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Unformatted text preview: 10A Midterm 1 Review
Martha Yip
Disclaimer: there may be typos! Functions: to each input x, assigns a unique output f (x) Any vertical line intersects the graph of a function in at most one point. Domain: all possible inputs into f Range: all possible outputs of f , (after the domain is speciﬁed) Example: . . ....... ....... . 4. . .... ..... . .... ..... . . . ... . .. . ... .. . . . ... . . . . . . . ◦... ....• Domain = [1; 7] 3. . ... . . .. . . ..... .... . .. . .. . . . . . . . . Range = [0; 1] (2; 4] ◦ 2. . . . . . . . . . .. . • 1. ...• . . .. . . ..... . . .. . . ... . . . .. . ..... . . . . ................•............................................................................. .............................................................................................. ..... . . . . . . 1 3 6 7 1 Functions: to each input x, assigns a unique output f (x) Any vertical line intersects the graph of a function in at most one point. Domain: all possible inputs into f Range: all possible outputs of f , (after the domain is speciﬁed) Example: . . ....... ....... . 4. . .... ..... . .... ..... . . . ... . .. . ... .. . . . ... . . . . . . . ◦... ....• Domain = [1, 7] 3. . ... . . .. . . ..... .... . .. . .. . . . . . . . . Range = [0, 1] ∪ (2, 4] ◦ 2. . . . . . . . . . .. . • 1. ...• . . .. . . ..... . . .. . . ... . . . .. . ..... . . . . ................•............................................................................. .............................................................................................. ..... . . . . . . 1 3 6 7
1a Inverse Functions
If y = f (x) is a function, then any vertical line intersects the graph in at most one point. If f (x) is a function with domain D and range R, then the inverse function of f (x), denoted by f −1(x) if it exists, is a function with domain R and range D. f (x) has an inverse if any horizontal line intersects the graph in at most one point. Procedure for ﬁnding the inverse function for y = f (x). (1) Switch y and x. (2) Solve for y. 2 Functions check your understanding 1. The graph of f −1 is a reﬂection of f in the line
x . 2. True or False? x is an even function.
x 3. True or False? x is an odd function. 4. Can a function be both odd and even? 5. True or False? f (x) = x2 has an inverse. 6. True or False? −1 (x) = −√x. f f (x) = x2 with domain (−∞, 0] has an inverse 3 Functions check your understanding 1. The graph of f −1 is a reﬂection of f in the line y = x.
x 2. True or False? x is an even function.
x 3. True or False? x is an odd function. 4. Can a function be both odd and even? 5. True or False? f (x) = x2 has an inverse. 6. True or False? −1 (x) = −√x. f f (x) = x2 with domain (−∞, 0] has an inverse 3a Functions check your understanding 1. The graph of f −1 is a reﬂection of f in the line y = x.
x 2. True or False? x is an even function.
x 3. True or False? x is an odd function. 4. Can a function be both odd and even? 5. True or False? f (x) = x2 has an inverse. 6. True or False? −1 (x) = −√x. f f (x) = x2 with domain (−∞, 0] has an inverse 3b Functions check your understanding 1. The graph of f −1 is a reﬂection of f in the line y = x.
x 2. True or False? x is an even function.
x 3. True or False? x is an odd function. 4. Can a function be both odd and even? 5. True or False? f (x) = x2 has an inverse. 6. True or False? −1 (x) = −√x. f f (x) = x2 with domain (−∞, 0] has an inverse 3c Functions check your understanding 1. The graph of f −1 is a reﬂection of f in the line y = x.
x 2. True or False? x is an even function.
x 3. True or False? x is an odd function. 4. Can a function be both odd and even? If this was true, then −f (x) =
f (−x) = f (x), so 2f (x) = 0, so f (x) must be the zero function. 5. True or False? f (x) = x2 has an inverse. f (x) = x2 with domain (−∞, 0] has an inverse 6. True or False? − 1 ( x) = − √ x. f 3d Functions check your understanding 1. The graph of f −1 is a reﬂection of f in the line y = x.
x 2. True or False? x is an even function.
x 3. True or False? x is an odd function. 4. Can a function be both odd and even? If this was true, then −f (x) =
f (−x) = f (x), so 2f (x) = 0, so f (x) must be the zero function. 5. True or False? f (x) = x2 has an inverse. f (x) = x2 with domain (−∞, 0] has an inverse 6. True or False? − 1 ( x) = − √ x. f 3e Functions check your understanding 1. The graph of f −1 is a reﬂection of f in the line y = x.
x 2. True or False? x is an even function.
x 3. True or False? x is an odd function. 4. Can a function be both odd and even? If this was true, then −f (x) =
f (−x) = f (x), so 2f (x) = 0, so f (x) must be the zero function. 5. True or False? f (x) = x2 has an inverse. f (x) = x2 with domain (−∞, 0] has an inverse 6. True or False? − 1 ( x) = − √ x. f 3f Transformations of functions
1. Composition: f (g (x)) or g (f (x)) 2. Vertical shift: g (x) = f (x) + k 3. Horizontal shift: g (x) = f (x − k) 4. Vertical stretch/shrink: g (x) = kf (x) 5. Reﬂection in the xaxis: g (x) = −f (x) 6. Reﬂection in the yaxis: g (x) = f (−x) 4 f (x) = x2. Sketch g (x) = −f (x + 1) + 4. . . . . . . . . . . . . . . . . . . . . . . . . . . − . . . . . . . . . . . . . . . . . . . . . . . . . . . . − . . . . . . . . . . . . . . . . . . . . . .. . . .. . . − . . .. . .. .. . . . . .. .. . .. . . . .. − .. . .. . . .. .. . . . .. . .. . ... . .. ... . .. . ....................................................... . ................................................................................................. ............................................ . . . . . . .     . . . . . . . . . . − . . . . . . . . . . . − . . . . . . . . . . − . . . . . . . . . . . − . . . .
5 f (x) = x2. Sketch g (x) = −f (x+1) + 4. . . . . . . . . . . . . . . . . . . . . . . . . . . −. . . . . . . . . .. . . . . .. . . . .. . . . .. . . −. . .. . .. . . . .. . .. . . . .. .. .. . .. . − .. . .. .. .. .. . .. .. .. . .. . . . .. . − .. .. . .. .. ... .. . . ... ... . ... .. . . ............................................................... ................................................................................................. ................................... . . . . . . . . .    . . . . . . . . . . − . . . . . . . . . . . − . . . . . . . . . . − . . . . . . . . . . . − . . . .
5a f (x) = x2. Sketch g (x) = −f (x + 1) + 4. . . . . . . − . . . . . . . . . . . − . . . . . . . . . . − . . . . . . . . . . . − . . . . . . . ................................................................................................... ............................................................................................... . . .. . .. . . .. . . . . . .  . ... . . .. . . ... ... .. . .. . . .. . .. . . − . ... . ... . .. .. .. .. .. . .. . . .. . .. −. . .. . . .. . .. .. . .. . .. . . . .. .. − .. . . .. . . . . . . . . . . . . . . . . . . . . . − .. . . . . . . . . . . . . . . . . . . .
5b f (x) = x2. Sketch g (x) = −f (x + 1)+4. . . . . ........ . ......... − . .. ... .. . . . .. . .. ... . . ... . . . . . . .. . .. .. . − .. .. . .. .. . .. .. . . . . .. . .. − .. . . .. . .. . .. .. . .. . .. . .. . . .. .. −. .. . . . . . . . . . . . . . . . ................................................... . ................................................................................ ................................................. ................... . . . .. .     . . . . . . . . . . . . . . . . . . . . . . . . . . − . . . . . . . . . . . − . . . . . . . . . . − . . . . . . . . . . . − . . . .
5c Transformations of functions check your understanding 1. True or False? f (g (x)) means apply f to x, then apply g . 2. Let g (x) = 2f (x) − 1. To get the graph of g , ﬁrst do then do . , 6 Transformations of functions check your understanding 1. True or False? f (g (x)) means apply f to x, then apply g . 2. Let g (x) = 2f (x) − 1. To get the graph of g , ﬁrst do then do . , 6a Transformations of functions check your understanding 1. True or False? f (g (x)) means apply f to x, then apply g . 2. Let g (x) = 2f (x) − 1. To get the graph of g , ﬁrst stretch vertically by a factor of 2, then shift down by 1 unit. 6b Linear Functions
Linear functions are of the form y = mx + b, where m is the slope, b is the y intercept The graph of a linear function is a straight line. ..... ..... ..... ..... ..... . ..... . ...... ..... .... .. . •..... ∆y . ... . (0, b) . ...... . ..... m = ∆x . ..... . . ..... . ..... . . ..... . ..... . ..... . ..... . . ..... . .... . . .............................................................. ............................................................. .................... .................... . .... ... ... ... ..... ..... . . . . 7 Linear functions check your understanding 1. If the slope m is positive, the function is . 2. If the slope m is negative, the function is . 3. If the slope m = 0, the graph is a line. 4. A line through the points (p, q ) and (c, d) has slope 8 Linear functions check your understanding 1. If the slope m is positive, the function is increasing. 2. If the slope m is negative, the function is . 3. If the slope m = 0, the graph is a line. 4. A line through the points (p, q ) and (c, d) has slope 8a Linear functions check your understanding 1. If the slope m is positive, the function is increasing. 2. If the slope m is negative, the function is decreasing. 3. If the slope m = 0, the graph is a line. 4. A line through the points (p, q ) and (c, d) has slope 8b Linear functions check your understanding 1. If the slope m is positive, the function is increasing. 2. If the slope m is negative, the function is decreasing. 3. If the slope m = 0, the graph is a horizontal line. 4. A line through the points (p, q ) and (c, d) has slope 8c Linear functions check your understanding 1. If the slope m is positive, the function is increasing. 2. If the slope m is negative, the function is decreasing. 3. If the slope m = 0, the graph is a horizontal line. d−q . c−p 4. A line through the points (p, q ) and (c, d) has slope 8d Exponential Functions
Exponential functions are of the form P = P0at, where P0 is the initial quantity, if a > 1, have exponential growth if 0 < a < 1, have exponential decay Can also express the growth factor a as a = 1 + r, and write P = P0(1 + r)t, where r is the growth rate. if r > 0, have exponential growth if r < 0, have exponential decay 9 Exponential Functions Continuous rate of growth Keep in mind the example of compound interest: k nt B = P0 1 + → P0ekt n If k > 0, have exponential growth. If k < 0, have exponential decay. as n → ∞ 10 Half life, doubling time
Half life is the time it takes for the exponentially decaying quantity to be reduced to half.
1 Given P = P0at, ﬁnd t so that P = 2 P0. So, solve for t in 1 P0 = P0at 2 or 1 = at. 2 Doubling time: Given P = P0at, ﬁnd t so that P = 2P0, so solve 2 = at. 11 Exponential functions check your understanding 1. Consider P = 25(0.6)t. The growth factor a is factor r is . The function is exponentially . The growth ing. 2. Which grows faster? ex, or 10x? 3. Which grows faster? 1000(2x) or 3x? 12 Exponential functions check your understanding 1. Consider P = 25(0.6)t. The growth factor a is 0.6. The growth factor r is a − 1 = −0.4. The function is exponentially decaying. 2. Which grows faster? ex, or 10x? 3. Which grows faster? 1000(2x) or 3x? 12a Exponential functions check your understanding 1. Consider P = 25(0.6)t. The growth factor a is 0.6. The growth factor r is a − 1 = −0.4. The function is exponentially decaying. 2. Which grows faster? ex, or 10x? 3. Which grows faster? 1000(2x) or 3x? 12b Exponential functions check your understanding 1. Consider P = 25(0.6)t. The growth factor a is 0.6. The growth factor r is a − 1 = −0.4. The function is exponentially decaying. 2. Which grows faster? ex, or 10x? 3. Which grows faster? 1000(2x) or 3x? 12c Linear vs Exponential functions: charts
Linear, slope m x 2 1 0 1 2 f(x) Exp, growth factor a x 2 1
m f(x)  ♣ m ~ ¡ 1
a ♥ + +2 +3 0 1 2 m m ~¡ ~¡ ~¡ a a 2 3 a 13 Linear vs Exponential functions: charts
Linear, slope m x 2 1 0 1 2 f(x) ♣−m ♣ ♣+m ♣ + 2m ♣ + 3m Exp, growth factor a x 2 1 0 1 2 f(x) ♥ · a−1 ♥ ♥·a ♥ · a2 ♥ · a3 13a Logarithmic functions
The inverse of the exponential function ax is the logarithm loga(x): eln x = x, ln(ex) = x. Properties
ea+b = eaeb
a−b = ea e eb ln(ab) = ln(a) + ln(b) ln( a ) = ln(a) − ln(b) b ln(xa) = a ln(x) ln(1) = 0 eab = (ea)b e0 = 1 14 sin θ, cos θ, tan θ deﬁned on the unit circle . . . . . . ............... ................ ..... . . ..... . .... .... . . .. ... . (cos θ, sin θ) ...•. ... . .... ... . . .. . . .. ... . .. . .. .. . . .. . .. . .. . . .. . . . .. . .. . ... . . .. . . . .. . .. . . . . . . .. . . ... .. . . . ... . . .θ . . .. .. ... . . . . ... . . . ............................................................ . .......................................................... ........ ....... . .. . . . . . . . . . . . . . . . . . . . . . . .. . . .. . . . .. . .. .. . .. . . . . .. .. . .. . ... . . ... . . .. .... ... . .... ...... . ........ ...... . ....... . . ............. ............ . . . . . . . . . tan θ = sin θ cos θ A circle has 2π radians = 360 degrees. One radian is the angle whose arc has length equal to the radius. ( 360 ∼ 57 degrees) 2π
15 Trig functions, cont’d
Sine or Cosine functions are of the form y = A sin (Bx) + C, where A is the amplitude (half the distance between max and min) and 2π is the period (time it takes to complete one cycle) B C is the vertical shift and the graph oscillates about that value y = A cos (Bx) + C Identities: sin2(x) + cos2(x) = 1 cos x = sin (x + π ) 2 (Pythagoras on the unit circle) (horizontal shift of graphs) 16 Trig functions check your understanding 1. True or False? sin θ is an odd function. 2. True or False? cos θ is an odd function. 3. The range of sin θ is . 4. The domain of tan θ is . 5. The range of 7 sin(2θ) + 1 is . 17 Trig functions check your understanding 1. True or False? sin θ is an odd function. 2. True or False? cos θ is an odd function. 3. The range of sin θ is . 4. The domain of tan θ is . 5. The range of 7 sin(2θ) + 1 is . 17a Trig functions check your understanding 1. True or False? sin θ is an odd function. 2. True or False? cos θ is an odd function. 3. The range of sin θ is . 4. The domain of tan θ is . 5. The range of 7 sin(2θ) + 1 is . 17b Trig functions check your understanding 1. True or False? sin θ is an odd function. 2. True or False? cos θ is an odd function. 3. The range of sin θ is [−1, 1]. 4. The domain of tan θ is . 5. The range of 7 sin(2θ) + 1 is . 17c Trig functions check your understanding 1. True or False? sin θ is an odd function. 2. True or False? cos θ is an odd function. 3. The range of sin θ is [−1, 1]. 4. The domain of tan θ is x = . . . , − π , π , 3π , 5π , . . .. 22 2 2 5. The range of 7 sin(2θ) + 1 is . 17d Trig functions check your understanding 1. True or False? sin θ is an odd function. 2. True or False? cos θ is an odd function. 3. The range of sin θ is [−1, 1]. 4. The domain of tan θ is x = . . . , − π , π , 3π , 5π , . . .. 22 2 2 5. The range of 7 sin(2θ) + 1 is [−6, 8]. 17e Polynomial functions
Polynomial functions of degree n are of the form p(x) = anxn + an−1xn−1 + · · · + a1x + a0. If n is even, the graph of p(x) is even. p(x) has at most n zeros, and may not have any zeros at all. If n is odd, the graph of p(x) is odd. p(x) has at most n zeros, and has at least 1 zero. If r is a zero of p(x), then p(x) has the factor (x − r): p(x) = (x − r)(an n − 1 degree polynomial). 18 Rational functions
Rational functions are of the form f ( x) = p ( x) , q (x) where both p(x) and q (x) are polynomials. To ﬁnd the zeros of f (x): ﬁnd the zeros of p(x). To ﬁnd where f (x) is not deﬁned: ﬁnd the zeros of q (x). Note: f (x) has a vertical asymptote x = r if r is a zero of q (x) (assuming it isn’t also a zero of p(x)) To ﬁnd any horizontal asymptotes, check the behaviour of f (x) as x → ∞ or −∞. Remember, the terms with the highest degrees dominate. 19 Rational functions
Let −2(x + 1)(x − 3)(x − 7) −2x3 + 18x2 − 22x − 42 = f ( x) = . 3 − 6x2 − 15x + 100 2 x (x + 4)(x − 5) The zeros are x = −1, 3, 7. The vertical asymptotes are x = −4, 5. As x → ±∞, −2x3 = −2, f (x) behaves like 3 x so the horizontal asymptote is y = −2. 20 Polynomial and rational functions check your understanding 1. As x → −∞, x13 goes to
1 2. As x → −∞, x2 goes to . . 3. Does x2 have a horizontal asymptote? 21 Polynomial and rational functions check your understanding 1. As x → −∞, x13 goes to −∞.
1 2. As x → −∞, x2 goes to . 3. Does x2 have a horizontal asymptote? 21a Polynomial and rational functions check your understanding 1. As x → −∞, x13 goes to −∞.
1 2. As x → −∞, x2 goes to 0. 3. Does x2 have a horizontal asymptote? 21b Polynomial and rational functions check your understanding 1. As x → −∞, x13 goes to −∞.
1 2. As x → −∞, x2 goes to 0. 3. Does x2 have a horizontal asymptote? No. 21c Limits
If as x approaches c, the values of f (x) approach the number L, then L is the limit of the function f (x) as x approaches c. Notation:
x→c lim f (x) = L. This is understood to be a twosided limit. The lefthand and righthand limits are
x→c− lim f (x) and x→c+ lim f (x) where we only look at approaching c from the left or the right. The twosided limit exists if and only if the onesided limits exist and are equal.
22 Limits at inﬁnity
The concept of limits at inﬁnity is the same as “end behaviour” and horizontal asymptotes. If as x approaches ∞, f (x) approaches a number L, then
x→∞ x2 −5 Example: f (x) = 3x2−1x lim f (x) = L. 3x2 − 5x lim x→∞ x2 − 1 3x2 − 5x 3−5 2 2 x = lim x = lim 3 − 0 = 3. = lim x 2 x→∞ x x→∞ 1 − 1 x→∞ 1 − 0 1 − x2 2 2 x x 23 Continuity of a function
f (x) is continuous at x = c if
x→c lim f (x) = f (c). Example: Consider the function x2 − 4 (x − 2)(x + 2) f ( x) = = . x+2 x+2 Its domain is (−∞, −2) ∪ (−2, ∞). Away from x = −2, its graph is the straightline y = x − 2. f (x) is continuous everywhere except at x = −2. In order to make this function continuous everywhere, we should deﬁne f (−2) = (−2 − 2) = −4. Intuitively, the graph is ”unbroken” at x = c.
24 Limits laws
If lim f (x) and lim g (x) each exist, then • Sums: lim f (x) ± g (x) = lim f (x) ± lim g (x) • Mul. by constant: lim kf (x) = k lim f (x). • Product Law: lim f (x)g (x) = lim f (x) lim g (x). • Quotient Law: if lim g (x) = 0, then lim lim f (x) f ( x) = . g ( x) lim g (x)
25 Limits and Continuity
f ( x) . . . . . . . . . . . . . ... . . . . . ... . . . . . ... . . . ◦. 3 −− . . . . . . ... . . . . . . ... . . . . . ... . . • . 2 −− . . ... . . . . . . . ... . . . . . ... . . . . ... . . ... .. .. .. .. .. .. .. .. . .......................................................... ......................................................... . .  . . . . . . 3 Examples to keep in mind g ( x) . . . . . . . . . . . . . ... . . . . . ... . . . . . ... . . . •. 3 −− ... . . . . . ... . . . . . . ... . . . . . ... . . . . . ... . . . . . ... . . . . . ... . . . . . ... . . .... .. .. .. .. .. .. .. .. . .............................................................. .............................................................. . .  . . . . . . 3 As long as the function is continuous at c, then we can “plug in c” to ﬁnd the limit.
26 Limits check your understanding 1. lim f (x) = L means that f (x) gets near to x→c ciently close to (but is diﬀerent from) . when x gets suﬃ 2. If lim f (x) = −2 and lim f (x) = −2, then lim f (x) =
x→c+ x→c− x→c . 3. If lim f (x) = 1 and lim g (x) = 2, then lim 2f (x) + g (x) =
x→2 x→2 x→2 . 4. 1 = x→2+ x − 2 lim . 27 Limits check your understanding 1. lim f (x) = L means that f (x) gets near to L when x gets suﬃciently x→c close to (but is diﬀerent from) c. 2. If lim f (x) = −2 and lim f (x) = −2, then lim f (x) =
x→c+ x→c− x→c . 3. If lim f (x) = 1 and lim g (x) = 2, then lim 2f (x) + g (x) =
x→2 x→2 x→2 . 4. 1 = x→2+ x − 2 lim . 27a Limits check your understanding 1. lim f (x) = L means that f (x) gets near to L when x gets suﬃciently x→c close to (but is diﬀerent from) c. 2. If lim f (x) = −2 and lim f (x) = −2, then lim f (x) = −2.
x→c+ x→c− x→c 3. If lim f (x) = 1 and lim g (x) = 2, then lim 2f (x) + g (x) =
x→2 x→2 x→2 . 4. 1 = x→2+ x − 2 lim . 27b Limits check your understanding 1. lim f (x) = L means that f (x) gets near to L when x gets suﬃciently x→c close to (but is diﬀerent from) c. 2. If lim f (x) = −2 and lim f (x) = −2, then lim f (x) = −2.
x→c+ x→c− x→c 3. If lim f (x) = 1 and lim g (x) = 2, then lim 2f (x) + g (x) = 4.
x→2 x→2 x→2 4. 1 = x→2+ x − 2 lim . 27c Limits check your understanding 1. lim f (x) = L means that f (x) gets near to L when x gets suﬃciently x→c close to (but is diﬀerent from) c. 2. If lim f (x) = −2 and lim f (x) = −2, then lim f (x) = −2.
x→c+ x→c− x→c 3. If lim f (x) = 1 and lim g (x) = 2, then lim 2f (x) + g (x) = 4.
x→2 x→2 x→2 4. 1 = +∞, the limit does not exist. x→2+ x − 2 lim 27d Example
. . ..... ........ . 4. . .... ..... . .... ..... . . . . . .. ... . ... .. . . . ... . . . . . . ◦... ....• 3. ... . . . .. . . ..... .. . . . .. .. . . . . . . . . ◦ 2. . . . . . . . . . . . • ... 1. . . .. . • . ..... . . .. . . ... . . . .. . ..... . .. . . ................•............................................................................. ............................................................................................... .. .. . . . . . . 1 3 6 7 What are f (2), f (3), f (6)? Does the function have a limit at x = 2, x = 3, x = 6? Is the function continuous at x = 2, x = 3, x = 6?
28 Speed, rates of change
The average rate of change of f (x) over the time interval a ≤ t ≤ b is the diﬀerence quotient f (b) − f (a) . b−a The instantaneous rate of change of f (x) at time t is f ( x) − f ( t ) , x→t x−t lim or, equivalently, f (t + h) − f (t) lim , h→0 h 29 Rates of change check your understanding f (b) − f (a) is the slope of the line b−a . 1. The average rate of change through the points and 2. The instantaneous rate of change lim of the f (t + h) − f (t) , is the slope h→0 h line through the point . 30 Rates of change check your understanding 1. The average rate of change f (b) − f (a) is the slope of the line b−a through the points (a, f (a)) and (b, f (b)). f (t + h) − f (t) 2. The instantaneous rate of change lim , is the slope h→0 h of the line through the point . 30a Rates of change check your understanding f (b) − f (a) 1. The average rate of change is the slope of the line b−a through the points (a, f (a)) and (b, f (b)). f (t + h) − f (t) , is the slope h→0 h of the tangent line through the point (t, f (t)). 2. The instantaneous rate of change lim 30b Example Let f (t) = t2.
The average rate of change from t = 0 to t = 2 is f (2) − f (0) 22 − 02 = = 2. 2−0 2−0 The instantaneous rate of change at t = 0 is f (t + h) − f (t) (t + h)2 − t2 lim = lim h→0 h→0 h h (t2 + 2th + h2) − t2 = lim h→0 h 2th + h2 = lim h→0 h = lim 2t + h = 2t.
h→0 31 Study hard and Good luck! 32 ...
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This note was uploaded on 03/28/2011 for the course MATH 10A taught by Professor Arnold during the Winter '07 term at UCSD.
 Winter '07
 Arnold

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