Mat137 study sheet for T1

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online