The First Derivative Test - PDF Free Download

The First Derivative Test We have already looked at this test in the last section even though we did not put a name to the process we were using. We use a y number line to test the sign of the first derivative on either side of the CVs. If a CV is a local extremum, then they will be the highest or lowest points in that vicinity. If it is an absolute extrema (what we did in lesson 4-), they will be the highest or lowest points anywhere on the curve or within a closed interval. When you do the problems, be sure to be aware of the difference between the two types of extrema! We can find local maxima or minima by looking at the number line and examining the signs of the slopes of the tangent lines. This method is called the First Derivative Test. In the graph below, the slope of the tangent line goes from positive slope to zero slope (at the critical value where the tangent is horizontal) to negative slope. This can only happen at a maximum. Look at the graph below to visualize this: In this next graph the slopes of the tangent lines go from negative to zero to positive. This can only happen at a minimum. Looking at the slopes of the tangent lines will tell us a lot about the behavior of the curve.

We can take the information from this figure and put it on a first derivative number line. The critical values are at all of the c-values, where f is either undefined or equal to zero. From the number line we can indicate the slopes of the curve between the CVs. f + + + c c c 3 c 4 c 5 Usually critical values will give you a maximum or minimum, but this does not always happen as you can see at points c and c 5 where there is no change in sign of the first derivative. The slope in those cases does not change sign so we have a horizontal tangent but no max or min. Thus you have to check on your number line to be sure that the curve is in fact going from increasing - zero - decreasing or vice-versa. When you are asked on an exam to find the extrema, you have to show your reasoning on the number line like the one we just used. Just finding the extrema with your calculator will not receive full credit (as in only getting point out of 4, not good). If you are checking for local maxima or minima, this is all the work you have to do. You have found the places with the peaks and valleys of the graph. However, you will have to present your answer with the proper wording justification or you will lose all the credit! Number line Information Summary: Here I have some diagrams that show you the different situations that you might encounter, both graphically and on the number line. a) Relative Minimum:

Signs of f on either side of c With a relative minimum, the slope will fall and then rise. f (x) Number line label to indicate the derivative which is being tested _ + c Critical value where f (c) = 0 b) Relative Maximum: Signs of f on either side of c For a relative maximum, the slope increases and then decreases. f (x) + Number line label to indicate the derivative which is being tested c _ Critical value where f (c) = 0 c) Neither a minimum or maximum found: If the slope does not change sign, there is no relative maximum or minimum. 3

The First Derivative Test - An example Note: There is a lot more here than what you actually need to show on your paper. Look at the Summary of the First Derivative Test to see a "stripped down version" of the process with this same example, showing what you need to show me. A: Find the extrema of f if f '(x) = (x - )(x + ). This is a good test and AP question!! It makes it impossible to use your calculator to find the answer for the extrema because we are not given the equation of the function, only of the derivative, f '(x) = (x - )(x + ). We must go through the steps to find the local extrema. a) We need to find the critical values. To do that we have to set the first derivative equal to zero and also to look for values of x for which the derivative is undefined. You should have everything on the left side of the table written on your paper to ensure you get full credit. You must show that you have set the derivative = 0 and what values of x will make that equation true. f '(x) = (x - ) (x + ) f '(x) = 0 when x =, - Our critical values are at x = and - No values of x will make the equation undefined. No values of x will make the derivative undefined. CVs: x =, - b) To find the intervals on which f is increasing or decreasing we must look at the behavior of f '(x). When the slope of the tangent is positive, the tangent and the curve are going up/increasing. When the slope is negative, the tangent and curve are decreasing/going down. Set up a number line, using our critical values that we found above as the dividing points on the number line. Check a value of x in each interval of the number line (we have three intervals; - to -, - to, and to ) to see if the slope is positive or negative. I expect to see this on your paper! This is your justification. f '(x): + - + - You must have a wrap up in words of the information you got from the number line! It is essential. You will receive no credit on the AP exam for a number line left by itself with no answer clearly written (I was told this during my workshop put on this summer by the chief grader of the AP exam). Therefore I will give you no credit on our test or homework either. You also are required to label the number line as being that for the first derivative (also an AP requirement). Never have unlabeled number lines scattered about! So, your wrap up for this part is (this extra wording is part of the new rules for AP Exam as of 005): The curve is increasing on the intervals (-, -) and (, ) because f '(x) is positive

and decreasing on the interval (-, ) because f '(x) is negative c) From the number line we can find our local minima and maxima. When a curve goes up, levels off, and then goes down we have a local maximum. On our number line we see this happening at x = -. When a curve goes down, levels off, and then goes back up we will get a local minimum. This is happening at x =. Below I will draw (or at least attempt to draw) the number line with arrows above it showing what the curve is doing. If it helps you visualize what is happening, you may want to do this on your paper until you get comfortable with the procedure. Pos m t Local max Local min + - + Neg m t - Pos m t That is it. Practice these until you understand the relationship between the derivative, your slope of the tangent line, the number line showing the sign of the first derivative (and therefore the sign of the slope of the tangents), and the critical values of the first derivative (possible extrema). Every AP has one of these and we will come back to this concept continually. If you need to, make a chart of these relationships and paste them on your bathroom mirror, in your locker, wherever you will see if over and over! The wrap up for this part (also part of the new rules): f (x) has a local maximum at x = - because the f '(x) changes from positive to negative at x = -. f (x) has a local minimum at x = because the f '(x) changes from negative to positive at x =. ** Note: Do NOT use the charts that they use in your book examples!! I want you to use the labeled number lines. They are quick, efficient, and easy to read at a glance. You just need the () labeled #lines, with the signs of f above each interval (none of the arithmetic you went though to get the signs is required - leave it off your paper and use scratch paper if you want), and the () justification wording below the chart.

The First Derivative Test - The Summary I am going to do the same problem as before, but without the extra explanations. Use the steps that I have below to see what I am doing in each step. The basic steps of using the First Derivative Test to find extrema are:. Find the derivative. Set it equal to zero. Solve for x. Find any values of x for which the derivative is undefined. Show these answers clearly. They are your CVs. Draw and label your y ' number line. Mark the CVs under the #line. Choose a number in each interval and put it into the derivative to find its sign. Mark the sign above the #line. 3. Draw your conclusions, including the required words of justification. For increasing intervals: The function is increasing on (a, b) because y ' is positive on that interval. For decreasing intervals: The function is decreasing on (c, d) because y ' is negative on that interval. For a max: There is a max at the point (x, y) because y ' changes from positive to negative at x = For a min: There is a min at the point (x, y) because y ' changes from negative to positive at x = When there are no extrema: There are no extrema because y ' does not change sign. Note: I know that is a lot of writing, but make yourself do it. It is required and you need to practice it enough so that you do it automatically. Note #: For the intervals, make sure not to include the endpoints in your interval. (remember that at the CVs the derivative has a value of zero so the function is neither increasing nor decreasing.)

Here is that problem, with just the required work showing. In this case we have been given the derivative already so we are saved some work. Find the extrema of f if f '(x) = (x - )(x + ). Also find the intervals in which the function is increasing and decreasing. f ' (x) = (x - )(x + ) f ' (x) = (x - ) (x + ) = 0 f ' (x) = 0 when x =, - No values of x will make the derivative undefined. CVs: x =, - f '(x): + - + - f (x) has a local maximum at x = - because the f '(x) changes from positive to negative at x = -. f (x) has a local minimum at x = because the f '(x) changes from negative to positive at x =. f (x) is increasing on the intervals, and, because f ' is positive on those intervals. f (x) is decreasing on the interval, because f ' is negative on the interval. That's it - you are done! Note: On this problem I could not give the point for the max and min because I don't have the equation of the function so there is no way to find y. I could only give the x-values of the extrema. Usually, though, you should give the coordinates of the extrema unless you are specifically asked just for the x-coordinates.

Concavity Now let s look at what we can find out about our curve from the second derivative. One of the first things you will need to do is resist the temptation to use your calculator immediately. Use it after you have made your sketch, to check yourself. Don't use it to copy from, because you will end up handicapping yourself. You need to be able to interpret the information that the derivatives give you and also understand what the graph tells you about the derivatives. Struggle with it for a while first and give yourself a chance to learn. There are lots of places on the AP exam and on our chapter test where the calculator will not be allowed or will not do you any good. You must learn to survive without it. The concavity of a curve is the "way it is bending". If it is curving upward, it is said to be concave up. It "holds water". If the graph of f lies above its tangent lines, it is said to be concave upward. The slope, f, is increasing. If it is curving downward, it is said to be concave down. It "spills water".

If the graph of f lies below its tangent lines, it is said to be concave downward. The slope, f, is decreasing. Technically, we say that the curve is concave up when f ' is increasing and the curve is concave down when f ' is decreasing. If something is increasing, that means its rate of change, its slope, is positive. Thus the derivative of f ' will be positive. That idea gives us the Test for concavity: if f '' > 0 within an interval, then the graph of f is concave upward in that interval. Conversely, if f '' < 0 within an interval, then the graph of f is concave downward in that interval. To find the concavity of a curve, you simply have to find the second derivative and set it equal to zero. Then use the number line test like we did to find extrema, to test the intervals on either side of where f '' = 0. Remember to label the number line with a y ''. You can't have number lines running around your paper without labels. The AP folks and I want to know that you know which derivative you are testing to get your information.

Testing for Concavity - The Summary The basic steps to follow when you want to find the concavity of a function are:. Find the second derivative. Set it equal to zero. Solve for x. Find any values of x for which the second derivative is undefined. Show these answers clearly. They are your CVs. Draw and label your y '' number line. Mark the CVs on the #line. Choose a number in each interval and put it into the derivative to find its sign. Mark the sign above the #line. 3. Draw your conclusions, including the required words of justification. For intervals of upward concavity: The function is concave up on (a, b) because y '' is positive on that interval. For intervals of downward concavity: The function is concave down on (c, d) because y ' is negative on that interval. Note: I know that is a lot of writing, but make yourself do it. It is required and you need to practice it enough so that you do it automatically. Note #: For the intervals, make sure not to include the endpoints in your interval. (remember that at the CVs the derivative has a value of zero so the function is neither concave up nor down.)

Here is a problem, with just the required work showing. Find the concavity of f if f (x) = (/3) x 3 + ½ x - x f ' (x) = x + x - f '' (x) = x + = 0 f '' (x) = 0 when x = - ½ No values of x will make f '' undefined. f (x) is concave down on the interval, because f '' is negative on the interval. f (x) is concave up on the interval, because f '' is positive on the interval. That's it - you are done!

Inflection Points Inflection points are pivot points where the curve changes its concavity (the relative position of its "cup"). If a curve is concave up it "holds water" and looks like a parabola opening up. When a curve is concave down it "spills water"; it looks like a parabola which is opening down. There is always an inflection point between two extrema because a max and a min have opposite concavity and there must be a transition point somewhere between them. However there does not have to be extrema on either side of an inflection point. The curve y = x 3 is a case in point. There is an infection point at (0,0) but no extrema. It looks like this: an inflection point does. See how the concavity changes at (0,0)? That is what To find a point of infection we will use the second derivative. The inflection point is possible where the second derivative is either equal to zero or undefined. However, just because the second derivative equals zero or is undefined at a point, it still might not be an inflection point. To test and make sure, we use the second derivative number line again, this time plotting the signs of the second derivative around its critical values (the places where it is equal to zero or undefined). There must be a sign change if it is in fact an inflection point - and you must show this in order to get credit for inflection points. In fact, whenever the second derivative is positive, the curve is concave up and wherever it is negative the curve is concave down. If there is no sign change, you will have a cusp: Notice how the concavity does not change here at the cusp. You also have to give the proper words of justification under the number line when giving the points of inflection.

Let s look at a hypothetical problem, where y = 0 at x =. First we would make the #line for y, put the x = under it, and then check the signs of y on either side of. Case : There is a point of inflection at (, 4) because at x = y changes signs. Case : There is no POI because y does not change signs at x =. Here is a chart to summarize all we know about graphing curves with the help of the first and second derivatives: Critical Max at points x = c y ' = 0 y '(c) = 0 and change of sign from + to - at this point y '' = 0 y '' (c) < 0 there is a max at the top of an upside down U Min at x = c y '(c) = 0 and change of sign from - to + at this point y ''(c) > 0 there is a min at the bottom of a U Increasing Function y ' > 0 slope of tangent is positive Decreasing function y ' < 0 slope of tangent is negative Concave Up Concave Down Inflection Point y ' y ' y ' may be increasing decreasing equal to 0, (but we (but we but does don t use don t use not have to this this be definition definition to to show show concavity) concavity) y '' > 0 y '' < 0 y '' = 0 and change of sign in y '' at this point

Testing for POIs - The Summary The basic steps to follow when you want to find the points of inflection of a function are:. Find the second derivative. Set it equal to zero. Solve for x. Find any values of x for which the second derivative is undefined. Show these answers clearly. They are your CVs. Draw and label your y '' number line. Mark the CVs on the #line. Choose a number in each interval and put it into the derivative to find its sign. Mark the sign above the #line. 3. Draw your conclusions, including the required words of justification. There is a POI at (x, y) because y '' changes sign at that point. Or: There is no POI because y '' does not change sign. Note: I know that is a lot of writing, but make yourself do it. It is required and you need to practice it enough so that you do it automatically.

Here is a problem, with just the required work showing. Find the points of inflection of f if f (x) = (/3) x 3 + ½ x - x f ' (x) = x + x - f '' (x) = x + = 0 f '' (x) = 0 when x = - ½ No values of x will make f '' undefined. There is a POI at 3, because f '' changes sign at x = - ½ That's it - you are done! Note: I found the point by taking the x = - ½ and putting it back into the equation for the function. Make sure you put it into f (x), not one of the derivatives!

The Second Derivative Test We have already talked about extrema (maxima and minima). In review, a maximum (peak) or minimum (valley) will only occur where there is a horizontal tangent. A horizontal tangent has slope of zero. The first derivative gives the slope of a tangent line, so the first derivative will equal zero at these points. These are also called critical points/values (CVs). To check if we had a max or min we used a number line and plotted the sign of the first derivative (this is called the first derivative test). If there was a sign change at a critical value, we had a local extrema. Now we have another way to check the critical points to see if they are a max or min. It is called the Second Derivative Test. To use this, find the second derivative. Then substitute the x-values of the critical points (they are called "c" in the chart) into the second derivative. The following chart summarizes the possible outcomes: f '' (c) > 0 the critical point is a minimum (because at a min, the curve is concave up, U ) f '' (c) < 0 the critical point is a maximum (because at a max, the curve is concave down, ) f '' (c) = 0 the critical point might be an inflection point. You have to look at the f number line to be sure what is happening. Here is a smiley face summary of the second derivative test: + +

The second derivative test is easy to use. It is not, however, conclusive if the second derivative is either equal to zero or undefined at a critical value of c. In either of these two cases, you will have to resort to the first derivative test and the first derivative number line to be sure about what is happening at the critical value. If all else fails - remember the smiley faces! A smile is concave up (where y '' > 0), and there is a minimum at the chin. A frown is concave down (where y '' < 0), and there is a maximum at the nose. When you do the homework, you need to find all of the above points and characteristics of the curve and then put together a sketch. It also helps to know the intercepts (0, y) and (x, 0). Sometimes you will be asked to sketch the curve without the function - only knowing the derivatives, so you have to use the above table and your knowledge of what the derivatives tell you to sketch an unknown curve. It will be a challenge, but if you work through carefully it this will become much easier. Hang in there and contact me if you have questions.

Using the Second Derivative Test - Examples A: y = -x 3 + 6x - 3 a) Find critical values, the rising or falling of the curve, and local extrema. To do this we will need to take the first derivative. y ' = -6x + x = -6x (x - ) = 0 y ' = 0 when x = 0 or We set the first derivative equal to zero to find our critical values. These are the places where there might be local extrema. They are also possible dividing places between rising and falling sections of the curve. Possible extrema at x = 0, Now we use the number line to explore the behavior of the first derivative (slope of tangent line) in our intervals formed by the critical values. You do NOT need the arrows above the #line (I included them here just as a visual). You only need the CVs below the line, the label to the left of the line to tell what derivative you are testing, and the signs above the line. y : + 0 Local min Local max To confirm the local extrema we could also use the second derivative test. y '' = -x +. At x = 0, y ''> 0 and at x =, y '' < 0. Now, to give the answer, we have to use words and proper justification. The number line or the second derivative work alone will NOT give you any credit! The curve is decreasing on the intervals x < 0 and x > because y ' is negative. The curve is increasing on the interval 0 < x < because y ' is positive. There is a local min at (0, -3) because y ' changes from negative to positive at x = 0 There is a local max at (, 5) because y ' changes from positive to negative at x =. OR There is a local min at (0, -3) because y ' = 0 and y '' > 0 at x = 0 There is a local max at (, 5) because y ' = 0 and y '' < 0 at x = 0 Remember - this word wrap up is required! A number line by itself is not considered an answer; it is merely your exploration/justification. I know I said that twice, but it is important to

remember!! Why do all the work and then lose credit for it just because you were too lazy to write the answers out properly?! Note: When finding the extrema, use one test or the other, but not both! You might be told to use one or the other but usually it is up to you. I use whichever is easier. If I need to use the y ' number line to find intervals of increasing or decreasing, I will usually use the first derivative test because I already have the y ' #line. But other times, the second derivative is so easy to find it is easier to use the second derivative test. b) Find the inflection points and concavity. To do this we have to look at the second derivative. y '' = -x + = - (x - ) = 0 y '' = 0 when x = possible inflection point at x = Take the second derivative and set it equal to 0. The only possible inflection we have found is at x =. To check to see if it is an inflection point we have to look at the number line to see if there is a change in concavity at x =. You do not need the smile/frown on the #line. You only need the possible POI below the line, the label to the left of the line to tell what derivative you are testing, and the signs above the line. y : + Inflection point Look at your number line work and give the answers with the correct wording. There is a change of sign in the second derivative at x =, so there is an inflection point at (, ). The curve is concave up when x < because y '' >. The curve is concave down when x > because the second derivative is negative. c) Use the information we have found to draw a sketch of the curve. You will frequently have to do this when you are not allowed to use your calculator! We have the concavity, the rise and fall of the curve, the extrema at (0, -3) at (, 5), and the inflection point at (, ).

Max (, 5) Inflection point (,) Min (0, -3) B: In this problem, we are only given the equation of the first derivative to work with. This is a good example of a problem in which your calculator will not do you much good. We have to find the relative extrema. a) Find the extrema. We will find the critical values from setting the first derivative equal to zero. y ' = x - x - 6 = (x - 3) (x + ) = 0 y ' = 0 when x = 3, - these are possible extrema We find two critical values, x = - and x = 3. They are possible local extrema, our CVs. Now we use the y number line to explore the slope of the curve in the intervals (-, -), (-, 3), and (3, ). y : + + - 3 From the number line information we can answer the question. We have a local max at x = - because y ' changes from positive to negative There is a local min at x = 3 because y ' changes from negative to positive. The curve is rising when x < - and x > -3 because y ' is positive. The curve is falling in the interval - < x < 3 because y ' is negative. 3

We could also have used the second derivative test to locate the extrema. If we find the second derivative we have: y = x - At x = -, the second derivative is less than zero and at x = 3 the second derivative is greater than zero. Our answer would then read: There is a local min at x = 3 because y ' = 0 and y '' > 0 at x = 3 There is a local max at x = because y ' = 0 and y '' < 0 at x = b) Find the concavity and inflection points. To do this we have to look at the second derivative. y '' = x - = 0 y '' = 0 when x = / possible inflection point at x = / Take the second derivative and set it equal to 0. The point at x = / is a possible inflection point. We will have to look at the number line to be sure that there is a change in concavity at x = /. y : / + Use the number line to get the answers. There is an inflection point at x = / because there is a change of sign in y'' The curve is concave down when x < / because y '' is negative The curve is concave up when x > / because y '' is positive. c) Sketch the curve. We have a max at x = -, an inflection point at x = /, and a min at x = 3. We can't find the exact points of the extrema or inflection point because we don't have an equation for the function itself, only its derivative. Thus, there are many ways to draw this curve, but the max, min, and inflection points will always be at the same x-values. 4

sketch!! A very rough The better you get at this the easier the rest of the year will be! You can do it if you will learn the proper steps and memorize the required wording for the answers. In a matter of days it will be second nature and these will be easy! 5

The Second Derivative Test - The Summary The basic steps of using the Second Derivative Test to find extrema are:. Find the derivative. Set it equal to zero. Solve for x. Find any values of x for which the derivative is undefined. Show these answers clearly. They are your CVs. Find the second derivative. 3. Take the CVs from step one and put them into the equation for y ''. Determine the sign of y '' in each case. 4. Draw your conclusions, including the required words of justification. For a max: There is a max at (x, y) because y ' = 0 and y '' < 0 at x = For a min: There is a min at (x, y) because y ' = 0 and y '' > 0 at x = Note: I know that is a lot of writing, but make yourself do it. It is required and you need to practice it enough so that you do it automatically.

Here is that problem, with just the required work showing. In this case we have been given the derivative already so we are saved some work. Find the extrema of f if f (x) = (/3) x 3 + ½ x - x Use the second derivative test. f ' (x) = x + x - f ' (x) = (x + )(x - ) = 0 CVs: x = -, f '' (x) = x + = 0 testing the CVs: f '' (-) < 0 f '' () > 0 f (x) has a local maximum at x = - because the f '(x) = 0 and f '' (-) < 0 at x = -. f (x) has a local minimum at x = because the f '(x) = 0 and f '' () > 0 at x =. That's it - you are done! Note: On this problem I could not give the point for the max and min because I don't have the equation of the function so there is no way to find y. I could only give the x-values of the extrema. Unusually, though, you should give the coordinates of the extrema unless you are specifically asked just for the x-coordinates.

Using Derivatives to find the Characteristics of a Curve First Derivative work: Find the CVs of a function by finding the first derivative, setting it equal to zero, and solving for x. Be sure to identify your CVs at this step. Also check for values for which the derivative is undefined. y ' =. = 0 x = a the function is undefined at x = c CVs: x = a, c Draw and label a first derivative number line and place the CVs under it 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. The arithmetic you do here does not have to shown. 4 Give your results for increasing/decreasing intervals and extrema underneath your work, using the required wording. If there are no extrema or the function never increases or decreases, make sure to say so and tell why. Unless you are specifically asked for only the x-coordinates of the extrema, give the coordinates of the points by putting your CV back into the original function to find your y- coordinate and then give the point. Remember that all intervals must be open. At the CVs the function is neither increasing nor decreasing. The function is increasing on the interval ( ) because y is positive on that interval. The function is decreasing on the interval ( ) because y is negative on that interval. There is a max at (a, b) because y ' changes from positive to negative at x = a. There is a min at (c, d) because y ' changes from negative to positive at x = c. If you prefer to use the Second Derivative test to find your max or min: 5 - Find the second derivative. 6 - Put the CVs into the second derivative and evaluate to find out if it is positive or negative. 7 - give the answers with the correct wording: There is a max at (a, b) because f '(a) = 0 and f ''(a) < 0 There is a min at (c, d) because f '(a) = 0 and f ''(a) > 0

Second Derivative work: Find the second derivative, set it equal to zero, and solve for x. These are your possible Points Of Inflection (POIs). Also check for values for which the derivative is undefined. y '' =. = 0 x = e the function is undefined at x = f possible POIs: x = e, f Draw and label a second derivative number line and place the CVs under it. 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. The arithmetic you do here does not have to shown. 4 Give your results for intervals of upward and downward concavity and POIs underneath your work, using the required wording. If there are no POIs or the concavity never changes, make sure to say so and tell why. Unless you are specifically asked for only the x-coordinates of the POIs, give the coordinates of the points by putting your CV back into the original function to find your y- coordinate. Remember that all intervals must be open. At the CVs the function is neither concave up nor down. The function is concave up on the interval ( ) because y ' is positive on that interval. The function is concave down on the interval ( ) because y ' is negative on that interval. There is a POI at (e, g) because y '' changes sign at x = e. Other possible questions: Find the zeros of the function (these need no justification) : Graphically - graph the function and use the calculator to find the zeros. Analytically - put zero if for x, and then for y, and solve each to find the intercepts.

An Example: Find the intercepts, intervals of increasing/decreasing/concavity, the 3 extrema, and the POIs for the function f( x) = x + x 0x+. Justify your answers. 3 f x x x x 3 3 ( ) = + 0 + Intercepts: y =, x = 0.05, -8.554, 7.0 f ( x) = x + x 0 f ( x) = x+ 5 x 4 = 0 ( )( ) x = 4, -5 the function is never undefined CVS: x = -5, 4 f (x) is increasing on the intervals (, 5) and ( ) f (x) is decreasing on the interval ( 5, 4) 4, because f ' is positive on those intervals. because f ' is negative on the interval. f (x) has a max at (-5, 7.833) because f '(x) changes from positive to negative at x = -5. f (x) has a min at (4, -49.67) because f '(x) changes from negative to positive at x = 4. Note: for the max and min you can use the second derivative test if you prefer: CVS: x = -5, 4 f ( x) = x+ f ( 5) < 0 f (4) > 0 There is a max at (-5, 7.833) because f ' (-5) = 0 and f ( 5) < 0 There is a min at (4, -49.67) because f ' (-5) = 0 and f (4) > 0 3

f ( x) = x+ = 0 x = - ½ f '' is never undefined possible POI: x = - ½ f (x) is concave down on the interval, because f '' is negative on the interval. f (x) is concave up on the interval, because f '' is positive on the interval. There is a POI at,.083 because f '' changes sign at x = - ½ 4

Homework Examples #, 3 #: Use the graph of the function f to estimate where f ' and then f '' are zero, positive, and negative. Look at the sketch. The curve comes in from the left, levels off at a max, goes down to a min, and then goes up to infinity. y ' = 0 at x = -, y ' > 0 when x < - and x > y ' < 0 when - < x < y '' = 0 at x = 0 y '' > 0 when x > 0 y '' < 0 when x < 0 The first derivative is zero whenever the tangent line is horizontal. All max and mins have horizontal tangents. The first derivative is positive when the curve is rising and negative when the curve is falling. When concavity changes, the second derivative is zero. This happens at all inflection points. The second derivative is positive when the curve concave up and negative when the curve concave down.

#3: Use the graph of f ' to estimate where the curve is increasing, decreasing, or has a local extreme. ** Remember we are looking at a graph of the first derivative! It is showing us the signs of f ', not the shape of the original function. I think it is much less confusing if you draw a quick y ' number line on your paper so you can see the signs of the first derivative the way we are used to using them. ** Remember to give your answers with the correct justification (practice now so that you won t forget to do it right on the AP exam)! Look at the sketch. The curve comes in from the left, levels off at a min, goes up to a max, and then goes down to infinity. Let s translate to a y number line. + + y - 0 You might want to change the graph to a number line so you won't get confused at what you are looking at: Increasing when x < - and on 0 < x < because y > 0 in those intervals. Decreasing on x 0 and when x > because y < 0 in those intervals. Local maxima at x = -, because y changes from positive to negative at those points. Local min at x = 0 because y changes from negative to positive at those points. The curve is increasing when the slope of the tangent lines is positive (y ' > 0). The curve is decreasing when the slope of the tangent lines is negative (y ' < 0). We have a local min when the sign of y ' changes from negative to positive (the curve goes down then up). We have a local max when the sign of y ' changes from positive to negative (the curve goes up then down).

Homework Examples #9, 3, 33 #9: Use the derivative of the function and analytical methods to find (a) the local maxima, (b) the local minima, and (c) the points of inflection. y x x y x x 0 when x =, CV: x =, The derivative is never undefined. y ' There is no max There is a min at x = because y changes from negative to positive at x =. + () ( ) ( ) ( ) ( ) ( 4) ( ) 3 5 0 y x x x y x x x y x x x y x x When x =, 5/3 possible POIs y '' + + 5/3 This is our derivative. Find the CVs by setting the derivative equal to zero and solving for x and/or finding where it is undefined. This derivative is never undefined. Now draw and label the y ' number line. As we can see from the number line, our curve is decreasing as it comes from the left, levels off at but keeps going down. Then at it goes back up again. Down - down - up means we only have a min. We can't give a point here because we don't have the original equation so we can't find the corresponding y-coordinate of our min. To get information about the POI we need the second derivative. We have to use the product rule. When we find the derivative of the first factor we use the general power rule. I simplified by factoring out a x -. Now look at the y'' number line to see if we do have POIs. Remember that there must be a change in sign of y '' in order for the POI to exist. There are POIs at x = and x = 5/3 because there is a change in sign of y '' at those points.

#3: Use the graph of the first and second derivatives of the function y = f(x) to find a possible sketch of f (x) that goes through point P. Since f ' is quadratic and f '' is linear, we know that f will be cubic. y ' + + a b The graphs are in your book. The second derivative is linear and the first derivative is quadratic. Transfer the signs of f ' and f '' to a number line so you can easily see what is happening. Looking at the curves can sometimes be confusing. I will call the CVs of f ' x = a, b. f '' = 0 at the origin. Thus we have a max at x = a and a min at x = b. y '' 0 + Thus we have a POI at x = 0. f P f ' Using our information, sketch a possible curve, making sure to position it so that the max and min and POI are in the right place and that the curve goes through P. f '' a b

#33: Use the information from the charts to (a) find the absolute extrema of f and where they occur, (b) find any points of inflection, and (c) sketch a possible graph of f. From the first row of the first chart we are given the following points on f (x): (0, 0), (, ), (, 0), and (3, -) From the first row of the second chart we know that the curve is above the x-axis between 0 and, and below the x-axis between and 3. f ' We have a local max at x = because y ' changes sign from positive to negative. Since the function is rising from 0 to, we know we have an absolute max at (, ). Because the function is continually going down from x = to the end of the interval, the absolute min will be at (3, -). f '' + 0 3 f '' is always negative, there is no change in sign, so there are no POIs. The curve is always concave down. (, ) (, 0) Look at the information for the function. You might want to start your graph and add to it as you go along. This gives us almost enough information to answer the first question, but you can't be sure about the max and min because there might be more than one local extrema. We need to look at the derivatives to be sure. Look at the first derivative information. I have put it on the number line. We know we have CVs at x = 0 and from the information in the first chart. We also know that there will be a corner at (, 0) where the derivative does not exist. Look at the second derivative information. I have put it on the number line. We know we have possible POIs at x = 0,, 3 from the information in the first chart. The second chart gives us the sign of f ''. Now sketch a graph. Use your points, the extrema, the slopes between them, the sharp corner where f ' does not exist, and the fact that the curve is always concave down to make your sketch. (3, -) 3

Homework Examples #9, 7, 9 Note: in these problems, do not do the problem in the order that they ask the questions. I want you to get used to doing all of the y ' work together, give those answers, then do the y '' work, and give those answers. So, give the answers in this order: first find: increasing and deceasing intervals and extrema, then find: intervals of concavity and POIs Note #: If you don't like the algebra involved here, use your calculator to find the CVs and possible POIs. Graph the equation and find its roots (zeros). Take the decimals correct out to 3 places (do not round). #9: Use analytical methods to find () intervals where the curve is increasing and/or decreasing, () any local extrema, (3) the intervals on which the function is concave up or down, and (4) any inflection points. 4 y x 4x This is our function. We will need to take the derivative in order to answer the questions. 3 y 8x 8x8xxx Take the derivative with the power rule and find the CVs by setting it equal to zero and solving for y ' = 0 when x = 0,, - x. This derivative is never undefined. CV: x = -, 0, (a) y is increasing on - < x < 0 and when x > because y ' > 0 in those intervals. (b) y is decreasing when x < - and on 0 < x < because y ' < 0 in those intervals. (a) there is a max at (0, ) because y ' changes from positive to negative at x = 0 (b) there are mins at (-, -) and (, -) because y ' changes from negative to positive at x = - and y x x y 0 when x 3 y ' y - 3 4 8 8 3 + + 0 + + 3 3 You need a number line to show the signs of the first derivative in order to get the answers to (a), (b), and (e). Be sure to give the answers under the #line with all the required words of justification. *** put all of the first derivative work together: solve y ' = 0, show the CVs, do the #line work, give the answers! Then, after you finish, do the same with the y '' work. Do NOT mix them up together. It makes it confusing to follow your work and the reader will only spend a second or two grading you paper. It needs to be clear and concise. You need a y '' number line to show the signs of the second derivative in order to get the answers to (c), (d), and (f). You can always use decimals correct to three places instead of the exact radical answers that I have here. Be sure to give the answers under the #line with all the required words of justification. (3a) the curve is concave up when

x or x because y '' > 0 on 3 3 those intervals. (3b) the curve is concave down on x because y '' < 0 on those 3 3 intervals. (4) there are POIs at, 3 9 and, 3 9 because y '' changes sign at x 3 #7: Use analytical methods to find () intervals where the curve is increasing and/or decreasing, () any local extrema, (3) the intervals on which the function is concave up or down, and (4) any inflection points. y e y y x x 5e 3e 0.8x x 0.8x e 3e e 3e 5e 5e e.4e e x 0.8x x x x 0.8x x 3e.8x 3e 0.8x 0 for all x y ' is never undefined there are no CVs a) y increasing for all x because y ' is always positive. b) y is never decreasing because y ' is never negative. ) there are no extrema because y ' never changes signs y x 0.8 x.8 x.8 x x.8 x.6 x e 3e 5.4e 3e e 0.8e 4.4e x.8x.6x e 6e 9e This is our function. We will need to take the derivative in order to answer the questions. Take the derivative with the quotient rule and find the CVs by setting it equal to zero and solving for x. This derivative is never undefined. You don't need a number line here because we don't have any CVs. Be sure to give the answers with all the required words of justification. You need a y '' number line to show the signs of the second derivative in order to get the answers to (c),

e 0.6 3 e y x 3e 0. 3e x 0 0.x e x 0.x.6x 0.8 e 3 0.x ln 3 x 5ln3 possible POI + y ' 3 0 5 ln3 (d), and (f). I skipped some steps here to save space (an ugly derivative for sure that required a bunch of algebra to reduce and simplify). I had to use the quotient rule, but I also had to multiply out the denominator so that I could get its derivative. Be sure to give the answers under the #line with all the required words of justification. 3a) the curve is concave up when x < 5 ln 3 because y '' > 0 on that interval 3b) the curve is concave down when x > 5 ln 3 because y '' < 0 on that interval. 4) there is a POI at (5 ln 3, 5/) because y '' changes sign at x = 5 ln 3. 3

#9: Use analytical methods to find () intervals where the curve is increasing and/or decreasing, () any local extrema, (3) the intervals on which the function is concave up or down, and (4) any inflection points. x, x y x, x, x y x, x When x <, y ' > 0 for all x When x, y ' < 0 for all x y ' undefined at x = a) y increasing for all x < because y ' is positive. b) y is decreasing for all x because y ' is negative. ) there are no extrema because where y ' changes signs (at x = ) the function is undefined 0, x y, x 3a) the curve is never concave up because y '' is never positive 3b) the curve is concave down when x > because y '' < 0 on that interval. This is our function. We will need to take the derivative in order to answer the questions. Take the derivative with the power rule. Even though it looks like we might have a CV at x = 0, x = 0 is not part of the domain of that piece. This derivative is undefined where the pieces join because function has a break there. Also, the slopes of the two pieces don't match at x =. You don't need a number line here because we can see the signs by inspection. Be sure to give the answers with all the required words of justification. You don't need a number line here because we can see the signs by inspection. Be sure to give the answers with all the required words of justification. 4) there are no POIs because y '' never changes signs 4