Mat137 study sheet for T1

Mat137 study sheet for T1 - MATH 137 (U of T) Study Sheet...

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Unformatted text preview: MATH 137 (U of T) Study Sheet Properties of Real Numbers Properties of real numbers: associativity, commutativity, existence of identity, existence of inverses Subsets of real numbers: natural numbers (e.g. 1,2,3,…), integers (e.g. -2,-1,0,1,2,…), rational numbers (i.e. xx= p q for p, q integers, q ≠ 0 ). Strong (or complete) induction: assume that the statement holds for n < k , and then prove it true for n=k. EXAMPLE: Prove that 1 + 2 + ... + n = n(n + 1) (i.e. the sum of 2 the first n positive integers is equal to n(n + 1) .) 2 SOLUTION: Properties of Functions Domain is the set of values of x for which f(x) is defined Range is the set of all possible values f(x) Even functions are functions that have the property f − x = f x for all x () () Odd functions are functions that have the property f (− x ) = − f ( x ) for all x Invertible function: a function f is invertible if there exists a function g such that y = f (x ) if and only if g ( y ) = x Increasing function: a function is increasing on an interval I if for all x, f ( x1 ) < f ( x 2 ) whenever x1 < x 2 on I Decreasing function: a function is decreasing on an interval I if for all x, f ( x1 ) > f ( x 2 ) whenever x1 < x 2 on I EXAMPLE: Suppose f(x) is even and x2 f (x ) − 5 f x 3 g (x ) = () (− x ) 3 f (− x ) − 5 f ([− x] ) is even. Therefore, since k ⋅ (k + 1) 1 + 2 + ... + k = 2 1 + 2 + ... + k = b such that for all k ⋅ (k + 1) , we can re-write it as 2 2 Collecting the (k + 1) terms, we find: ) since k (k + 1 ) + (k + 1 ) 2 ⎛k ⎞ ( k + 1 ) ⋅ (k + 2 ) = (k + 1 ) ⋅ ⎜ + 1 ⎟ = 2 ⎝2 ⎠ (( k + 1) + 1) = (k + 1 ) 2 (1 + 2 + ... + k ) + k + 1 = x∈ A, x ≤ b. Least Upper Bound: b is the least upper bound of A iff b is an upper bound of A and if for all other upper bounds a of A, b ≤ a . Greatest Lower Bound: b is the greatest lower bound of A iff b is a lower bound of A and if for all lower bounds a of A, a ≤ b . Completeness of real numbers: If A is a non empty set of real numbers and A has an upper bound then A has a least upper bound If A is a non empty set of real numbers and A has a lower bound then A has a greatest lower bound EXAMPLE: Solve the inequality x2 + 7 x − 8 < 0 . SOLUTION: First, lets factor the expression x 2 + 7x − 8 as x + 7 x − 8 = ( x + 8)( x − 1) . (x + 8)(x − 1) < 0 . For x < − 8 , both x + 8 and x − 1 are negative and, therefore F ( x ) = ( x + 8)( x − 1) is positive. follows: We now need to show the statement is true for n = k + 1. That means, we are concerned with the sum of the first k + 1 integers, i.e. with the sum 1 + 2 + ... + k + k + 1 . But we can write this as follows: 1 + 2 + ... + k + k + 1 = (1 + 2 + ... + k ) + k + 1 . (1 + 2 + ... + k ) + k + 1 = k (k + 1) + (k + 1) 2 x2 x2 = = 3 f (− x ) − 5 f − x f ( x ) − 5 f (x 3 ) f (x ) Thus, the left side is equal to the right side and the expression is valid for the case n = 1. Induction Step: Our induction hypothesis is that the expression holds for the case n = k, and thus the following expression holds: a≤x 2 This gives us: When we pass through x = −8, x + 8 changes sign and, therefore, so does F(x). But when we later pass through x = 1, x − 1 changes sign and F(x) changes back to being positive. Thus, F(x) is negative for −8 < x < 1. follows: SOLUTION: ( 1⋅ (1 + 1) 2 = = 1. 2 2 But since our induction hypothesis is . Determine whether g(x) is even, odd, or neither. g (− x ) = We’ll use induction to prove the above equality. Base case: When n = 1, the left side of the equality is equal to 1. When n = 1, the right side of the equality is equal to has a lower bound if there is a number a such that for all x ∈ A , . Bounded: A is bounded iff A has either an upper bound and a lower bound; iff there exists a number Logarithmic Functions Properties of logarithms: log b (mn) = log b m + log b n , ⎛ m⎞ log b ⎜ ⎟ = log b m − log b n , ⎝ n⎠ 1 log b mr = r log b m , log b = − log b m , m log b 1 = 0 , log b b = 1 , log b b r = r , More free study sheet and practice tests at: g ( x ) = g (− x ) , g ( x ) is Therefore, we know: even. Mathematical Induction INDUCTION STEP BY STEP Step 1—Base case. Show that the statement holds in the simplest case (normally for n = 1 ). n = 0 and/or Step 2—Induction Hypothesis. Assume the statement holds for an arbitrary k. Step 3. Prove that it holds for above induction hypothesis. k + 1 given the (1 + 2 + ... + k ) + k + 1 = (k + 1) ((k + 1) + 1) This is 2 what we need to show. Therefore, the statement is true for all positive integers n. b logb m = m , 10 = x , e ln x = x , log a m . log b m = log a b log x EXAMPLE: Inequalities and Absolute Values Absolute Value Rules: a−b ≥ a − b a+b ≤ a + b , a ⋅b = a ⋅ b Solve for x: ln(x2 + x) = 1 + ln x SOLUTION: ln(x2 + x) = 1 + ln x ln(x2 + x) − ln x = 1 ⎛ x2 + x ⎞ ⎟ =1 ⎜x⎟ ⎠ ⎝ , Upper/lower bound: A has an upper bound if there ln ⎜ is a number a such that for all ln(x + 1) = 1⇒ x + 1 = e ⇒ x = e − 1. x∈ A, x ≤ a . A More free Study Sheets and Practice Tests at: www.prep101.com www.prep101.com More free study sheets and practice tests at Limits and Continuity L’HOPITAL’S RULE: Formal Definition: The function f(x) has limit L, as x approaches a, denoted lim f x = L if given any () x→a ε > 0 , there exists a δ > 0 such that |f(x) - L| < ε for all x satisfying 0 < |x – a|< δ. lim[ f ( x ) ± g ( x )] = lim f ( x ) ± lim g (x ) , x→a x→a x→a lim x→a [ n x→a x→a ε >0. Since ε Then x − a < δ 1 , then f ( x ) − L < ε Also, since such . Let x−a < δ 2 , then g (x ) − K < δ = minimum (δ 1 , δ 2 ) . Hence, if g (x ) − K < Therefore, if then 5x 3 − 4 x + 2 lim = lim x→∞ 3 + x 2 − 6 x 3 x→∞ 4 2⎞ ⎛ x3 ⎜ 5 − 2 + 3 ⎟ x x⎠ ⎝ 1 3⎞ ⎛ x3⎜− 6 + + 3 ⎟ x x⎠ ⎝ 5−0+0 5 =− . x→∞ − 6 + 0 + 0 6 Continuity: a function is continuous at point a if and only if the left hand and right hand limits exist and left hand limit = right hand limit = f(a) (i.e. if lim f ( x) = f (a) ). = lim . ε 2 =ε (by the triangle x →a as required. and lim g ( x ) = K x→a lim[ f (x ) + g ( x )] = L + K . x→a By the I.V.T, there is a ⎛ π⎞ ⎜ 0, ⎟ ⎝ 2⎠ such that f(c) = 1, SQUEEZE THEOREM: If f (x ) ≤ g (x ) ≤ h( x ) near a and lim x ) = lim h( = L More free study sheet and practicef (testsx)at: 2 lim f ( x ) = L x →a π ). 2 i.e. such that c is the solution to the expression x + sin x = 1. , then ≤ f (x ) − L + g (x ) − K 2 Thus, f(0) < 1 < f( π )= π + sin π = π +1. 22 22 5x − 4 x + 2 3 + x 2 − 6 x3 . SOLUTION: x →∞ 2 + f(0) = 0 + sin 0 = 0 and f( ⎡ π⎤ . ⎢0, 2 ⎥ ⎣ ⎦ number c in the interval lim 2 [ f (x ) + g (x )] − (L + K ) = [ f (x ) − L] + [g (x ) − K ] ε Let f(x) = x + sin x. Then f is continuous on 3 . So, inequality) < f (c ) = k Determine the following limit: f (x ) − L < ε and Use the Intermediate theorem to show that the equation x + sin x = 1 has a solution in the interval EXAMPLE: x − a < δ 1 and ε and , and therefore, x−a <δ x − a < δ2 ε f (a ) and f (b ) then there exists a c such that ⎡ π⎤ . ⎢0, 2 ⎥ ⎦ ⎣ SOLUTION: First try substitution, factoring If you’re taking the limit as x → ∞ of a quotient, and substitution yields an invalid answer, then try first to divide both the numerator and the denominator of the limit expression by its highest power of x If the limit expression contains absolute values, then try breaking the limit up into two one-sided limits. If the limit expression is a quotient, and if factoring doesn’t work, then try l’Hopital’s Rule x →a such that if a<c<b TECHNIQUES FOR FINDING LIMITS: 2 lim g ( x ) = K , there exists δ 2 > 0 , ⇒ EXAMPLE: x →a that if ) 3 x + 6 x − 10 14 = =7 x→2 3x 2 − 6 x + 2 2 > 0. 2 lim f ( x ) = L , there exists δ 1 > 0 x →1 is continuous on [a, b ] and if k is between If = lim x→a Let f (x ) ) 2 lim[ f ( x ) + g ( x )] = L + K . = lim+ 2 x = 2 Therefore: 1 + eK = 2 K = 0. Thus, if f ( x) is continuous at x = 1, then K = 0. ( ε − δ proof of the addition property of If lim f ( x ) = L and lim g ( x ) = K , then x →a x →a 2x 2 − 2x x−1 x−1 ( n ) x →1+ INTERMEDIATE VALUE THEOREM: d3 x + 3 x 2 − 10 x x 3 + 3 x 2 − 10 x lim dx = lim x → 2 x 3 − 3x 2 + 2 x x→2 d x 3 − 3x 2 + 2 x dx Give an SOLUTION: x →1 (x − 1) ⋅ 2 x x →1 0 ( x →1 K = 1 + e K = f (1) = lim+ . Note first that this limit is in 0 form. Therefore, use EXAMPLE: limits: x 3 + 3 x 2 − 10 x L’Hopital’s theorem: We find lim[ f ( x )] = lim f ( x ) x →1− lim f (x ) = lim− x + e x →1 − x →1+ SOLUTION: x→a lim f (x ) = lim f ( x ) = f (1) We require: lim f (x ) = lim+ 3 2 Evaluate x → 2 x − 3x + 2 x lim[ f ( x ) ⋅ g ( x )] = lim f ( x ) ⋅ lim g ( x ) , ⎡ f ( x ) ⎤ lim f ( x ) , lim⎢ = x →a x →a g ( x ) ⎥ ⎣ ⎦ lim g ( x ) x →a f (x ) and g ( x ) are of types 0 or ∞ , and if g ′(a ) 0 ∞ is not 0, then lim f ( x ) = lim f ′( x ) x→a g ( x) x → a g ′( x ) EXAMPLE: PROPERTIES OF LIMITS: x→a SOLUTION: If the limit of the quotient of differentiable functions EXAMPLE: , If ⎧x + eK if x ≤ 1 is continuous at x = 1, ⎪ f (x ) = ⎨ 2 x 2 − 2 x if x > 1 ⎪ ⎩ x −1 then determine the value of K. x →a then x→a lim g ( x ) x →a exists and is equal to L. EXAMPLE: If, for all values of x in the interval (0, 5), 1 + x ≤ f (x) ≤ 3 + sin πx, find lim f x . x →2 () SOLUTION: lim 1 + x =1+2=3 x→2 lim 3 + sin πx x→2 = 3 + sin 2π = 3. ⇒ lim f ( x ) = 3 by the squeeze theorem. x →2 Helping students since 1999 ...
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