Rolle s Theorem. The theorem states that if f (a) = f (b), then there is at least one number c between a and b at which f ' (c) = 0.

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1 Rolle s Theorem Rolle's Theorem guarantees that there will be at least one extreme value in the interior of a closed interval, given that certain conditions are satisfied. As with most of the theorems we deal with, you first must make sure that the conditions are met before you can apply the theorem and draw any conclusions. In order for Rolle's theorem to be true, the function must be continuous within a closed interval [a,b] and differentiable everywhere within this interval [i.e. within the open interval (a,b)]. Remember that no function is differentiable at the endpoints of its domain (because there is more than one tangent which can be drawn at an endpoint). The theorem states that if f (a) = f (b), then there is at least one number c between a and b at which f ' (c) = 0. In simple words, this tells us that if the y-values for two different x values, f (a) and f (b), are equal, then there has to be at least some other x-value between a and b where the slope of the tangent line [f ' (c) ] is equal to zero - which means that there is a max or min at that point. This makes sense. If you have a continuous curve going between two points which are both horizontally equal in position, you must have a peak or valley in between the two points where the curve turns around to go back to the y-value it was at before leaving. There is only one legal exception to that hump/valley scenario. If you know what that is, send me a Message with the heading Rolle's and tell me what it is (for points of course). Here is an illustration. If there is a slope of zero between two points on a curve (the secant line between the points has a slope of zero), then there must be a tangent line somewhere between x = a and x = b that also has a slope of zero. Note: The curve must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). That means that there are no discontinuities or cusps.

2 The Mean Value Theorem The Mean Value Theorem is related to Rolle's Theorem in that it will also give us conclusions about the slopes of tangents within the closed interval. This theorem also demands that the function be continuous everywhere within a closed interval [a,b] and everywhere differentiable within the open interior interval of (a,b). The theorem states there exists some number c at which Mean Value Theorem f b f a b a f c f b f a Remember that the left side of that equation, is the basic Algebra I definition of b a the slope between two points (a, f(a)) and (b, f(b)). (Remember that f (b) and f (a) stand for y- values of the points). It is the slope of the secant line between any two points on the curve. On the right side, f ' (c) stands for the value of the first derivative at the x-value "c"; or the slope of the tangent line at point x = c. Thus the theorem is telling us that somewhere between points where x = a and x = b we can find a point x = c that has a tangent line whose slope is the same as the slope of the secant line between the two endpoints of a and b. Mean Value Theorem Slope of secant = slope of tangent line for some point x = c We can also interpret this as saying: Mean Value Theorem Average change over interval = instantaneous change at some point c This too is common sense, and is well illustrated by this drawing:

3 f (c) = slope of secant line ** A last note before we look at a couple of examples: Remember when you look at these complicated-looking formulas to take the time to break it up into parts and think about the things we have already learned. Translate them into words - slowly. Read it two or three times until you know what it is saying. You can do it!

4 The Mean Value Theorem The MVT is telling us that somewhere between points where x = a and x = b we can find a point x = c that has a tangent line whose slope is the same as the slope of the secant line f b f a between the two endpoints of a and b. f c b a Now for some examples: Mean Value Theorem Slope of secant = slope of tangent line for some point x = c A: Find the value of c for which the Mean Value Theorem holds true with the function f(x) = x + /x in the closed interval [/, ]. Endpoints of interval: (/, 5/) and (, 5/). f x x x x x f x x f (x) is continuous on [ ½, ] because its only point of discontinuity, at x = 0, occurs outside the interval. f (x) is differentiable on ( ½, ) because the derivative only fails to exist at x = 0, which occurs outside the interval. Slope of secant = tangent slope 5 5 f f c 0 c c =, - x = - lies outside the interval c = We will need to know the points at the endpoints, because the MVT requires us to find the slope between the endpoints. To do this, put the x-values into our function. Find the derivative of the function. We must show that the conditions of the Mean Value Theorem are met. The function must be continuous on [ ½, ] and differentiable on ( ½, ) The Mean Value Theorem tells us that there will be a point (c, f(c) ) where the value of the derivative (the slope of the tangent line) will be equal to the slope of the secant line between the endpoints of the interval. Remember that f (b) just means the y-value at the x- value of b. Solve for c. We cannot use the value of c = -. Thus when x = the tangent has the same slope as that of the slope of the secant line between the endpoints.

5 B: The temperature, T, is changing with respect to time, t. We have two data points (t, T) of (0, 0) and (0, ) given. Show that the mercury had to be rising at a rate equal to 0. o F/s at some point in time, We have an interval, with some sort of curve between these two points. To prove this, we will use the Mean Value Theorem. Our temperature is continuous and differentiable by the nature of the physical constraints of the problem. Slope of secant = tangent slope 0 f f c 0 c Thus by the MVT there must be a value somewhere whose instantaneous rate of temperature change will also be equal to 0. First we must show that the conditions of the Mean Value Theorem are met. Neither time nor Temperature can skip values or make sharp changes. We use the theorem, subbing values in from the points. Thus, since the slope of the secant line connecting the endpoints of the interval is equal to 0., there must be (according to the Mean Value Theorem) a point (c, f(c) ) somewhere in the interval whose instantaneous rate of temperature change will also be equal to 0..

6 C: Show that the function has ]. x 4x 7 0 exactly one solution on the interval [-0,- 3 To do this, we first have to show (using the Intermediate Value Theorem) that there exists at least one root. Then we will use our knowledge of extrema to show that there are no extrema between the two endpoints and therefore no way for the curve to "turn around" and come across the x-axis more than once. Our function is continuous in the closed interval and differentiable on the open interval, thus the Intermediate Value Theorem applies. f (-3) < 0 and f (-) > 0 Thus, according to the Intermediate Value Theorem the function must cross the x-axis at least once. There will be at least one root. First we must state that the conditions of the Intermediate Value Theorem are met. We pick any two points within the interval and find two which are on opposite sides of the x-axis. Thus, according to the Intermediate Value Theorem the function must cross the x-axis at least once (have at least one root). Now we have to show that there will be no more than one crossing (the curve will not, for instance, go down, cross the axis, and then turn around and go back across again.). Were there such a "turnaround" then there would be local extrema (humps or valleys), so we will check out the first derivative to see if we have local extrema within our interval. 3 y x 4x 7 3 y 3x 8x 0 When x = (8/3) /5 The CV lies outside our interval so there are no relative extrema within the interval and there cannot be more than one root. We find the first derivative, set it equal to 0 and find the critical values of x. Our only critical values are outside of our interval. This tells us that there are no local extrema within the interval and the curve will not be able to turn around and come back over the x-axis. Note: You must use some mathematical reason for why the curve will not cross more than once. You cannot just say "I can tell from my calculator that it only crosses once"! When you do problems like B and C you must use words! You can't just go through some steps and expect the reader to interpret what it all means. That is your job and that is what is being tested! 3

7 Summary of Rolle s Theorem and The Mean Value Theorem - Show that the function satisfies the conditions of the theorem: a) tell that the function is continuous on [a, b] (this means that the function has no discontinuities on the closed interval) b) tell that the function is differentiable on (a, b) (this means that the function has no discontinuities or cusps on the closed interval) c) For Rolle s only: also show that the endpoints of the interval have a slope between them that is equal to zero (this means that the secant line is horizontal) d) State the theorem applies. (this is the conclusion of what you did in steps a-c) For both theorems: find the derivative of the function, properly labeled Then: For Rolles: 3 set the derivative (which is the slope of the tangent) = 0 (which is the slope of the secant in this case) 4 solve for x this is your answer For the MVT 3 find the slope of the secant line between the endpoints of the interval 4 set the derivative (which is the slope of the tangent) = the slope of the secant line that you just found 4 solve for x this is your answer

8 Homework Examples #3, 7, This first example is extra and not from your book. A: Use analytical methods to find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. f x 3x x 5 This is our function. f x 6x Find the derivative of f (x). f x 6x 0 When x = Now we will set it equal to zero to find critical values and check for values which make the function undefined. The function is never undefined. CV = Our critical value is x =. to find our absolute extrema. Draw the labeled number line and mark our CV on it. Test values on either side of our CV and put the sign of the derivative above the #line. There is a min at (, -7) because f changes from negative to positive at x =. f is decreasing when x < because f is negative on that interval. f is increasing when x > because f is positive on that interval. Make our conclusions, using the proper (and required) wording. #3: Use analytical methods to find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. h (x) = /x = x - h (x) = -x - never equals zero. h (x) is undefined when x = 0 CV: x = 0 There will be no extrema, however, because the function is also undefined at our CV. h (x) = -x - < 0 for all x. Thus the curve is decreasing for all x. This is our function. I rewrote it to make it ready to take the derivative. Take the derivative with the power rule and find the CVs by setting it equal to zero and solving for x and/or finding where it is undefined. Since the derivative is always negative, our curve is everywhere decreasing.

9 #7: Use analytical methods to find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. / y 4 x 4 x / y x x equals zero. y is undefined when x. never This is our function. I rewrote it to make it ready to take the derivative. Take the derivative with the power rule and find the CVs by setting it equal to zero and solving for x and/or finding where it is undefined. CV: x = - y < 0 for all x. Thus the curve is decreasing for all x. Since the derivative is always negative, our curve is everywhere decreasing. The domain of y is x and from x = - the function is always decreasing. Thus we have a max at (-, 4)

10 #: Use analytical methods to find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. x hx ( ) x 4 x 4() x() x h ( x) x 4 4 ( )( ) h ( x) 0 CV: x = -, h (x) x x x x 4 x 4 + There is a max at (-, ¼) because h changes from positive to negative. There is a min at (, -¼) because h changes from negative to positive. The curve is increasing on the intervals (, ) and (, ) because h is positive. The curve is decreasing on the interval (-, ) because h is negative. - + This is our function. Take the derivative with the quotient rule and find the CVs by setting it equal to zero and solving for x and/or finding where it is undefined. This derivative is never undefined, but will equal zero when the numerator equals zero. This time, since we have two CVs and our derivative is not always positive or negative, we will look at the number line so that we can answer the questions. Put the Cvs on a labeled number line, then check any number in each interval to find the sign of the derivative (I checked -3, 0, and 3). Then write that sign above each interval. As we can see from the number line, our curve is increasing as it comes from the left, levels off at and starts back down. Then at it goes back up again. Up - down - up means we have a max and then a min. Answer the questions, using the wording I have here (it is required for the AP exam, so learn it now.). 3

11 Increasing and Decreasing Functions If you remember that the derivative gives us the slope of the tangent to a curve, then this is an easy concept to grasp. An increasing function is where the curve is "going up" (i.e. the slope of the tangent line is positive); a decreasing function is when the curve is going down (i.e. the slope of the tangent line is negative). These areas will be easy to find using the first derivative, because the first derivative gives us the slope of the tangent line (I know I keep saying that over and over, but I want you all to know that relationship!). The easiest way to find the intervals where the function is increasing or decreasing is to first find the critical values. (Guessing is NOT a good way to do it and using the graphing calculator will get you no credit!) For example, let's use the function y = x 3-4x. To find the critical values, find the first derivative and set it equal to zero: y ' = 6x - 4 = 6 (x - ) (x + ) 0 = 6 (x - ) ( x + ) x =, - so the critical points are at the points where x = and - ** these are the x-coordinates of the points on the curve where the slope of the tangent line is equal to zero. That means that these are the places where we probably have a peak or valley on the curve. Now draw a number line for the first derivative, label it, and mark the critical values of x on it: y ': - We now have a number line with three sections on it. To the left of -, we have the interval (-, -), between the two critical values we have the open interval (-, ), and to the right of we have the interval (, ). We will check a number (any number except for the endpoint of a section itself) in each of those sections to see if the first derivative is going to be positive or negative: check x = -5, y ' = 6(5) - 4 > 0 check x = 0, y ' = 0-4 < 0 check x = 5, y ' = 6(5) - 4 >0 Note: Again, it does not matter what x-value you choose as long as it lies within the section of the number line you want to check, and is not equal to one of the critical values which divide the sections of the number line. This step, where you actually find the signs of the derivative does not have to be shown on your paper just move on to the next step. Arithmetic just clutters up your paper! Now mark these results on the number line (this must be shown):

12 + - + y ': - Write your results that we got from our number line that showed the signs of the slope of the tangent lines along the curve in the intervals: - the curve is increasing from - to - because y is positive. - the curve is decreasing in the interval from - to because y is negative. - the curve is increasing again from to + because y is positive. An alternate way to word the conclusions is to use interval notation. Notice that if you are talking about intervals we use the word ON, not AT (which is what we say when we are talking about a point) - the curve is increasing on (- -) because y is positive on that interval. - the curve is decreasing on the interval from (-,) because y is negative. - the curve is increasing on (, ) because y is positive. Looking ahead: What we just did is the process that we will be using to show where a curve is increasing or decreasing, as well as the process we will use in the next section to find max and mins along our curve. I am sure you can already see how that will work. Look again at the last number line we had: y ': - If the curve is going up until it gets to - and then starts going down again we must have a maximum. And if it continues down until it gets to and then starts going back up, then we must have a minimum at x =. Without our calculator we already know what the curve will look like! ** Note: I expect you to use the labeled number lines like I do in the examples here. I will not accept the charts that the book uses! The number lines are quicker and easier to read at a glance.

13 Summary of using the y number line to determine the increasing and decreasing intervals of a function Find the CVs of a function by finding the derivative, setting it equal to zero, and solving for x This works because the derivative gives us the slope of a function, so by setting it equal to zero we will find the places where there is a horizontal tangent line. Note if you are not good at factoring, you can use your calculator to find the values of x. Any algebra teacher can help, or contact me. (what you do is graph the derivative then use the calculator function to find its zeros) Draw and label a number line and place the CVs under it This divides the number line into sections. Note: The #line must be labeled with the name of the derivative (y or dy/dx or A, ) or you will lose your justification points on the AP and on our HW and on our tests. 3 Test a number within each section of the number line by putting it into the derivative. Determine if the derivative will be positive or negative. Place either a + or a above each interval according to your results. Since the derivative gives the slope, by finding the sign of the derivative we know if the slope of the tangent line is positive or negative within each interval. Note: the arithmetic that you do here does not have to be put on your paper in fact I prefer that you don t; it just clutters things up. Note #: Pick easy numbers from each interval! For the example below I used the factored derivative and the numbers x = -, -, and. If you pick easy numbers then you should be able to do this work in your head. 4 Give your results underneath your work, using the required wording. The wording is required by the AP memorize it now so that it becomes second nature to do it correctly. 4 Note: you can use interval notation,,0 3, or inequalities, 4 x 0 3 sure not to include the endpoints because that is where y = 0, so the slope is zero and the function is neither increasing nor decreasing. An example:, for this step. Just be For the function y = x 3 + 4x find the increasing and decreasing intervals of the function. Note: The stuff on the left side of the table is what is required to be shown.

14 y = x 3 + 4x y = 6x + 8x = 0 x(3x + 4) = 0 x = 0, -4/3 The CVs are x = 0, -4/3 y The function is increasing on the intervals 4, 3 on those intervals /3 0 and 0, because y is positive This is the function. Take the derivative. Set it equal to zero. Solve for x. Give your CVs. Draw the number line and label it. Mark the CVs under it. Test a number in each section to see if y will be positive or negative. Mark the results above the #line. Give the results with the required wording. The function is decreasing on the interval 4,0 3 interval. because y is negative on that PS I know it seems like a pain to have to write all the words! But, it is required so make yourself do it. No sense in learning it wrong

15 Homework Examples #7, 33 #7: Find all possible functions f with the given derivative. f ( x) 3x x This is our derivative. 3 f ( x) x x x C To find the function from the derivative, we reverse the power rule. Add one to the exponent and then divide by that new exponent. It is easy to check to see if you did it right by taking the derivative to see if you get back to the original derivative. We need the + C to take care of any constants that were in the original function. Their derivatives are zero, so they drop out when we take the derivative of the function. #33: Find the function f with the given derivative whose graph passes through the given point. f ( x) where x > - x f (x) = ln x + + C P (-, 3) 3 = ln C 3 = ln + C C = 3 f (x) = ln (x + ) + 3 This is our derivative. This time we are going to reverse the derivative of the natural log. We need the + C to take care of any constants that were in the original function. We have a particular point that this function contains, so we will be able to find the value of that C. Put the point into the equation and solve for C. Write your final answer. We can write the ln without the absolute value signs because we were told that x > -, so x + will always be positive.

16 Antiderivatives and Integration Even if you are not thrilled at the prospect, anti-derivatives are very important in the real world. Derivatives are functions which tell us how variables are changing with respect to each other or with respect to time. Many times in science or business we can more easily measure how things are changing than to see what the original relationship was before the change started. With derivatives we can measure this change and write an equation expressing it and then, using anti-derivatives, we can then find an equation which equates how the variables were originally related to one another. This chart shows some of the relationships we have discovered already. Up to now we could only move across the chart to the right by taking the derivative. Now we will be able to move back to the left by taking the anti-derivative. original function antiderivative position velocity equation of curve cost revenue derivative of original function changing relationship velocity acceleration slope of tangent line marginal cost marginal revenue Now the big question: how do we find the anti-derivative? It happens that all we have to do is reverse the process that we used to find the derivative in the first place. As you can imagine, some of our rules for taking the derivatives will not be real easy to reverse (i.e. the product, chain, and quotient rules). For right now, all we have to do is reverse the simple power rule and the basic trig functions. We also have to pay attention to any constants that may have been lost while taking the derivative.

17 Antidifferentiation - Reversing the Power Rule Power Rule: The power rule had us multiply out in front by the original exponent, then subtract one from the exponent. To reverse this process, first add one to the exponent, and then divide by this new exponent. In summary: Find the derivative with the power rule: * Multiply any coefficient by the old exponent. * Subtract one from the exponent. Find the antiderivative by reversing the power rule: * Add one to the exponent. * Divide any coefficient by the new exponent. Examples: ( The + C will be explained later) Differential Equation Antiderivative What we did: y ' = 0x y = 5x + C Add one to the exponent of, and then divide by the. y ' = 4x y = 8x 3 + C Add one to the exponent of, and then divide by the 3. 3 y 6 4x 3x 5x 3 5 Go term by term, same process as 4 y 6x x x x C in the above examples. / y x x 3/ y x C 3 y x 4 3 y x C 3 y x 4 Add / + = 3/ and then divide by the 3/. Add one to -4 to get -3, and then divide by the -3.!!!!***** Ahhhhhhh! We can't do this yet! If we try to add one to the exponent we get 0, and division by 0 is not defined. Be patient - we will get to it later. The wonderful thing about antiderivatives is that you can always check yourself by taking the derivative of your answer to see if you end back where you started.

18 Antidifferentiation Reversing the Trig Derivatives Basic trig functions: Just go backwards. But remember that if the angle is a multiple of x, you had to multiply by that coefficient when taking the derivative (chain rule), so we will have to divide when finding the antiderivative. *** Taking the antiderivative of a trig function seems to be tricky for many people, probably because of the negative sign in the derivative of cos x. Always stop and check that you have done it correctly. I suggest keeping a chart of the trig derivatives handy until you get good at this. I have one below for reference: When you take the derivative, move right -> When you find the antiderivative, move left <- sin x cos x tan x sec x Trig function Derivative Function cos x -sin x sec x sec x tan x Examples: Differential Antiderivative How we did it: Equation y ' = sin x y = - cos x + C The derivative of cos x is - sin x. y ' = cos x y = sin x + C The derivative of sin x is cos x. y ' = 3 cos 3x y = sin 3x + C The derivative of sin 3x is 3 cos 3x : the function 3x embedded in the function of sine. Since 3x is an embedded function we had to divide by its derivative to find the antiderivative.

19 Disappearing Constants and Indefinite Integrals The disappearing constants: Whenever you take the derivative of a function, the derivative of any constant is zero. Thus, when the antiderivative is taken, we have to remember that there may have been a constant added to the end of the original function which disappeared when we first took the derivative. That is why we have the + C at the end of the antiderivative; to hold the place of that mystery constant. It must be there - points will also disappear if it is left off! Indefinite integral is a synonym for antiderivative and if we want to solve a differential equation (i.e. find the original equation), we use this notation: F x dx F x C. The left side is our differential equation, in an integral sign (that s-like swish). The right side is its antiderivative plus that mystery constant. One of the most important things you can do to make your life easier is to rewrite the differential equation before you try to take the integral. Or at least make sure that it can't be rewritten before you launch yourself into the process. Here are some examples: Original Integral Rewrite Integrate 3 3 dx x dx use negative x exponents to clear the singleterm denominator add one to the exponent to get - x C divide by new exponent / x dx x dx 3/ rewrite the radical with fractional exponents sinxdx sinx dx constants should be moved out in front 3 x C add one to the exponent to get 3/ divide by new exponent (- cos x) + C - cos x + C reverse the trig derivative - be careful of the sign Remember that you can always (and definitely should as you begin learning this process) take the derivative of your answer and see if you get back to the original integrand (the function in the integral that you started with).

20 Initial Conditions and Finding the Particular Solution Finally we get to the subject of Initial Conditions. This is where we are given some information which will allows us to find the value of the mystery constant. When we only find the antiderivative with the + C, we have a family of functions that all differ by a constant. With initial conditions we find a single solution that belongs to a unique function. Here are some examples: a) dy/dx = x - 7 This is the differential equation. y= 0 when x = Initial conditions have been given. y = x - 7x + C First take the antiderivative of dy/dx = x - 7. Remember that you can check by taking the derivative of the answer and comparing with the derivative you were given to start with. y = 0 when x = : Sub in the initial condition values you were given and 0 = C solve for C. C = 0 y = x - 7x + 0 Finally, rewrite the function with the value of C. You must do this step for full credit. We can do a quick check and see that the derivative of y = x - 7x + 0 is in fact the derivative which we were given: dy/dx = x - 7. b) If we are given velocity and asked to find the equation for position, we have to take the antiderivative. Remember that the derivative of position is velocity and the derivative of velocity is acceleration. Derivatives move us to the right from position -> velocity -> acceleration. The antiderivative will move us back to the left. v = 9.8t + 5 s = 0 when t = 0 s = 4.9t + 5t + C s = 0 when t = 0: 0 = C C = 0 s = 4.9t + 5t + 0 This is the derivative function for velocity. Initial conditions have been given. First take the antiderivative. Remember that the antiderivative of velocity is position. Sub in the values you were given and solve for C. Finally, rewrite the function with the value of C. A quick derivative check verifies our work: s = 9.8 t + 5

21 c) We are given a fourth derivative and have to find our way back to the original function. This means we will be taking the antiderivative 4 times, stopping each time to find the C. This last bit is very important! Don't clog your work up with tons of constants, because each time you take the antiderivative you will get a new one. It is much easier to stop each time and find the value of the constant, and then move on to the next step. y (4) = - sin t + cos t y ''' = 7, y '' = -, y ' = -, and y = 0 when t = 0 y ''' = cos t + sin t + C y ''' = 7 when t = 0 7 = cos 0 + sin 0 + C 7 = C C = 6 y ''' = cos t + sin t + 6 y '' = sin t - cos t + 6t + C y '' = - when t = 0 - = sin 0 - cos C - = C C = 0 y '' = sin t - cos t + 6t y ' = - cos t - sin t + 3t + C y ' = - when t = 0 - = -cos 0 - sin C - = C C = 0 y ' = -cos x - sin t + 3t y = - sin t + cos t + t 3 + C y = 0 when t = 0 0 = - sin 0 + cos C 0 = C C = - y = - sin t + cos t + t 3 - This is the fourth derivative function. Initial conditions have been given. Lots of them because we are to find the original function y, which is 4 steps back! We will have to pay careful attention to the notations so that we use the right numbers for each step. Find the first antiderivative. We stop and find the value of the first mystery constant (constant of integration is its official name). I put the 7 in for y ''', and the 0 in for t. Do not forget to stop and do this every time you go through the process of finding the antiderivative. Write the entire third derivative before moving on. This is the second antiderivative. Again we find the new constant of integration, using the values for t and y '' that we were given in the initial conditions. Write the entire second derivative before moving on. Take the antiderivative again. Again we find the new constant of integration, using the values for t and y ' that we were given in the initial conditions. Only one more time to go! This is the last antiderivative. Whew! Again we find the new constant of integration, using the values for t and y ' that we were given in the initial conditions. Now we are done. We have found y, the original function using antidifferentiation and the given initial conditions.

22 One last note: You can always check to see if your antiderivative is correct; just take the derivative of your answer and see if you end up with what you were given to start. It is easy to do and well worth doing! Don't forget the constants of integration! I have one other example I want to do before I get to the homework pages. As you go through all of these, if you don't understand something, let me know right away. It is your responsibility to let me know when something does not make sense - my ESP is dismal. d) We are to find the curve in the xy-plane that passes through the point (9,4) and whose slope at each point is 3 x. mt 3 x dy 3 dx x 3x 3/ y x C thru (9, 4): 4 9 3/ C / First we have to translate the second part of the sentence into math. We know the slope of a tangent line or the curve comes from the first derivative of the equation of the curve, so I have rewritten the equation with derivatives. Take the antiderivative. Adding one to / gives a new exponent of 3/. Then divide by that new exponent. Use our initial conditions to find C. We have a point (9, 4) so we know that x = 9 and y = 4. 4 = 54 + C C = -50 3/ y x 50 Finally we write the final equation, with the value of C included. Not too bad, I hope. Just be careful to do a quick check when you are done (especially with the trig functions) and to stop to use your initial conditions to find the constant of integration every time you take the antiderivative. 3

23 Homework Examples #5, 9 #5: Show that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find the value of c that satisfies the theorem. f (x)=x + x - [0, ] f (x) is continuous on [0, ] and differentiable on (0, ) Slope of secant = slope of tangent Slope of secant = derivative Secant between (0, -) and (, ) f (x)= x + c 0 3 = c + c = / This is our function and interval. Show that the function is both continuous on the close interval and differentiable on the open interval. Find the value of x where the slope of the tangent equals the slope of the secant line. Set the slope of the secant line equal to the derivative and solve for c. #9: Write the equation for the secant line AB. Then write the equation of the tangent line that is parallel to AB. f( x) x x 0.5 x Secant line between (0.5,.5) and (,.5) m = 0 secant line: y =.5 Tangent line ( ) f x x x Tangent slope will be zero: 0 when x = -, x x = - does not lie in the interval Point: (, ) m = 0 Tangent line: y = This is our function and interval. Find the slope of the secant line and write its equation. Since the slope is zero, we know the line is horizontal. To write the equation of the tangent line, we need to know where the derivative will equal the slope of the secant line.

24 Homework Examples #39, 4, 49 #39: A trucker handed in a ticket at a toll booth showing that in hours she had covered 59 miles on a toll road with a speed limit of 65 mph. The trucker was cited for speeding. Why? The average speed across the road: 59/ = 79.5 mph The path of the truck and its speed are continuous and differentiable, so the Mean Value Theorem applies to this situation. Find the average speed (essentially the secant slope of this interval). Use the MVT to show that somewhere during the trip there had to be a time when the trucker was going faster than 65 mph. Thus, there must be at least one instant when the trucker was exceeding the 65 mph speed limit and going 79.5 mph. #4: Classical accounts tell us that a 70-oar trireme once covered 84 sea miles in 4 hours. Explain why at some point during this feat the trireme s speed exceeded 7.5 knots. (and for a point of EC, send me a message with the heading Trireme and tell me what it is and why it has the tri in its name) The average speed of the boat: 84/4 = knots The path of the boat and its speed are continuous and differentiable, so the Mean Value Theorem applies to this situation. Find the average speed (essentially the secant slope of this interval). Use the MVT to show that somewhere during the trip there had to be a time when the trireme was going faster than 7.5 knots. Thus, since 7.5 < 7.667, there must be at least one instant when the boat was exceeding 7.5 mph.

25 #49: Show that the equation has exactly one solution in the given interval. x + ln (x + ) = 0 0 x 3 Let y = x + ln (x + ) The function is continuous and differentiable in our interval. At x = 0, y = 0 so there must be at least one root. x y x x only equals zero at x = -, which is out of our interval. Since y cannot equal zero, there are no extrema within our interval and thus there will only be one root to the equation. This is our equation and interval. There are two steps to the process. First we show that there must be at least one solution. Usually we would have to find one point above the x-axis and one below, and then cite the intermediate value theorem that says if a continuous function goes from positive to negative it must cross the x-axis at least once, but in this case when I tested one end of the interval I immediately found a root. Now we have to show that there cannot be more than one solution. To do this, we show that there are no extrema (in order for the function to cross the axis once and come back again there has to be a max or min).