Math 1131Q Section 10
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1 Math 1131Q Section 10 Review Oct 5, 2010
2 Exam 1 DATE: Tuesday, October 5 TIME: 6-8 PM Exam Rooms Sections 11D, 14D, 15D CLAS 110 Sections12D, 13D, 16D PB 38 (Physics Building) Material covered on the exam: Chapter 1 Chapter 2 Sections Review (Oct 5, 2010) Math 1131Q Section 10 2 / 24
3 Review (Oct 5, 2010) Math 1131Q Section 10 3 / 24 Question Find the inverse of f(x) = x + 1 x 1 y = f 1 (x).. and write the answer in the form y = f(x) and x = f 1 (y) are equivalent relationships between x and y. Given the relationship y = f(x), we need to solve for x in terms of y to find x = f 1 (y).
4 Question Find the inverse of f(x) = x + 1 x 1 y = f 1 (x).. and write the answer in the form y = f(x) and x = f 1 (y) are equivalent relationships between x and y. Given the relationship y = f(x), we need to solve for x in terms of y to find x = f 1 (y). y = x + 1 x 1 y(x 1) = x + 1 yx y = x + 1 yx x = y + 1 x(y 1) = y + 1 x = y + 1 y 1 f 1 (y) = y + 1 y 1 f 1 (x) = x + 1 (x 1 Review (Oct 5, 2010) Math 1131Q Section 10 3 / 24
5 Review (Oct 5, 2010) Math 1131Q Section 10 4 / 24 Instructions for True/False questions: If the statement is always true, choose TRUE. If the statement is sometimes false, choose FALSE. In either case, prepare a justification or invent a counterexample. The values of x and arcsin sin (x) are always equal. (a) TRUE (b) FALSE
6 Review (Oct 5, 2010) Math 1131Q Section 10 4 / 24 Instructions for True/False questions: If the statement is always true, choose TRUE. If the statement is sometimes false, choose FALSE. In either case, prepare a justification or invent a counterexample. The values of x and arcsin sin (x) are always equal. (a) TRUE (b) FALSE FALSE - arcsin (y) is the inverse of y = sin x, π/2 x π/2. Note that sin (3π/4) = 1/ 2 = sin (π/4)
7 Review (Oct 5, 2010) Math 1131Q Section 10 4 / 24 Instructions for True/False questions: If the statement is always true, choose TRUE. If the statement is sometimes false, choose FALSE. In either case, prepare a justification or invent a counterexample. The values of x and arcsin sin (x) are always equal. (a) TRUE (b) FALSE FALSE - arcsin (y) is the inverse of y = sin x, π/2 x π/2. Note that sin (3π/4) = 1/ 2 = sin (π/4) arcsin (sin (3π/4)) is the angle θ in [ π/2, π/2] for which sin θ = sin (3π/4) = 1/ 2 θ = π/4 and we have arcsin (sin (3π/4)) = π/4
8 Review (Oct 5, 2010) Math 1131Q Section 10 4 / 24 Instructions for True/False questions: If the statement is always true, choose TRUE. If the statement is sometimes false, choose FALSE. In either case, prepare a justification or invent a counterexample. The values of x and arcsin sin (x) are always equal. (a) TRUE (b) FALSE FALSE - arcsin (y) is the inverse of y = sin x, π/2 x π/2. Note that sin (3π/4) = 1/ 2 = sin (π/4) arcsin (sin (3π/4)) is the angle θ in [ π/2, π/2] for which sin θ = sin (3π/4) = 1/ 2 θ = π/4 and we have arcsin (sin (3π/4)) = π/4
9 f(x) = L if the values of the function f(x) get closer and x a closer to L as x approaches a from both sides. SUM A Limit Tool Box - The Limit Laws f(x) + g(x) = f(x) + g(x) x a x a x a CONSTANT MULTIPLE PRODUCT cf(x) = c f(x) x a x a f(x)g(x) = [ f(x)][ g(x)] x a x a x a f(x) QUOTIENT x a g(x) = x a f(x) x a g(x) Provided you do not divide by 0 POWER [f(x)] n = [ f(x)] n x a x a FRACTIONAL POWER [f(x)] n/m = [ x a provided that when m is even we have f(x) 0 for x near a x a f(x)] n/m Review (Oct 5, 2010) Math 1131Q Section 10 5 / 24
10 Review (Oct 5, 2010) Math 1131Q Section 10 6 / 24 We can evaluate the one-sided it as x 3 by using the it tool box. x 3 x 3 2 x [x 3 x 3 x 3 2 x ]1/2 = [ x 3 2 x ]1/2 FRACTIONAL POWER x 3 x 3 = [ x]1/2 2 x 3 QUOTIENT = 0 1 = 0 LINEAR FUNCTIONS
11 Review (Oct 5, 2010) Math 1131Q Section 10 7 / 24 By the definition of it, if x 1 f(x) = 2 then for each value of ɛ > 0 one can produce a value for δ > 0 such that If 0 < x 1 < δ, then f(x) 2 < ɛ. From the graph, determine the largest value of δ > 0 from the list satisfying the statement when ɛ = y If 0 < x 1 < δ, then f(x) 2 < x (a) 0 (b) (c).01 (d).02 (e).03
12 Review (Oct 5, 2010) Math 1131Q Section 10 8 / y x If 0 < x 1 < 0.01, then 1.9 < f(x) < 2.1, that is f(x) 2 < 0.1 f(1.098) < 1.9 so f(1.098) 2 > 0.1. Taking δ = 0.02 or larger will not work.
13 Review (Oct 5, 2010) Math 1131Q Section 10 9 / 24 A function f is continuous at a number x = a if x a f(x) = f(a). If this condition is not met, then a is a point of discontinuity. To say that x a f(x) = f(a) means 3 things are true: Continuity Check List 1. f is defined at x = a 2. The it exists 3. The it is actually equal to f(a)
14 Review (Oct 5, 2010) Math 1131Q Section / 24 Polynomials are continuous at every point a because of the direct substitution rule x a p(x) = p(a) (when p(x) is a polynomial). Informally, a function f is continous at a point x = a if the graph of f contains no holes or breaks at x = a (that is, can be draw without lifting the pencil off the paper). Since sin (x), cos (x) and e x have unbroken graphs and are continuous for all x, we can evaluate these its by direct substitution: sin (x) = sin (a) x a cos (x) = cos (a) x a x a ex = e a
15 Review (Oct 5, 2010) Math 1131Q Section / 24 Consider a function f(x) with the property that f(x) = 6 and x a another function g(x) such that g(x) = 0. Then x a f(x) x a g(x) =. (a) TRUE (b) FALSE
16 Review (Oct 5, 2010) Math 1131Q Section / 24 Consider a function f(x) with the property that f(x) = 6 and x a another function g(x) such that g(x) = 0. Then x a f(x) x a g(x) =. (a) TRUE (b) FALSE FALSE - Justification: Take f(x) = 3x and g(x) = x 2. Then f(x) = 3x = 6 x 2 x 2 g(x) = x 2 = 0 x 2 x 2
17 Review (Oct 5, 2010) Math 1131Q Section / 24 Consider a function f(x) with the property that f(x) = 6 and x a another function g(x) such that g(x) = 0. Then x a f(x) x a g(x) =. (a) TRUE (b) FALSE FALSE - Justification: Take f(x) = 3x and g(x) = x 2. Then f(x) = 3x = 6 x 2 x 2 g(x) = x 2 = 0 x 2 x 2 f(x) but x 2 g(x) = 3x 3x = since x 2 x 2 x 2 < 0 as x 2.
18 Review (Oct 5, 2010) Math 1131Q Section / 24 Consider a function f(x) with the property that f(x) = 6 and x a another function g(x) such that g(x) = 0. Then x a f(x) x a g(x) =. (a) TRUE (b) FALSE FALSE - Justification: Take f(x) = 3x and g(x) = x 2. Then f(x) = 3x = 6 x 2 x 2 g(x) = x 2 = 0 x 2 x 2 f(x) but x 2 g(x) = 3x 3x = since x 2 x 2 x 2 < 0 as x 2. f(x) x 2 + g(x) = 3x x 2 + x 2 = since 3x x 2 > 0 as x 2+.
19 Review (Oct 5, 2010) Math 1131Q Section / 24 f(x) If f(x) = 0 and g(x) = 0 then x 6 x 6 x 6 g(x) (a) TRUE (b) FALSE does not exist.
20 Review (Oct 5, 2010) Math 1131Q Section / 24 f(x) If f(x) = 0 and g(x) = 0 then x 6 x 6 x 6 g(x) (a) TRUE (b) FALSE FALSE - Counterexample: If f(x) = x 6 x + 1 f(x) g(x) = and g(x) = x 6 then x 6 x+1 x 6 = (x 6) (x 6)(x + 1) does not exist.
21 Review (Oct 5, 2010) Math 1131Q Section / 24 f(x) If f(x) = 0 and g(x) = 0 then x 6 x 6 x 6 g(x) (a) TRUE (b) FALSE FALSE - Counterexample: If f(x) = x 6 x + 1 f(x) g(x) = hence x 6 f(x) g(x) = x 6 and g(x) = x 6 then x 6 x+1 x 6 = (x 6) (x 6)(x + 1) (x 6) (x 6)(x + 1) = x 6 does not exist. 1 x + 1 = 1/7
22 f(x) If f(x) = 0 and g(x) = 0 then x 6 x 6 x 6 g(x) (a) TRUE (b) FALSE FALSE - Counterexample: If f(x) = x 6 x + 1 f(x) g(x) = hence x 6 f(x) g(x) = x 6 and g(x) = x 6 then x 6 x+1 x 6 = (x 6) (x 6)(x + 1) (x 6) (x 6)(x + 1) = x 6 does not exist. 1 x + 1 = 1/7 Note: If f(x) = x 6 x + 1 and g(x) = (x 6)2 then the it would not exist. Review (Oct 5, 2010) Math 1131Q Section / 24
23 Review (Oct 5, 2010) Math 1131Q Section / 24 The function f(x) = x3 + x 2 + x + 1 2x 3 + 3x 2 + 5x + 7 asymptote of y = 1/7. has a horizontal (a) TRUE (b) FALSE
24 Review (Oct 5, 2010) Math 1131Q Section / 24 The function f(x) = x3 + x 2 + x + 1 2x 3 + 3x 2 + 5x + 7 asymptote of y = 1/7. has a horizontal (a) TRUE (b) FALSE FALSE -Justification: f(x) = [ (x 3 + x 2 + x + 1)/x 3 x x (2x 3 + 3x 2 + 5x + 7)/x ] 3 = [ (1 + 1/x + 1/x2 + 1/x 3 ) x (2 + 3/x + 5/x 2 + 7/x 3 ) ]
25 Review (Oct 5, 2010) Math 1131Q Section / 24 The function f(x) = x3 + x 2 + x + 1 2x 3 + 3x 2 + 5x + 7 asymptote of y = 1/7. has a horizontal (a) TRUE (b) FALSE FALSE -Justification: f(x) = [ (x 3 + x 2 + x + 1)/x 3 x x (2x 3 + 3x 2 + 5x + 7)/x ] 3 = [ (1 + 1/x + 1/x2 + 1/x 3 ) x (2 + 3/x + 5/x 2 + 7/x 3 ) ] = (1 + 1/x + x 1/x2 + 1/x 3 ) (2 + 3/x + x 5/x2 + 7/x 3 ) = 1/2
26 The function f(x) = x3 + x 2 + x + 1 2x 3 + 3x 2 + 5x + 7 asymptote of y = 1/7. has a horizontal (a) TRUE (b) FALSE FALSE -Justification: f(x) = [ (x 3 + x 2 + x + 1)/x 3 x x (2x 3 + 3x 2 + 5x + 7)/x ] 3 = [ (1 + 1/x + 1/x2 + 1/x 3 ) x (2 + 3/x + 5/x 2 + 7/x 3 ) ] = (1 + 1/x + x 1/x2 + 1/x 3 ) (2 + 3/x + x 5/x2 + 7/x 3 ) = 1/2 In a similar fashion, we find that f(x) = 1/2 x Review (Oct 5, 2010) Math 1131Q Section / 24
27 Review (Oct 5, 2010) Math 1131Q Section / 24 The line y = 1 is the only horizontal asymptote of the function f(x) = x 1 x2 4. (a) TRUE (b) FALSE
28 Review (Oct 5, 2010) Math 1131Q Section / 24 The line y = 1 is the only horizontal asymptote of the function f(x) = x 1 x2 4. (a) TRUE (b) FALSE FALSE - For the right-hand horizontal asymptote, we take the it as x. (x 1)/x f(x) = x x ( x 2 4)/x (1 1/x) = x ( x 2 4)/ x = (1 1/x) 2 x ( x 2 4)/x 2
29 Review (Oct 5, 2010) Math 1131Q Section / 24 The line y = 1 is the only horizontal asymptote of the function f(x) = x 1 x2 4. (a) TRUE (b) FALSE FALSE - For the right-hand horizontal asymptote, we take the it as x. f(x) = x x = x = x (x 1)/x ( x 2 4)/x (1 1/x) ( x 2 4)/ x = 2 x (1 1/x) ( x 2 4)/x 2 (1 1/x) ( 1 4/x = x (1 1/x) 2 = 1 x 1 4/x 2 y = 1 is the horizontal asymptote to the right.
30 Review (Oct 5, 2010) Math 1131Q Section / 24 For the left-hand horizontal asymptote, we take the it as x. f(x) = x x (x 1)/x ( x 2 4)/x (x 1)/x) = x ( x 2 4)/( x = (1 1/x) 2 x ( x 2 4)/x 2
31 Review (Oct 5, 2010) Math 1131Q Section / 24 For the left-hand horizontal asymptote, we take the it as x. f(x) = x x (x 1)/x ( x 2 4)/x (x 1)/x) = x ( x 2 4)/( x = (1 1/x) 2 x ( x 2 4)/x 2 (1 1/x) = x ( 1 4/x = x (1 1/x) 2 = 1 x 1 4/x 2 y = 1 is the horizontal asymptote to the left.
32 Review (Oct 5, 2010) Math 1131Q Section / 24 x 0 tan (2x) sin (18x) (a) 0 (b) 1/9 (c) 9 (d) 1/36 (e) does not exist
33 Review (Oct 5, 2010) Math 1131Q Section / 24 x 0 tan (2x) sin (18x) (a) 0 (b) 1/9 (c) 9 (d) 1/36 (e) does not exist Answer = b tan (2x) x 0 sin (18x) = sin (2x) x 0 cos (2x) sin (18x)
34 Review (Oct 5, 2010) Math 1131Q Section / 24 x 0 tan (2x) sin (18x) (a) 0 (b) 1/9 (c) 9 (d) 1/36 (e) does not exist Answer = b tan (2x) x 0 sin (18x) = sin (2x) x 0 cos (2x) sin (18x) = x 0 = 1 1 cos (2x) = sin (2x) [ 2x x 0 18 sin (18x) 18x ] = sin (2x) x 0 2x x 0 sin (18x) 18x
35 Review (Oct 5, 2010) Math 1131Q Section / 24 Rules of Differentiation Product Rule: (fg) = f g + fg Quotient Rule: ( f g ) = f g fg g 2 Shorthand version Shorthand version Derivative of e kx : d dx [ekx ] = ke kx for any constant k Derivative of sin x and cos x: d d (sin x) = cos x dx (cos x) = sin x dx
36 Review (Oct 5, 2010) Math 1131Q Section / 24 By the Power Rule, the derivative of f(x) = π 5 is f (x) = 5π 4 (a) TRUE (b) FALSE
37 Review (Oct 5, 2010) Math 1131Q Section / 24 By the Power Rule, the derivative of f(x) = π 5 is f (x) = 5π 4 (a) TRUE (b) FALSE FALSE - Justification: π is a constant, so π 5 is also a constant and has derivative 0.
38 Review (Oct 5, 2010) Math 1131Q Section / 24 The function f(x) = ex x has a horizontal tangent when (a) x = 0 (b) x = 1 (c) x = 1 (d) all of the above (e) none of the above
39 Review (Oct 5, 2010) Math 1131Q Section / 24 The function f(x) = ex x has a horizontal tangent when (a) x = 0 (b) x = 1 (c) x = 1 (d) all of the above (e) none of the above Answer = c f (x) = (ex ) (x 2 + 1) e x (x 2 + 1) = (ex )(x 2 + 1) e x (2x) (x 2 + 1) 2 (x 2 + 1) 2 = (ex )(x 2 2x + 1) (x 2 + 1) 2 = (ex )((x 1) 2 ) (x 2 + 1) 2 = 0
40 The function f(x) = ex x has a horizontal tangent when (a) x = 0 (b) x = 1 (c) x = 1 (d) all of the above (e) none of the above Answer = c f (x) = (ex ) (x 2 + 1) e x (x 2 + 1) = (ex )(x 2 + 1) e x (2x) (x 2 + 1) 2 (x 2 + 1) 2 = (ex )(x 2 2x + 1) (x 2 + 1) 2 = (ex )((x 1) 2 ) (x 2 + 1) 2 = 0 f (x) = 0 only when (e x )((x 1) 2 ) = 0 which is when the factor x 1 = 0. Review (Oct 5, 2010) Math 1131Q Section / 24
41 Review (Oct 5, 2010) Math 1131Q Section / 24 Write the derivative of e 3x f(x) in terms of f(x) and f (x). (a) 3e 3x f (x) (b) e 3x f(x) + e 3x f (x) (c) e 3x f(x) e 3x f (x) (d) 3e 3x f(x) + e 3x f (x) (e) e 3x f(x) + 3e 3x f (x)
42 Review (Oct 5, 2010) Math 1131Q Section / 24 Write the derivative of e 3x f(x) in terms of f(x) and f (x). (a) 3e 3x f (x) (b) e 3x f(x) + e 3x f (x) (c) e 3x f(x) e 3x f (x) (d) 3e 3x f(x) + e 3x f (x) (e) e 3x f(x) + 3e 3x f (x) Answer = d (e 3x f(x)) = (e 3x ) f(x)) + e 3x f (x)) PRODUCT RULE = (3e 3x )f(x)) + e 3x f (x)) EXPONENTIAL FUNCTION RULE
43 Review (Oct 5, 2010) Math 1131Q Section / 24 Find values of m and b so that the function mx + b if x < 2 f(x) = is continuous and differentiable at 2x 3 if x 2 all x. (a) b = 0, m = 8 (b) b = 8, m = 12 (c) b = 16, m = 0 (d) Any b, m where 2m + b = 16 (e) Such values do not exist
44 Review (Oct 5, 2010) Math 1131Q Section / 24 Find values of m and b so that the function mx + b if x < 2 f(x) = is continuous and differentiable at 2x 3 if x 2 all x. (a) b = 0, m = 8 (b) b = 8, m = 12 (c) b = 16, m = 0 (d) Any b, m where 2m + b = 16 (e) Such values do not exist Answer = none of the answers are correct. We must have m = 24 and b = -8. x = 2 is the only point in question.
45 Review (Oct 5, 2010) Math 1131Q Section / 24 mx + b if x < 2 f(x) = 2x 3 if x 2 For continuity, we must have x 2 f(x) = f(2) = 16. x 2 x 2 f(x) = (mx + b) = 2m + b = 16.
46 mx + b if x < 2 f(x) = 2x 3 if x 2 For continuity, we must have x 2 f(x) = f(2) = 16. x 2 x 2 f(x) = (mx + b) = 2m + b = 16. For differentiability, we must have f f(2 + h) f(2) (2) =. h 0 h This means that the following one-sided its must be equal: f(2 + h) f(2) h 0 h = d dx (mx + b) x=2 = m. Review (Oct 5, 2010) Math 1131Q Section / 24
47 mx + b if x < 2 f(x) = 2x 3 if x 2 For continuity, we must have x 2 f(x) = f(2) = 16. x 2 x 2 f(x) = (mx + b) = 2m + b = 16. For differentiability, we must have f f(2 + h) f(2) (2) =. h 0 h This means that the following one-sided its must be equal: f(2 + h) f(2) h 0 h f(2 + h) f(2) h 0 + h = d dx (mx + b) x=2 = m. = d dx (2x3 ) x=2 = 6x 2 x=2 = 24. Thus m = 24 and then b = 8 since we also must have 2m + b = 16 Review (Oct 5, 2010) Math 1131Q Section / 24
48 Review (Oct 5, 2010) Math 1131Q Section / 24 Which of the following is a true statement about the function f? f(x) a b c x (a) f is continuous at a, b and c. (b) f is differentiable at a and b (c) f is continuous at b but not differentiable at b (d) f is differentiable but not continuous at c (e) f (a) = 0 and f (x) exists at b but not at c
49 Which of the following is a true statement about the function f? f(x) a b c x (a) f is continuous at a, b and c. (b) f is differentiable at a and b (c) f is continuous at b but not differentiable at b (d) f is differentiable but not continuous at c (e) f (a) = 0 and f (x) exists at b but not at c Answer = c Review (Oct 5, 2010) Math 1131Q Section / 24
50 Review (Oct 5, 2010) Math 1131Q Section / Compute t 4 t 2 16 t + 4 and t + 7 t 5 t 2 25
51 Review (Oct 5, 2010) Math 1131Q Section / Compute t 4 t 2 16 t + 4 and t + 7 t 5 t Find an equation of the tangent line to y = x 3 + 2x at the point (2, 12).
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