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# Exam 3 Study Guide Math 210

University of Louisiana at Lafayette

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This 18 page Study Guide was uploaded by Cynthia Ndulaka on Tuesday July 19, 2016. The Study Guide belongs to Math 210 at University of Louisiana at Lafayette taught by Professor Bryant in Summer 2016. Since its upload, it has received 52 views. For similar materials see Calculus in Math at University of Louisiana at Lafayette.

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Date Created: 07/19/16

Calculus 1 (Chapter 4) Exam # 3 Study Guide This studyguide covers all topics with examples from chapter four and will include all step by step explanations of each problems. The topics that will be covered are: o Absolute and Relative Extrema o Rolle’s Theorem / Mean Value Theorem o The 1 and 2 Derivative Test o Limits at Infinity o Sketching Graphs o L’Hopital’s Rule/ Indeterminate Form o Optimization (Max/Min Problems) 4 3 1. ???? ???? = 3???? − 4???? , [-1.2] (a.) Find the absolute extrema of the function on the closed interval, [-1.2] st o Step: 1- Take the 1 derivative of f(x) ???? ???? = 3???? − 4???? 3 ???? ???? = 12???? − 12???? 2 o Step 2: Set the 1 derivative of f(x) = 0 and find the critical numbers. ′ 3 2 ???? ???? = 12???? − 12???? = 0 (Note: you can factor the 1 derivative to see the zeros more clearly.) ′ 2 ???? ???? = 12???? ???? − 1 = 0 ) 12???? =0 and ???? − 1 = 0 ???? = 1 12 2 0 12 ???? = 12 ???? = 0 ???? = 0 Now, we can see that are critical numbers are 0 and 1. Step 3: Make a table and use the critical numbers and numbers in the interval as the x-values in 4 3 the table and find the y-values by plugging in the x-values into???? ???? = 3???? − 4???? Y-values: Table: Note: (our absolute max and min is the highest and lowest y-value) ( ) 4 ( )3 f(-1)=???? ???? = 3(−1) − 4 −1 = 7 x f(x) ( ) 4 ( )3 f(0)= ???? ???? = 3(0) − 4 0 = 0 -1 7 ( ) 4 ( )3 0 0 f(1)= ???? ???? = 3(1) − 4 1 = −1 1 -1 (absolute 4 3 f(2)= ???? ???? = 3(2) − 4 2 ( ) = 16 min) 2 16 (absolute max) 3 3 2 2. ???? ???? = ???? − ???? 2 (a). find all relative extrema, if it exists. o Step 1: Take the 1 derivative of f(x) 3 ???? ???? = ???? − ???? 2 ′ 2 2 ???? ???? = 3???? − 3???? o Step 2: Set the 1 derivative of f(x)=0 and find the critical points ???? ???? = 3???? − 3???? = 0 o (Note: Factor out the function to see the critical numbers or zeros easily.) ???? ???? = 3???? ???? − 1 = 0) 3???? = 0 and ???? − 1 = 0 ???? = 0 ???? = 1 Now, we can see that are critical numbers are 0 and 1. o Step 3: Perform the 1 derivative test to find the relative extrema and to find the intervals of where it is increasing or decreasing by forming a number line. -Here is the number line with the critical numbers on it and in order to find where the intervals that increasing and/ or decreasing and find relative extrema. Increasing: (-∞, 0) and (1,∞) Decreasing: (0, 1) (Note: To see which value is max or mint. Look at the image above where I circled where arrows meet. The relative max is where 2 arrows are pointed upward and the relative min is where 2 arrows are pointed downward. Also remember plug in the x-values in the original function, f(x), in order to find the y-values.) f(0)= ???? ???? = (0) −3 3(0)2 = 0 3 1 f(1)= ???? ???? = (1) −3 (1)2 = − 2 2 Relative maximum: (0, 0) Relative minimum: (1,-1/2) 3. ???? ???? = ???? − 4???? 3 (a.) Find all points of inflection and discuss the concavity of the graph of the function. o Step 1: Take the 1 derivative of f(x). 4 3 ???? ???? = ???? − 4???? ????′ ???? = 4???? − 12???? 2 nd o Step 2: Now take the 2 derivative of f(x). ????′ ???? = 4???? − 12???? 2 ′ 2 ???? ???? = 12???? − 24???? o Step 2: Set the 2 derivative of f(x) =0 and find the critical points. ???? ???? = 12???? − 24???? = 0 o (Note: Factor out the function to see the critical numbers or zeros easily.) ???? ???? = 12???? − 24???? = 0 ′ ???? ???? = 12????(???? − 2) = 0 12???? = 0 and ???? − 2 = 0 ???? = 0 ???? = 2 Now, we can see that are critical numbers are 0 and 2. o Step 3: Perform the 2 derivative test to find the possible inflections points and to find the intervals of where it is concave up or down by forming a number line. -Here is the number line with the critical points on it and possible inflection points in order to find where the intervals that are concave up or down. Concave Up :( -∞, 0) and (2, ∞) Concave Down: (0, 2) (Note: To find out if the two critical numbers are possible inflection points are to look at the number line and see at each critical value if it alters from concave up to concave down or vice versa.) As you seefrom theimageabovethatat 0and2thenumberlineconsist oftheshift between concave up and concave down and back to concave up. So both critical values 0 and 2 are inflection points. Hint: To find the y-value you will need to plug in the critical numbers (the x-value) into the original function, f(x). 4 3 ???? 0 = (0) −44 0 ( ) 3= 0 ???? 2 = (2) − 4 2 ( ) = −16 Inflection Points: (0, 0) and (2,-16) 2 4. ???? ???? = ???? − 2 ???? + 3 ) (a.) Determine whether Rolle’s Theorem can be applied to f on the closed interval [- 3,2]. If Rolle’s Theorem can be applied, find all values of c in the open interval (-3,2) such that f '(c)= 0. o Step 1: Find if the function above is continuous and differentiable to find out if Rolle’s Theorem can be applied. -We can see from the function that it is continuous because it does not have a restricted domain because there is no real number that makes the function undefined. -We can find if the function is differentiable if f (-3) =f (-2): 2 ???? −3 = ( −3 − 2)( −3 + 3) = 0 2 ???? 2 = ( 2 − 2)( 2 + 3) = 0 -As you can see f (-3) =f (-2), so f(x) is differentiable because it’s proven above. st o Step 2: Find the c-values of f(x) such that f’(c) =0 by finding the 1 derivative. (Hint: The chain rule and product rule are a necessity of finding of finding the first derivative.) ′ ′ For refreshers the product rule is: ???? ???? × ???? ???? + ???? ???? × ???? ???? and chain rule is apparent in taking the first derivative. ???? ???? = ???? − 2 ???? + 3 )2 Use Product Rule and chain rule ????′(????) = 1 ???? + 3 )2 + (???? − 2)(2(???? + 3) × 1) 2 Multiply and distribute 2????′ ???? = (???? + 3) + (2 − 2)(2???? + 6) Distribute ????′ ???? = ???? + 6???? + 9 + (2???? + 6???? − 4???? − 12) Combine like terms and solve????′ ???? = ???? + 6???? + 9 + (2???? + 2???? − 12) 2 The first derivative of f(x)????′(????) = 3???? + 8???? − 3 st o Step 2: Set the 1 derivative of f(x) =0 and find the critical points. ???? ???? = 3???? + 8???? − 3 = 0 (Note: Factor out the function to see the critical numbers or zeros easily.) ???? ???? = (3???? − 1)(???? + 3) = 0 3???? − 1 = 0 and ???? + 3 = 0 3???? = 1 ???? = −3 1 ???? = 3 -As you can see, we have two solutions but our c-value will only be one because one of the values above are not in between the interval [-3,2], which is x=-3 so the only possible solution is x=1/3. So our c value is x=1/3. 5. ???? ???? = 1 ???? (a). Determine whether the Mean Value Theorem can be applied to f on the closed interval [1,4]. If Mean Value Theorem can be applied, find all values of c in the open interval (1,4) ′ ???? ???? −????(????) such that ???? ???? = ????−???? . Step 1: Find if the Mean Value Theorem is applicable by finding if the function is continuous and differentiable. 1 ???? ???? = = 0 (0) In this case, you need to set the f(x)= 0 and see if the value that makes the whole function is in the interval [1,4], which you can see if you set it equal to zero. The value is zero which is not in the interval meaning it is differentiable on the open interval of (1,4) and also continuous on the close interval,[1,4]. ???? ???? −????(????) o Step 2: Use the Mean Value Theorem to find out if ???? ???? =????−???? . ???? ???? −????(????) Mean Value Theorem Formula: ???? ???? = ) ????−???? -In order to find the f (b) and f(a), we need to plug in both 1 and 4 in f(x). ???? ???? = ???? 1 =) 1 = 1 (1) 1 1 ???? ???? = ???? 4 =) = (4) 4 Now that we have found the values for f(b) and f(a), let’s out now plug it in the Mean Value Theorem. 1 3 3 ′ ???? ???? − ????(????) 4 − 1 − 4 − 4 3 3 3 1 ???? ???? = ???? − ???? = 4 − 1 = 3 = 3 − 4 × 1 = − 12 = − 4 1 -As you can see, we found the f’(c) which is -1/4(simplified). Now we need to set it equal to f’(x) to find our c value(s). st Step 3: Find the 1 derivative of f(x). 1 −2 −2 1 ???? ???? = ???? = 1 × −1???? )= −???? ???????????????? ???????????????????????????? ???????? → − ???? 2 Note: You can use the Quotient Role, but I figured it is much easier since it divided by 1 term in thedenominator wecould just bringit uptothenumeratorand makeit theexponent negative because it brought out of denominator( x) = the term. o Step 4: Set f’(x) =f’(c) to find our c value(s). 1 1 1 1 2 2 − ????2 = − 4 → ????2 × 4→ ???? = 4 → ) √???? = 4√→ ???? = 2 We find out our c value is 2 which is between the interval [1, 4]. 6. We need to enclose a rectangular field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the fence that will enclose the largest area. (a.)Draw up a picture a situation of the situation. (Step 1) o Step 2: Write the primary equation and the secondary equation for the problem. (Note: 500 feet of material states we will be using perimeter and also area, so the area formula based on the picture is ???? = 2???? + ???? = 500 (secondary equation) because our perimeter is given in our equation, so we are going to use it to find our area. Our area equation ???? = ???????? (based on this situation) Primary Equation Secondary Equation ???? = ???????? ???? = 2???? + ???? = 500 o Step 3: Let all values be of (x) by using the secondary equation to solve for y and substitute in our primary equation. st o Step 4: Write the 1 derivative of A(x). o Step 5: Set the 1 derivative of A(x)=0 and solve for the maximum area. Optional: It is recommended check to see if your number is the maximum by performing the first derivative test. ???? 0 = 500 − 4 0 = 500 − 0 = 500 ???? 500 = 500 − 4 500 = 500 − 2000 = −1500 o Step 6: Find the y value so you can answer the question of finding the dimensions which is putting in the x-value in the equation of y: 7. Find the Limits. 5???? − 3???? 3 5???? − 3???? 3 5 5???? − 3???? 3 3 3 3 3 − 3 0 − 3 (????. lim 2 = lim ????2 ???? = lim ???? 2 ???? = ∞ = ????→∞ 4???? + 1 ????→∞ 4???? + 1 ????→∞ 4???? + 1 4 + 1 0 − 0 ????3 ???? 3 ????3 ????3 ∞ ∞ 3 = 0 −???? − ∞) ( ) ( ) (????.????→∞im 13 − 4???? = 13 − 4???? = 13 − 4 0 = 13 − 0 = 13 (Note: e to negative infinity will always equal 0.) 4???? 4???? 4 ( ) 4???? ????2 ???? 2 ???? 4 ????. = l????→∞???? 2= ????→∞ ????2 = ????→∞ ???? 2= ????→∞ 1 = ∞ = 0 ????2 ???? 2 (d.) lim ???????????????? = cos ∞ = ????????????,???????????????????????????? ???????? ???????? ???????????????? ???????????????????????????????????????????? ????→∞ ????−???? 8. ???? ???? = ????+???? , find all the following for this function: (a.)Domain x − 2 ???? x = x + 7 (Note: look at denominator. In rational functions the domain is found by setting the denominator equal to zero.) ???? + 7 = 0 → ???? = −7 Therefore our domain is −∞,−7 ???????????? −7,∞ .) (b.) X-intercepts (Hint: Set the numerator equal to zero to find the x-int(s).) ???? − 2 = 0 → ???? = 2 Therefore, point will be (2,0) (c.) Y-intercepts (Note: Evaluate f(x) at 0 to find the y-intercept) ( 0 − 2 2 ???? 0 = ((0) + 7= − 7 2 Therefore the point will be (0, − ) 7 (d.)Vertical asymptotes (Note: How you find the vertical asymptotes is by setting the denominator to 0.) ???? + 7 = 0 → ???? = −7 (e.) Limits at +/- ∞ and the horizontal asymptotes x − 2 x − 2 x − 2 1 − 2 1 − 0 lim = lim ???? ????= lim ???? ????= ∞ = = 1 ????→∞x + 7 ????→∞ x + 7 ????→∞ x + 7 1 + 7 1 + 0 ???? ???? ???? ???? ∞ Therefore at ???? → ∞,????ℎ???? ???????????????????? ???????????? ????ℎ???? ℎ???????????????????????????????????? ???????????????????????????????????? ???????? 1 x 2 x 2 2 x − 2 − − 1 − 1 − (−0) lim = lim ???? ????= lim ???? ???? = −∞ = = 1 ????→∞ x + 7 ????→∞ x + 7 ????→∞x + 7 1 + 7 1 + (−0) ???? ???? ???? ???? −∞ Therefore at ???? → −∞,????he limit and the horizontal asumptote is 1 as well. Meaning our Horizontal asymptote is y=1 (f.) Critical values (Note: to find the critical values, you have to set the 1 derivative to 0.) -Use the Quotient Rule to take the first derivative. ′ ???? ???? × ???? ???? − ????(????) × ????′(????) ????′(????) ???? (????) =???? − 2 = (???? + 7 1 − ???? − 2 1)( )= (???? + 7 − ???? − 2 )= (???? + 7 − ???? + 2 ???? + 7 (???? + 7) (???? + 7)2 ???? + 7)2 (???? + 7 − ???? + 2 9 = 2 = 2 (???? + 7) (???? + 7 9 ???? ???? = (???? + 7) (Note: The numerator ≠ 0) = (???? + 7) = ???? + 7 = 0 → ???? = −7=Critical # (g.) Intervals of increasing/decreasing behavior (i) Possible inflections points -(Note: Possible Inflections Points can be found by taking the second derivative.) 9 ???? ???? = 2 (???? + 7) ((???? + 7) )× 0 − (9)(2(???? + 7) × 1) (0 − (9)(2(???? + 7) × 1) −18 ????′(???? = = = (???? − 7)4 (???? + 7)4 (???? + 7)3 (Note: The numerator ≠ 0) = (???? + 7) = ???? + 7 = 0 → ???? = −7=Critical#andalsoaverticalasymptotemeaningthat ???? = −7 ≠ a inflection point, but we will still do our 2 derivative test. (a.) Intervals of upward/downward concavity (j.) graph 9. A sheet of cardboard 24 ft. by 24 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with the largest volume? o Step 1:Draw a picture of the situation described above. Primary Equation ???? = ???? × ???? × ℎ ????(????) = 24 − 2???? 24 − 2???? ???? )( ) o Step 2: Find the 1 derivative. ????(????) = 24 − 2???? 24 − 2???? ???? )( ) (Hint: Use the product rule ???? ???? = −2 24 − 2???? ???? + 24 − 2???? −2 ???? + 24 − 2???? 24 − 2???? 1 )( )( ) ????′ = −48 + 4???? 2)+ −48 + 4???? 2 )+ 576 − 96???? + 4???? 2) ????′ = 12???? − 192???? + 576 (Note: Factor if it needs to be simplified) ????′ = 12???? − 192???? + 576 12 ????′ = ???? − 16 + 48 st o Set the 1 derivative to 0 to find the maxiumum, if simplified. ???? = ???? − 4 ???? − 12 = 0 ) ???? − 4 = 0 ???? − 12 = 0 ???? = ???? ???? = 12 Okay, so we have two solutions but based on our domain, 0 ≤ ???? ≤ 12, x=12 cannot be our maximum so x=4 is our maximum and to get our dimensions we need to plug in four into ????(????). ???? ???? = (24 − 2 4 )(24 − 2 4 ) 4 = 16 × 16 × 4 ???????? (Dimensions) 10. A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. a. Draw a picture of the situation described above. b. How fast is the top of the ladder moving down the wall when its base is 7 feet away from the wall? c. Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. 11. An air traffic controller spots two planes at the same altitude converging to a point as they fly at right angles to each other. One plane is 750 miles from the point moving 250mph. The other plane is 1800 miles from the point moving 600 mph. a. Draw a picture of this situation. b. At what rate is the distance between the planes decreasing? -----------------------------------------END OF STUDY GUIDE----------------------------------- .

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