line to be horizontal? More on this later. Remember your cubics? 19A: #1ef,2cd,3d,4c,5c,6 24 even last one (Tangents and Normals) Lots of cool problems involving these ideas. Think them through - what does it all mean? 9/28 19A:Tangents and Normals 19

20 19A: #1ef,2cd,3d,4c,5c,6 24 even last one (Tangents and Normals) Present questions 19B: #2dhi,4,6,8df (Increasing/Decreasing intervals) 19C: #1,2hi,3 15 odd (Stationary points) 1. Use derivatives to understand increasing and decreasing portions of curves. Derivatives can be used to find intervals of the domain in which a function is increasing or decreasing. In what interval(s) is the function increasing? 2.7 < x < 0.5 and 2.9 < x < + Decreasing? < x < 2.7 and 0.5 < x < 2.9 Formally we define increasing and decreasing as: What is true about the derivative in these intervals? Increasing and Decreasing Intervals f(x) is increasing on S <=> f'(x) 0 for all x in S f(x) is strictly increasing on S <=> f'(x) > 0 for all x in S f(x) is decreasing on S <=> f'(x) 0 for all x in S f(x) is strictly decreasing on S <=> f'(x) < 0 for all x in S If a function is increasing or decreasing for all real numbers we call it monotone. So functions can be monotone increasing or monotone decreasing. Analysis of the signs of the derivative thus becomes important in understanding the direction of slope of a curve in some interval (or at some value of x). Sign diagrams come in handy. For example, the function above is a quartic. It's derivative is a cubic. Using the numbers estimated above for the extremes, draw a sign diagram of the derivative Do part a) only: Support your conclusion with a sign diagram Now for something more interesting... 9/30 19B: Increasing/Decreasing functions 20

21 Hand out Stationary point practice 1. Use derivatives to find stationary points and use them to identify important points on a curve. A stationary point is a place where the function is neither increasing nor decreasing. Using derivatives, this means that: A stationary point is a point such that f'(x) = 0 Stationary points come in three flavors. To understand them, consider the following curve: Local max: f'(x) = 0 Local min: f'(x) = 0 Notice that when analyzing extremes of functions with a restricted domain you must also consider the Global max: Global min: These are not stationary points. Notice that the slope can be flat and f'(x) = 0 at places that are not local extremes. Such places are called horizontal inflection points or stationary inflection points. (British: inflexion) Big idea: For a point to be a local extreme (min or max) f'(x) = 0 at the point and f'(x) must have different signs on either side of the point. In other words, the slope must change sign (+ to or to +). This is one reason to draw the sign diagram. Try one: a) Identify the stationary points b) Draw the sign diagram of the gradient function below the graph shown Summary A local minimum of f(x) occurs at x = a if f'(a) = 0 and f'(x) changes from to + at a A local maximum of f(x) occurs at x = a if f'(a) = 0 and f'(x) changes from + to at a If f'(a) = 0 but f(x) does not change sign at x = a, then a is a point of horizontal inflection Some practice Always check endpoints if you are working with a closed interval. Don't expect to have a calculator on these types of problems. 19B: #2dhi,4,6,8df (Increasing/Decreasing intervals) 19C: #1,2hi,3 15 odd (Stationary points) 9/30 19C:Stationary Points 21

22 19B: #2dhi,4,6,8df (Increasing/Decreasing intervals) Questions? 6?, 8d? 19C: #1,2hi,3 15 odd (Stationary points) Questions? 11 (2 things to show!), 13 Hand out Stationary point practice 19D.1: #3g,7,9,12,13 (Inflection points) 19D.2: #1,2 (Graphing) QBApps: #3,4,8,9 (IB Practice) 1. Use derivatives to find inflection points and describe the shape of a curve. We have seen that we can take second and higher derivatives. What do they mean? The Second Derivative f''(x) describes the rate of change of the slope of a curve. Let's take a look... Curve is concave up f'(x) is increasing f''(x) > 0 Curve is concave down f'(x) is decreasing f''(x) < 0 Now that we understand the meaning of the second derivative, notice that we can use it to characterize a stationary point: If f'(a) = 0 and: f''(a) > 0 => Concave up => f(a) is a local minimum f''(a) < 0 => Concave down => f(a) is a local maximum f''(a) = 0 => Flat (no curvature) => f(a) could be a horizontal inflection point or a flat spot! Let's look further at f''(a) = 0 If f'(x) = 0: Stationary or horizontal point of inflection (flat) If f'(x) 0: Non stationary or nonhorizontal point of inflection (not flat) A point of Inflection means that the function is changing curvature! Consider f(x) = x 4 at x = 0: f(x) = 0 f'(x) = 4x 3 f'(0) = 0 stationary point f''(x) = 12x 2 f''(0) = 0 = f'(x) flat. f''(x) does not change sign at 0! not an inflection point From inspection, try to sketch the first and second derivatives of the function above g(x) g'(x) g''(x) Using sign diagrams to characterize a function: f(x) f'(x) + + f''(x) Let's look more closely at the case of f''(x) = 0. f'(x) is flat Point of Inflection! f(x) is changing curvature from up to down. f''(x) = 0 and is changing from + to Here is the other interesting place. Identify what is going on with f, f', and f'' f''(x) = 0 and is changing from + to f(x) is changing curvature from down to up and is flat. Horizontal or Stationary Point of Inflection! f'(x) is flat and is = 0 19D.1: #3g,7,9,12,13 (Inflection points) 10/3 19D:Inflection & Shape 22