DOCSLIB.ORG

AP Calculus BC - Sequences and Series Chapter 11- AP Exam Problems Solutions

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
AP Calculus BC - Sequences and Series Chapter 11- AP Exam Problems Solutions AP Calculus BC - Sequences and Series Chapter 11- AP Exam Problems solutions 100 15⎛⎞+ n 1 1 1. A ssnn==⎜⎟,lim ⋅1= 54⎝⎠+ n n→∞ 5 5 2. C I. convergent: p-series with p = 2 >1 II. divergent: Harmonic series which is known to diverge 1 III. convergent: Geometric with r = <1 3 3. A I. Converges by Alternate Series Test n 1⎛⎞3 II Diverges by the nth term test: lim ⎜⎟≠ 0 n→∞ n ⎝⎠2 ∞ 1 L III Diverges by Integral test: dx = lim ln (ln x)=∞ ∫2 xxln L→∞ 2 4. A I. Compare with p-series, p = 2 6 II. Geometric series with r = 7 III. Alternating harmonic series 5. B I. Divergent. The limit of the nth term is not zero. II. Convergent. This is the same as the alternating harmonic series. III. Divergent. This is the harmonic series. 6. A This is the integral test applied to the series in (A). Thus the series in (A) converges. None of the others must be true. 7. D The first series is either the harmonic series or the alternating harmonic series depending on whether k is odd or even. It will converge if k is odd. The second series is geometric and will converge if k < 4 . ∞ 8. A Take the derivative of the general term with respect to x: ∑()−1 n+1 x22n− n=1 23 uu23 ()33xx() 9. E Since euu =+1 + + +" , then ex3x =1++3 + +" 2! 3! 2! 3! 33 9 The coefficient we want is = 3! 2 tt35 10. B The Maclaurin series for sin t is t − +−" . Let tx= 2 . 3! 5! (2xx )35(2 ) (−1)n−1 (2x)2n−1 sin(2x) =2x−+ −"+ +" 3! 5! (2n −1)! xx3 5 (x23) (x25) x6x 10 11. A sin xx=− + −"" ⇒sin x22=x− + − =x2− + −" 3! 5! 3! 5! 3! 5! xx35 1315 1 1 12. E sin xx≈− + ; sin1 ≈1− + =−+ 1 3! 5! 3! 5! 6 120 ∞∞∞ nn−1n−1 13. D If f(x)==∑axnn , then f′(x) ∑nax=∑ naxn. nnn===001 ∞∞ n−1 f ′(1) ==∑nan1 ∑ nan nn==11 14. A The series is the Maclaurin expansion of e−x. Use the calculator to solve ex−x = 3 . 15. D The center is x =1, so only C, D, or E are possible. Check the endpoints. ∞ (−1)n At x = 0: ∑ converges by alternating series test. n=1 n ∞ 1 At x = 2: ∑ which is the harmonic series and known to diverge. n=1 n ∞ (−1)n 16. C Check x =−1, ∑ which is convergent by alternating series test n=1 n ∞ 1 Check x =1, ∑ which is the harmonic series and known to diverge. n=1 n x −1 17. C This is a geometric series with r = . Convergence for −11<<r . Thus the series is 3 convergent for −<24x < . n 18. B You may use the ratio test. However, the series will converge if the numerator is (1)− and diverge if the numerator is 1n . Any value of x for which x +21 > in the numerator will make the series diverge. Hence the interval is −31≤<x −. 19. 1990 BC5 Solution (a) Taylor approach Geometric Approach f ()2=1 11= − f ′()2=− (2 − 1 )2 =− 1 x−11+−()x 2 23 n n − f ′′ ()2 =−1uu+ −u +"" +−()1 u+ f ′′ ()22=− (212; )3 = = 1 2! where ux=− 2 − f ′′′ ()2 f ′′′ ()26216;=− ( − )4 =− =−1 3! 1 23 nn Therefore =−1 ()xx −2+()− 2−()x − 2+" +−( 1)()x− 2 +" x −1 (b) Antidifferentiates series in (a): nn+1 1213 14()(−−1x 2) ln x−=1 C+x−() x −+2()x −− 2()x−++ 2 "+" 23 4 n +1 0ln21=−⇒=−C 2 1 4 Note: If C ≠ 0 , “first 4 terms” need not include −()x −2 4 35 1112113 (c) ln=− ln 1 =−+−" 22 22232 111 =−+ −" 2824 1111 since <, −=0.375 is sufficient. 24 20 2 8 Justification: Since series is alternating, with terms convergent to 0 and decreasing in absolute value, the truncation error is less than the first omitted term. n+1 =<1 11 <5 RCn + , where 2 Alternate Justification: (C −1)n 1 n +12 2 < 11 n +12n+1 1 < when n ≥2 20 20. 1992 BC6 Solution 11 (a) 0 << for ln ()nn>1, for ≥3 nnpln () np 1 by p-series test, ∑ converges if p > 1 n p ∞ 1 and by direct comparison, ∑ p converges. n=2 nnln () 1 ∞ (b) Let fx()= , so series is ∑ fn() xxln n=2 ∞ ⌠ 1 b dx=lim ln ln x =lim[ln(ln(b))− ln(ln 2)] =∞ b→∞ b→∞ ⌡ 2 xxln 2 Since f (x) monotonically decreases to 0, the integral test shows ∞ 1 ∑ diverges. n=2 nnln 11 (c) >>0 for p <1, nnp ln nnln ∞ 1 ≤< so by direct comparison, ∑ p diverges for 01p n=2 nnln Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 21. 1995 BC4 Let f be a function that has derivatives of all orders for all real numbers. Assumeff (1) 3, cc(1) 2, fc(1) 2 , and f ccc(1) 4 . (a) Write the second-degree Taylor polynomial for f about x 1 and use it to approximate f (0.7) . (b) Write the third-degree Taylor polynomial for f about x 1 and use it to approximate f (1.2). (c) Write the second-degree Taylor polynomial for f c , the derivative of f , about x 1 and use it to approximate f c(1.2) . 1995 BC4 Solution 2 2 (a) Tx2 3 2 x1 x1 2 f 0.7 |3 0.6 0.09 3.69 24 3 (b) Tx3 32 x1 x 1 x1 6 2 f 1.2 |3 0.4 0.04 0.008 2.645 3 c 2 (c) Tx3 2 2 x12 x 1 f c 1.2 |2 0.4 0.08 1.52 22. 1997 BC2 Let Px() 73(4)x 5(4)x 232(4x)6(4)x 4 be the fourth-degree Taylor polynomial for the function f about 4. Assume f has derivatives of all orders for all real numbers. (a) Find ff(4) and ccc(4). (b) Write the second-degree Taylor polynomial for f c about 4 and use it to approximate f c(4.3) . x (c) Write the fourth-degree Taylor polynomial for gx() f(t)dt about 4. ³ 4 (d) Can f (3) be determined from the information given? Justify your answer. 1997 BC2 Solution (a) fP 4 4 7 f ccc 4 2, f ccc 4 12 3! 23 (b) Px3 37x4 5x 4 2 x 4 2 Px3c 301464 x x 2 f c 4.3| 3 10 0.3 6 0.3 0.54 x (c) Pg43 ,x Ptd()t ³4 x 23 ªº7345544 tt tdt ³4 ¬¼ 325314 x474xx 4 x 4 23 2 (d) No. The information given provides values for ff 4, cc 4,ffcc 4, cc 4 andf 4 4 only. 23. 1998 Calculus BC Scoring Guidelines 24. BC{4 1999 0 4. The function f has derivatives of all orders for all real numb ers x. Assume f (2) = 3, f (2) = 5, 00 000 f (2) = 3, and f (2) = 8. (a) Write the third{degree Taylor p olynomial for f ab out x =2anduseittoapproximate f (1:5). (4) (b) The fourth derivativeoff satis es the inequality jf (x)j3 for all x in the closed interval [1:5; 2]. Use the Lagrange error bound on the approximation to f (1:5) found in part (a) to explain why f (1:5) 6= 5. 2 (c) Write the fourth{degree Taylor p olynomial, P (x), for g (x) = f (x +2) ab out x = 0. Use P to explain why g must have a relativeminimum at x =0. 8 8 3 2 3 (x 2) (x 2) (a) T (f; 2)(x)=3+5(x 2) + 3: T (f; 2)(x) > 3 3 < 2 6 < 1 > each error 4 > : f (1:5) T (f; 2)(1:5) 3 1: approximation of f (1:5) 4 3 3 2 (0:5) (0:5) = 3+ 5(0:5) + 2 3 = 4:9583= 4:958 ( 3 4 1: value of Lagrange Error Bound j1:5 2j =0:0078125 (b) Lagrange Error Bound = 4! 2 1: explanation f (1:5) > 4:9583 0:0078125 = 4:966 > 5 Therefore, f (1:5) 6= 5. 8 2: T (g; 0)(x) (c) P (x)=T (g; 0)(x) > 4 4 > > < < 1 > each incorrect, missing, 3 3 4 2 2 or extra term x = T (f; 2)(x +2) = 3+5x + 2 > > 2 > : 1: explanation 0 The co ecientof x in P (x)isg (0). This co ecient 0 is 0, so g (0)=0. 00 (0) g 2 . This co ecient The co ecientof x in P (x)is 2! 00 is 5, so g (0) = 10 which is greater than 0. Therefore, g has a relative minimum at x =0. Note: < 1 > max for improp er use of + ::: or equality AP® CALCULUS BC 2001 SCORING GUIDELINES 25. Question 6 A function f is defined by 12 3 n 1 fx()x x2 Lxn L 3 3323 3n 1 for all x in the interval of convergence of the given power series. (a) Find the interval of convergence for this power series. Show the work that leads to your answer. 1 fx() (b) Find lim 3 . x l0 x (c) Write the first three nonzero terms and the general term for an infinite series that 1 represents ¨ fx()dx. 0 (d) Find the sum of the series determined in part (c).
Load more

Recommended publications
  • The Mean Value Theorem Math 120 Calculus I Fall 2015
    The Mean Value Theorem Math 120 Calculus I Fall 2015 The central theorem to much of differential calculus is the Mean Value Theorem, which we'll abbreviate MVT. It is the theoretical tool used to study the first and second derivatives. There is a nice logical sequence of connections here. It starts with the Extreme Value Theorem (EVT) that we looked at earlier when we studied the concept of continuity. It says that any function that is continuous on a closed interval takes on a maximum and a minimum value. A technical lemma. We begin our study with a technical lemma that allows us to relate 0 the derivative of a function at a point to values of the function nearby. Specifically, if f (x0) is positive, then for x nearby but smaller than x0 the values f(x) will be less than f(x0), but for x nearby but larger than x0, the values of f(x) will be larger than f(x0). This says something like f is an increasing function near x0, but not quite. An analogous statement 0 holds when f (x0) is negative. Proof. The proof of this lemma involves the definition of derivative and the definition of limits, but none of the proofs for the rest of the theorems here require that depth. 0 Suppose that f (x0) = p, some positive number. That means that f(x) − f(x ) lim 0 = p: x!x0 x − x0 f(x) − f(x0) So you can make arbitrarily close to p by taking x sufficiently close to x0.
    [Show full text]
  • Series: Convergence and Divergence Comparison Tests
    Series: Convergence and Divergence Here is a compilation of what we have done so far (up to the end of October) in terms of convergence and divergence. • Series that we know about: P∞ n Geometric Series: A geometric series is a series of the form n=0 ar . The series converges if |r| < 1 and 1 a1 diverges otherwise . If |r| < 1, the sum of the entire series is 1−r where a is the first term of the series and r is the common ratio. P∞ 1 2 p-Series Test: The series n=1 np converges if p1 and diverges otherwise . P∞ • Nth Term Test for Divergence: If limn→∞ an 6= 0, then the series n=1 an diverges. Note: If limn→∞ an = 0 we know nothing. It is possible that the series converges but it is possible that the series diverges. Comparison Tests: P∞ • Direct Comparison Test: If a series n=1 an has all positive terms, and all of its terms are eventually bigger than those in a series that is known to be divergent, then it is also divergent. The reverse is also true–if all the terms are eventually smaller than those of some convergent series, then the series is convergent. P P P That is, if an, bn and cn are all series with positive terms and an ≤ bn ≤ cn for all n sufficiently large, then P P if cn converges, then bn does as well P P if an diverges, then bn does as well. (This is a good test to use with rational functions.
    [Show full text]
  • Chapter 9 Optimization: One Choice Variable
    RS - Ch 9 - Optimization: One Variable Chapter 9 Optimization: One Choice Variable 1 Léon Walras (1834-1910) Vilfredo Federico D. Pareto (1848–1923) 9.1 Optimum Values and Extreme Values • Goal vs. non-goal equilibrium • In the optimization process, we need to identify the objective function to optimize. • In the objective function the dependent variable represents the object of maximization or minimization Example: - Define profit function: = PQ − C(Q) - Objective: Maximize - Tool: Q 2 1 RS - Ch 9 - Optimization: One Variable 9.2 Relative Maximum and Minimum: First- Derivative Test Critical Value The critical value of x is the value x0 if f ′(x0)= 0. • A stationary value of y is f(x0). • A stationary point is the point with coordinates x0 and f(x0). • A stationary point is coordinate of the extremum. • Theorem (Weierstrass) Let f : S→R be a real-valued function defined on a compact (bounded and closed) set S ∈ Rn. If f is continuous on S, then f attains its maximum and minimum values on S. That is, there exists a point c1 and c2 such that f (c1) ≤ f (x) ≤ f (c2) ∀x ∈ S. 3 9.2 First-derivative test •The first-order condition (f.o.c.) or necessary condition for extrema is that f '(x*) = 0 and the value of f(x*) is: • A relative minimum if f '(x*) changes its sign y from negative to positive from the B immediate left of x0 to its immediate right. f '(x*)=0 (first derivative test of min.) x x* y • A relative maximum if the derivative f '(x) A f '(x*) = 0 changes its sign from positive to negative from the immediate left of the point x* to its immediate right.
    [Show full text]
  • Critical Points and Classifying Local Maxima and Minima Don Byrd, Rev
    Notes on Textbook Section 4.1 Critical Points and Classifying Local Maxima and Minima Don Byrd, rev. 25 Oct. 2011 To find and classify critical points of a function f (x) First steps: 1. Take the derivative f ’(x) . 2. Find the critical points by setting f ’ equal to 0, and solving for x . To finish the job, use either the first derivative test or the second derivative test. Via First Derivative Test 3. Determine if f ’ is positive (so f is increasing) or negative (so f is decreasing) on both sides of each critical point. • Note that since all that matters is the sign, you can check any value on the side you want; so use values that make it easy! • Optional, but helpful in more complex situations: draw a sign diagram. For example, say we have critical points at 11/8 and 22/7. It’s usually easier to evaluate functions at integer values than non-integers, and it’s especially easy at 0. So, for a value to the left of 11/8, choose 0; for a value to the right of 11/8 and the left of 22/7, choose 2; and for a value to the right of 22/7, choose 4. Let’s say f ’(0) = –1; f ’(2) = 5/9; and f ’(4) = 5. Then we could draw this sign diagram: Value of x 11 22 8 7 Sign of f ’(x) negative positive positive ! ! 4. Apply the first derivative test (textbook, top of p. 172, restated in terms of the derivative): If f ’ changes from negative to positive: f has a local minimum at the critical point If f ’ changes from positive to negative: f has a local maximum at the critical point If f ’ is positive on both sides or negative on both sides: f has neither one For the above example, we could tell from the sign diagram that 11/8 is a local minimum of f (x), while 22/7 is neither a local minimum nor a local maximum.
    [Show full text]
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • Drainage Geocomposite Workshop
    January/February 2000 Volume 18, Number 1 Drainage geocomposite workshop By Gregory N. Richardson, Sam R. Allen, C. Joel Sprague A number of recent designer’s columns have focused on the design of drainage geocomposites (Richard- son and Zhao, 1998), so only areas of concern are discussed here. Two design considerations requiring the most consideration by a designer are: 1) assumed rate of fluid inflow draining into the geocomposite, and 2) long-term reduction and safety factors required to assure adequate performance over the service life of the drainage system. In regions of the United States that are not arid or semi-arid, a reasonable upper limit for inflow into a drainage layer can be estimated by assuming that the soil immediately above the drain layer is saturated. This saturation produces a unit gradient flow in the overlying soil such that the inflow rate, r, is approximately θ equal to the permeability of the soil, ksoil. The required transmissivity, , of the drainage layer is then equal to: θ = r*L/i = ksoil*L/i where L is the effective drainage length of the geocomposite and i is the flow gradient (head change/flow length). This represents an upper limit for flow into the geocomposite. In arid and semi-arid regions of the USA saturation of the soil layers may be an overly conservative assumption, however, no alternative rec- ommendations can currently be given. Regardless of the assumed r value, a geocomposite drainage layer must be designed such that flow is not in- advertently restricted prior to removal from the drain.
    [Show full text]
  • Calculus 120, Section 7.3 Maxima & Minima of Multivariable Functions
    Calculus 120, section 7.3 Maxima & Minima of Multivariable Functions notes by Tim Pilachowski A quick note to start: If you’re at all unsure about the material from 7.1 and 7.2, now is the time to go back to review it, and get some help if necessary. You’ll need all of it for the next three sections. Just like we had a first-derivative test and a second-derivative test to maxima and minima of functions of one variable, we’ll use versions of first partial and second partial derivatives to determine maxima and minima of functions of more than one variable. The first derivative test for functions of more than one variable looks very much like the first derivative test we have already used: If f(x, y, z) has a relative maximum or minimum at a point ( a, b, c) then all partial derivatives will equal 0 at that point. That is ∂f ∂f ∂f ()a b,, c = 0 ()a b,, c = 0 ()a b,, c = 0 ∂x ∂y ∂z Example A: Find the possible values of x, y, and z at which (xf y,, z) = x2 + 2y3 + 3z 2 + 4x − 6y + 9 has a relative maximum or minimum. Answers : (–2, –1, 0); (–2, 1, 0) Example B: Find the possible values of x and y at which fxy(),= x2 + 4 xyy + 2 − 12 y has a relative maximum or minimum. Answer : (4, –2); z = 12 Notice that the example above asked for possible values. The first derivative test by itself is inconclusive. The second derivative test for functions of more than one variable is a good bit more complicated than the one used for functions of one variable.
    [Show full text]
  • Summary of Tests for Series Convergence
    Calculus Maximus Notes 9: Convergence Summary Summary of Tests for Infinite Series Convergence Given a series an or an n1 n0 The following is a summary of the tests that we have learned to tell if the series converges or diverges. They are listed in the order that you should apply them, unless you spot it immediately, i.e. use the first one in the list that applies to the series you are trying to test, and if that doesn’t work, try again. Off you go, young Jedis. Use the Force. Remember, it is always with you, and it is mass times acceleration! nth-term test: (Test for Divergence only) If liman 0 , then the series is divergent. If liman 0 , then the series may converge or diverge, so n n you need to use a different test. Geometric Series Test: If the series has the form arn1 or arn , then the series converges if r 1 and diverges n1 n0 a otherwise. If the series converges, then it converges to 1 . 1 r Integral Test: In Prison, Dogs Curse: If an f() n is Positive, Decreasing, Continuous function, then an and n1 f() n dn either both converge or both diverge. 1 This test is best used when you can easily integrate an . Careful: If the Integral converges to a number, this is NOT the sum of the series. The series will be smaller than this number. We only know this it also converges, to what is anyone’s guess. The maximum error, Rn , for the sum using Sn will be 0 Rn f x dx n Page 1 of 4 Calculus Maximus Notes 9: Convergence Summary p-series test: 1 If the series has the form , then the series converges if p 1 and diverges otherwise.
    [Show full text]
  • Summary of Derivative Tests
    Summary of Derivative Tests Note that for all the tests given below it is assumed that the function f is continuous. Critical Numbers Definition. A critical number of a function f is a number c in the domain of f such that either f 0(c) = 0 or f 0(c) does not exist. Critical numbers tell you where a possible max/min occurs. Closed Interval Method Use: To find absolute mins/maxes on an interval [a; b] Steps: 1. Find the critical numbers of f [set f 0(x) = 0 and solve] 2. Find the value of f at each critical number 3. Find the value of f at each endpoint [i.e., find f(a) and f(b)] 4. Compare all of the above function values. The largest one is the absolute max and the smallest one is the absolute min. First Derivative Test Use: To find local max/mins Statement of the test: Suppose c is a critical number. 1. If f 0 changes from positive to negative at c; then f has a local max at c 2. If f 0 changes from negative to positive at c, then f has a local min at c 3. If f 0 is positive to the left and right of c [or negative to the left and right of c], then f has no local max or min at c Trick to Remember: If you are working on a problem and forget what means what in the above test, draw a picture of tangent lines around a minimum or around a maximum 1 Steps: 1.
    [Show full text]
  • Just the Definitions, Theorem Statements, Proofs On
    MATH 3333{INTERMEDIATE ANALYSIS{BLECHER NOTES 55 Abbreviated notes version for Test 3: just the definitions, theorem statements, proofs on the list. I may possibly have made a mistake, so check it. Also, this is not intended as a REPLACEMENT for your classnotes; the classnotes have lots of other things that you may need for your understanding, like worked examples. 6. The derivative 6.1. Differentiation rules. Definition: Let f :(a; b) ! R and c 2 (a; b). If the limit lim f(x)−f(c) exists and is finite, then we say that f is differentiable at x!c x−c 0 df c, and we write this limit as f (c) or dx (c). This is the derivative of f at c, and also obviously equals lim f(c+h)−f(c) by setting x = c + h or h = x − c. If f is h!0 h differentiable at every point in (a; b), then we say that f is differentiable on (a; b). Theorem 6.1. If f :(a; b) ! R is differentiable at at a point c 2 (a; b), then f is continuous at c. Proof. If f is differentiable at c then f(x) − f(c) f(x) = (x − c) + f(c) ! f 0(c)0 + f(c) = f(c); x − c as x ! c. So f is continuous at c. Theorem 6.2. (Calculus I differentiation laws) If f; g :(a; b) ! R is differentiable at a point c 2 (a; b), then (1) f(x) + g(x) is differentiable at c and (f + g)0(c) = f 0(c) + g0(c).
    [Show full text]
  • MATH 162: Calculus II Differentiation
    MATH 162: Calculus II Framework for Mon., Jan. 29 Review of Differentiation and Integration Differentiation Definition of derivative f 0(x): f(x + h) − f(x) f(y) − f(x) lim or lim . h→0 h y→x y − x Differentiation rules: 1. Sum/Difference rule: If f, g are differentiable at x0, then 0 0 0 (f ± g) (x0) = f (x0) ± g (x0). 2. Product rule: If f, g are differentiable at x0, then 0 0 0 (fg) (x0) = f (x0)g(x0) + f(x0)g (x0). 3. Quotient rule: If f, g are differentiable at x0, and g(x0) 6= 0, then 0 0 0 f f (x0)g(x0) − f(x0)g (x0) (x0) = 2 . g [g(x0)] 4. Chain rule: If g is differentiable at x0, and f is differentiable at g(x0), then 0 0 0 (f ◦ g) (x0) = f (g(x0))g (x0). This rule may also be expressed as dy dy du = . dx x=x0 du u=u(x0) dx x=x0 Implicit differentiation is a consequence of the chain rule. For instance, if y is really dependent upon x (i.e., y = y(x)), and if u = y3, then d du du dy d (y3) = = = (y3)y0(x) = 3y2y0. dx dx dy dx dy Practice: Find d x d √ d , (x2 y), and [y cos(xy)]. dx y dx dx MATH 162—Framework for Mon., Jan. 29 Review of Differentiation and Integration Integration The definite integral • the area problem • Riemann sums • definition Fundamental Theorem of Calculus: R x I: Suppose f is continuous on [a, b].
    [Show full text]
  • Maximum and Minimum Definition
    Maximum and Minimum Definition: Definition: Let f be defined in an interval I. For all x, y in the interval I. If f(x) < f(y) whenever x < y then we say f is increasing in the interval I. If f(x) > f(y) whenever x < y then we say f is decreasing in the interval I. Graphically, a function is increasing if its graph goes uphill when x moves from left to right; and if the function is decresing then its graph goes downhill when x moves from left to right. Notice that a function may be increasing in part of its domain while decreasing in some other parts of its domain. For example, consider f(x) = x2. Notice that the graph of f goes downhill before x = 0 and it goes uphill after x = 0. So f(x) = x2 is decreasing on the interval (−∞; 0) and increasing on the interval (0; 1). Consider f(x) = sin x. π π 3π 5π 7π 9π f is increasing on the intervals (− 2 ; 2 ), ( 2 ; 2 ), ( 2 ; 2 )...etc, while it is de- π 3π 5π 7π 9π 11π creasing on the intervals ( 2 ; 2 ), ( 2 ; 2 ), ( 2 ; 2 )...etc. In general, f = sin x is (2n+1)π (2n+3)π increasing on any interval of the form ( 2 ; 2 ), where n is an odd integer. (2m+1)π (2m+3)π f(x) = sin x is decreasing on any interval of the form ( 2 ; 2 ), where m is an even integer. What about a constant function? Is a constant function an increasing function or decreasing function? Well, it is like asking when you walking on a flat road, as you going uphill or downhill? From our definition, a constant function is neither increasing nor decreasing.
    [Show full text]