Exercise 2.5 P.128 20. Explain why the function is discontinuous at the given number a . Sketch the graph of the function. { x2 −x , if x ̸= 1 x2 −1 f (x) = a=1 1 , if x = 1 < pf > ∵ limx→1+ f (x) = limx→1− f (x) = limx→1 ∴ f (x) is discontinuous at x = 1 x2 −x x2 −1 = limx→1 x x+1 = 1 2 ̸= f (1) = 1 30. Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain. State the domain. tan x B(x) = √ 4 − x2 < pf > (2k−1)π (2k+1)π ),∀k∈Z (1) tan 2 √ x is continuous in any interval ( 2 , (2) x is continuous at every x ≥ 0 (3) polynomial 4 − x2 is continuous at every x ∈ R By (1),(2),(3) B(x) is continuous at every x ∈ (0, π2 ) 38. Use continuity to evaluate the limit lim arctan( x→2 x2 − 4 ) 3x2 − 6x < pf > 4 2 −4 lim arctan( 3xx2 −6x ) = arctan( lim (x−2)(x+2) 3x(x−2) ) = arctan( 6 ) = arctan( 3 ) 2 x→2 x→2 P.129 42. Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f . , if x ≤ 1 x+1 1 , if 1 < x < 3 f (x) = x √ x − 3 , if x ≥ 3 < pf > ∵(1) x + 1 is continuous everywhere (2) x1 is continuous at every x ̸= 0 √ (3) x − 3 is continuous at every x ≥ 3 ∴we only need to check x = 1 ∧ 3 lim f (x) = lim+ x1 = 1 ̸= lim− f (x) = lim− (x + 1) = 2 x→1+ x→1 x→1 x→1 √ lim f (x) = lim x1 = 13 ̸= lim f (x) = lim x − 3 = 0 x→3− x→3− x→3+ ⇒ f (x) is discontinuous at x = 1 ∧ 3 x→3+ 1 46. Find the values of a and b that make f continuous everywhere. 2 −4 , if x < 2 xx−2 f (x) = ax2 − bx + 3 , if 2 ≤ x < 3 2x − a + b , if x ≥ 3 < pf > It’s easy to check f (x) continuous at every x ̸= 2 ∧ 3 So we only need to check x = 2 ∧ 3 if f (x) is continuous at everywhere then lim f (x) = f (2)∧ lim f (x) = f (3) x→2 x→3 lim f (x) = lim ax2 − bx + 3 = 4a − 2b + 3 x→2+ x→2+ 2 lim f (x) = −4 x→2 lim f (x) = lim xx−2 = lim− (x−2)(x+2) =4 x−2 x→2− x→2− x→2 { lim f (x) = lim+ 2x − a + b = 6 − a + b x→3+ x→3 lim f (x) = lim− f (x) = lim− ax2 − bx + 3 = 9a − 3b + 3 x→3 x→3 x→3 { { 4a − 2b + 3 = 4 a = 12 ⇒ ⇒ 9a − 3b + 3 = 6 − a + b b = 12 54. Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. sin x = x2 − x, (1, 2) < pf > sin x and x2 − x are both continuous at (1, 2) Define f (x) = sin x − (x2 − x) then f (1) = sin 1 − (1 − 1) = sin 1 > 0 f (2) = sin 2 − (4 − 2) = sin 2 − 2 < 0 ∵ f (1)f (2) < 0 ∴By I.V.T there exist a c ∈ (1, 2) s.t f (c) = 0 64. For what values of x is g continuous { g(x) = 0 , if x is rational x , if x is irrational < pf > Claim : g(x) is continuous only at x = 0 Given ε > 0 take δ = ε > 0 if 0 <{|x − 0| < δ = ε |g(x) − 0| = 0 < δ = ε , if x is rational then |g(x) − 0| = |x| < δ = ε , if x is irrational ⇒ lim g(x) = 0 = g(0) x→0 For any c ̸= 0 (1) if c is irrational Take ε = 2c then∀ δ > 0 if 0 < { |x − c| < δ |g(x) − g(c)| = c > 2c = ε then |g(x) − g(c)| = |x − c| < δ (2) if c is rational , if x is rational , if x is irrational 2 Given ε > 0 take δ = ε if 0 < { |x − c| < δ |g(x) − g(c)| = 0 < ε , if x is rational then |g(x) − c| = |x − c| < δ = ε , if x is irrational ⇒ lim g(x) = 0 ̸= lim g(x) = c x→c, x∈Q / x→c, x∈Q By (1) , (2) lim g(x) does not exist at every c ̸= 0 x→c ⇒ g(x) is discontinuous at every x ̸= 0 65. Is there a number that is exactly 1 more than its cube? < pf > Yes, Define f (x) = x − (1 + x3 ) f (−1) = −1 ∧ f (−2) = 5 ∵ f (x) is continuous at everywhere ∴By I.V.T ∀ t ∈ (−1, 5) ∃c ∈ (−2, −1) s.t f (c) = t i.e ∃ c ∈ (−2, −1) s.t f (c) = 0 ⇒ c − (1 + c3 ) = 0 ⇒ c = 1 + c3 66. If a and b are positive numbers, prove that the equation x3 a b + 3 =0 2 + 2x − 1 x + x − 2 has at least one solution in the interval (−1, 1). < pf > x3 + 2x2 − 1 = (x + 1)(x2 + x − 1) ∧ x3 + x − 2 = (x − 1)(x2 + x + 2) a b Define f (x) = x3 +2x 2 −1 + x3 +x−2 b b lim f (x) = a2 + lim x3 +x−2 ⇒ lim x3 +x−2 = −∞ x→1− x→1− lim √ x→( 5−1 + ) 2 f (x) = lim √ x→( 5−1 + ) 2 x→1− a x3 +2x2 −1 + b √ 3 5−9 2 ⇒ lim √ x→( 5−1 + ) 2 Given ε > 0 ∃ x1 , x2 √ ∈ (−1, 1) s.t 1 − x1 < ε ∧ x2 − 5−1 <ε 2 satisfy f (x1 ) < 0 ∧ f (x2 ) > 0 ⇒ f (x) has at least one solution in the interval (−1, 1). 3 a x3 +2x2 −1 =∞
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