Dr. Emily Chan Page 1 GE1359 Problem Set 1 Solution Summer 2022 1. 𝐴 , 𝐵 , 𝐶 and 𝐷 are equal, since they only contain the elements 1, 2, 3. 𝐸 is not equal to 𝐴 , 𝐵 , 𝐶 or 𝐷 , since 4 ∈ 𝐸 but 4 is not an element of 𝐴 , 𝐵 , 𝐶 , 𝐷 2. 𝑌 ⊆ 𝐴 & 𝑌 ⊆ 𝐵 ⇒ 𝑌 ⊆ 𝐴 ∩ 𝐵 = {𝑑, 𝑒, 𝑓} 𝑌 ⊈ 𝐶 , so there is at least one element in 𝑌 which does not belong to 𝐶 Only the element 𝑒 in the set {𝑑, 𝑒, 𝑓} is not in 𝐶 , so 𝑌 must contain the element 𝑒 ∴ 𝑌 could be {𝑒} , {𝑑, 𝑒} , {𝑒, 𝑓} or {𝑑, 𝑒, 𝑓} ∴ Among the sets 𝐴 , 𝐵 , 𝐶 , 𝐷 , 𝐸 , 𝐹 , only 𝐷 and 𝐸 can equal 𝑌 3. (a) 𝐴 ∩ 𝐵 = {𝑏, 𝑑} , 𝐴 ∩ 𝐶 = {𝑐, 𝑑} , 𝐵 ∩ 𝐶 = {𝑑, 𝑓} (b) 𝐴 ∪ 𝐵 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑓, ℎ} , 𝐴 ∪ 𝐶 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓} , 𝐵 ∪ 𝐶 = {𝑏, 𝑐, 𝑑, 𝑒, 𝑓, ℎ} (c) 𝐴\𝐵 = {𝑎, 𝑐} , 𝐵\𝐴 = {𝑓, ℎ} , 𝐵\𝐶 = {𝑏, ℎ} , 𝐶\𝐵 = {𝑐, 𝑒} , 𝐴\𝐶 = {𝑎, 𝑏} , 𝐶\𝐴 = {𝑒, 𝑓} 4. (a) 𝐴 ∪ 𝐵 = [1, 4] (b) 𝐴 ∩ 𝐵 = (2, 3] (c) (ℝ\𝐴) ∩ 𝐵 = (3, 4] (d) (ℝ\𝐵) ∩ 𝐴 = [1, 2] (e) (ℝ\𝐴) ∩ (ℝ\𝐵) = (−∞, 1) ∪ (4, ∞) (f) (ℝ\𝐴) ∪ (ℝ\𝐵) = (−∞, 2] ∪ (3, ∞) (g) 𝐵 ∪ [𝐴 ∩ (ℝ\𝐵)] = [1, 4] (h) [(ℝ\𝐴) ∩ 𝐵] ∪ [(ℝ\𝐵) ∩ 𝐴] = (3, 4] ∪ [1, 2] 5. 𝑓(𝑥) = 𝑥 3 + 2 , 𝐷𝑜𝑚(𝑓) = ℝ 𝑔(𝑥) = 2 𝑥−1 , 𝐷𝑜𝑚(𝑔) = ℝ\{1} ℎ(𝑥) = √𝑥 , 𝐷𝑜𝑚(ℎ) = [0, ∞) (a) (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) = 𝑥 3 + 2 + 2 𝑥−1 𝐷𝑜𝑚(𝑓 + 𝑔) = ℝ\{1} (b) ( 𝑔 𝑓 ) (𝑥) = 𝑔(𝑥) 𝑓(𝑥) = 2 𝑥−1 𝑥 3 +2 = 2 (𝑥−1)(𝑥 3 +2) Dr. Emily Chan Page 2 𝑔 𝑓 is defined when (𝑥 − 1)(𝑥 3 + 2) ≠ 0 ⇒ 𝑥 − 1 ≠ 0 and 𝑥 3 + 2 ≠ 0 ⇒ 𝑥 ≠ 1 and 𝑥 ≠ √−2 3 = − √2 3 𝐷𝑜𝑚 ( 𝑔 𝑓 ) = ℝ\{− √2 3 , 1} (c) (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(𝑥 3 + 2) = 2 (𝑥 3 +2)−1 = 2 𝑥 3 +1 which is only defined when 𝑥 3 + 1 ≠ 0 , i.e. when 𝑥 ≠ √−1 3 = −1 𝐷𝑜𝑚(𝑔 ∘ 𝑓) = ℝ\{−1} (d) (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = 𝑓 ( 2 𝑥−1 ) = ( 2 𝑥−1 ) 3 + 2 𝐷𝑜𝑚(𝑓 ∘ 𝑔) = ℝ\{1} (e) (𝑓 ∘ 𝑔 ∘ ℎ)(𝑥) = 𝑓(𝑔(ℎ(𝑥))) = 𝑓 (𝑔(√𝑥 )) = 𝑓 ( 2 √𝑥 −1 ) = ( 2 √𝑥 −1 ) 3 + 2 which is only defined when √𝑥 − 1 ≠ 0 and 𝑥 ≥ 0 , i.e. when √𝑥 ≠ 1 and 𝑥 ≥ 0 ⇒ 𝑥 ≠ 1 and 𝑥 ≥ 0 𝐷𝑜𝑚(𝑓 ∘ 𝑔 ∘ ℎ) = [0, ∞)\{1} or written as [0, 1) ∪ (1, ∞) 6. 𝐹(𝑥) = 3 2+𝑥 , 𝐺(𝑥) = 1 1+ 1 𝑥 (a) 𝐹(𝑥) is defined when 2 + 𝑥 ≠ 0 ⇒ 𝑥 ≠ −2 𝐷𝑜𝑚(𝐹) = ℝ\{−2} 𝐺(𝑥) is defined when 𝑥 ≠ 0 and 1 + 1 𝑥 ≠ 0 ⇒ 𝑥 ≠ 0 and 1 𝑥 ≠ −1 ⇒ 𝑥 ≠ 0 and 𝑥 ≠ −1 𝐷𝑜𝑚(𝐺) = ℝ\{−1, 0} (b) (𝐹 ∘ 𝐺)(𝑥) = 𝐹(𝐺(𝑥)) = 𝐹 ( 1 1+ 1 𝑥 ) = 3 2+ 1 1+1 𝑥 which is defined when 𝑥 ∈ 𝐷𝑜𝑚(𝐺) and 2 + 1 1+ 1 𝑥 ≠ 0 ⇒ 𝑥 ∈ ℝ\{−1, 0} and 1 1+ 1 𝑥 ≠ −2 ⇒ 𝑥 ∈ ℝ\{−1, 0} and 1 + 1 𝑥 ≠ 1 −2 ⇒ 𝑥 ∈ ℝ\{−1, 0} and 1 𝑥 ≠ − 3 2 ⇒ 𝑥 ∈ ℝ\{−1, 0} and 𝑥 ≠ − 2 3 Dr. Emily Chan Page 3 𝐷𝑜𝑚(𝐹 ∘ 𝐺) = ℝ\ {−1, − 2 3 , 0} 7. (a) 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 3 = (𝑥 − 1) 2 − 4 𝐷𝑜𝑚(𝑓) = ℝ For every 𝑥 ∈ 𝐷𝑜𝑚(𝑓) = ℝ , (𝑥 − 1) 2 ≥ 0 ⇒ (𝑥 − 1) 2 − 4 ≥ −4 𝑅𝑎𝑛(𝑓) = [−4, ∞) (b) 𝑔(𝑥) = 𝑥−3 𝑥+2 = (𝑥+2)−5 𝑥+2 = 1 − 5 𝑥+2 𝑔 is defined when 𝑥 + 2 ≠ 0 ⇒ 𝑥 ≠ −2 𝐷𝑜𝑚(𝑔) = ℝ\{−2} For every 𝑥 ∈ 𝐷𝑜𝑚(𝑔) = ℝ\{−2} , − 5 𝑥−2 can be any real number except 0, thus 1 − 5 𝑥+2 ≠ 1 + 0 = 1 𝑅𝑎𝑛(𝑔) = ℝ\{1} (c) ℎ(𝑥) = √𝑥 2 + 4 ℎ is defined when 𝑥 2 + 4 ≥ 0 , which is always true for all real values of 𝑥 𝐷𝑜𝑚(ℎ) = ℝ For every 𝑥 ∈ 𝐷𝑜𝑚(ℎ) = ℝ , 𝑥 2 ≥ 0 ⇒ 𝑥 2 + 4 ≥ 0 + 4 = 4 ⇒ √𝑥 2 + 4 ≥ √4 = 2 𝑅𝑎𝑛(ℎ) = [2, ∞) 8. (a) 𝑓(𝑥) = 5 − 4𝑥 − 𝑥 2 𝐷𝑜𝑚(𝑓) = ℝ (b) 𝑔(𝑥) = 1 5−4𝑥−𝑥 2 𝑔 is defined when 5 − 4𝑥 − 𝑥 2 ≠ 0 ⇒ (1 − 𝑥)(5 + 𝑥) ≠ 0 ⇒ 𝑥 ≠ 1 and 𝑥 ≠ −5 𝐷𝑜𝑚(𝑔) = ℝ\{−5, 1} (c) ℎ(𝑥) = √5 − 4𝑥 − 𝑥 2 ℎ is defined when 5 − 4𝑥 − 𝑥 2 ≥ 0 ⇒ (1 − 𝑥)(5 + 𝑥) ≥ 0 𝑥 < −5 𝑥 = −5 −5 < 𝑥 < 1 𝑥 = 1 𝑥 > 1 Sign of (1 − 𝑥) + + + 0 − Sign of (5 + 𝑥) − 0 + + + Sign of (1 − 𝑥)(5 + 𝑥) − 0 + 0 − 𝐷𝑜𝑚(ℎ) = [−5, 1] Dr. Emily Chan Page 4 9. (a) If 𝑓(𝑥) is an odd function and 𝑔(𝑥) is an even function, then 𝑓(−𝑥) = −𝑓(𝑥) and 𝑔(−𝑥) = 𝑔(𝑥) for all 𝑥 ∈ 𝐷𝑜𝑚(𝑓) ∩ 𝐷𝑜𝑚(𝑔) (𝑓𝑔)(−𝑥) = 𝑓(−𝑥)𝑔(−𝑥) = −𝑓(𝑥)𝑔(𝑥) = −(𝑓𝑔)(𝑥) ∴ (𝑓𝑔)(𝑥) is an odd function. (b) If 𝑓(𝑥) and 𝑔(𝑥) are both odd functions, then 𝑓(−𝑥) = −𝑓(𝑥) and 𝑔(−𝑥) = −𝑔(𝑥) for all 𝑥 ∈ 𝐷𝑜𝑚(𝑓) ∩ 𝐷𝑜𝑚(𝑔) (𝑓𝑔)(−𝑥) = 𝑓(−𝑥)𝑔(−𝑥) = [−𝑓(𝑥)][−𝑔(𝑥)] = 𝑓(𝑥)𝑔(𝑥) = (𝑓𝑔)(𝑥) ∴ (𝑓𝑔)(𝑥) is an even function. (c) If 𝑓(𝑥) and 𝑔(𝑥) are both even functions, then 𝑓(−𝑥) = 𝑓(𝑥) and 𝑔(−𝑥) = 𝑔(𝑥) for all 𝑥 ∈ 𝐷𝑜𝑚(𝑓) ∩ 𝐷𝑜𝑚(𝑔) (𝑓𝑔)(−𝑥) = 𝑓(−𝑥)𝑔(−𝑥) = 𝑓(𝑥)𝑔(𝑥) = (𝑓𝑔)(𝑥) ∴ (𝑓𝑔)(𝑥) is an even function. 10. (a) 𝑓 1 (𝑥) = sin(𝑥 3 +𝑥) 𝑥 4 +3 𝑓 1 (−𝑥) = sin[(−𝑥) 3 + (−𝑥)] (−𝑥) 4 + 3 = sin(−𝑥 3 − 𝑥) 𝑥 4 + 3 = − sin(𝑥 3 + 𝑥) 𝑥 4 + 3 = −𝑓 1 (𝑥) ∴ 𝑓 1 (𝑥) is an odd function. (b) 𝑓 2 (𝑥) = |𝑥 5 + 1| 𝑓 2 (−𝑥) = |(−𝑥) 5 + 1| = |−𝑥 5 + 1| ≠ 𝑓 2 (𝑥) nor − 𝑓 2 (𝑥) ∴ 𝑓 2 (𝑥) is neither even nor odd. (c) 𝑓 3 (𝑥) = cos 3 (2𝑥) 𝑓 3 (−𝑥) = cos 3 [2(−𝑥)] = cos 3 (−2𝑥) = [cos(−2𝑥)] 3 = [cos(2𝑥)] 3 = cos 3 (2𝑥) = 𝑓 3 (𝑥) ∴ 𝑓 3 (𝑥) is an even function. 11. 𝐹(𝑥) = 𝑥 2 + 2𝑥 − 3 , 𝑥 ≥ 0 (a) 𝐹(𝑥) = 𝑥 2 + 2𝑥 − 3 = (𝑥 + 1) 2 − 4 = (𝑥 + 3)(𝑥 − 1) 0 1 −3 𝑥 𝑦 Dr. Emily Chan Page 5 By the Horizontal Line Test, no horizontal intersects the graph more than once, therefore 𝐹(𝑥) is one-to-one. 𝑅𝑎𝑛(𝐹) = [−3, ∞) (b) Since 𝐹(𝑥) is one-to-one, its inverse exists. Let 𝑦 = 𝑥 2 + 2𝑥 − 3 , 𝑥 ≥ 0 Then 𝑦 = (𝑥 + 1) 2 − 4 ⇒ 𝑦 + 4 = (𝑥 + 1) 2 ⇒ 𝑥 + 1 = √𝑦 + 4 or 𝑥 + 1 = −√𝑦 + 4 (rejected since 𝑥 ≥ 0 so that 𝑥 + 1 ≥ 1 > 0 ) ⇒ 𝑥 = √𝑦 + 4 − 1 ∴ 𝐹 −1 (𝑥) = √𝑥 + 4 − 1 𝐷𝑜𝑚(𝐹 −1 ) = 𝑅𝑎𝑛(𝐹) = [−3, ∞) 𝑅𝑎𝑛(𝐹 −1 ) = 𝐷𝑜𝑚(𝐹) = [0, ∞) 12. (a) 𝑝(𝑥) = 3𝑥 − 2 , 𝑥 ∈ ℝ Let 𝑝(𝑥 1 ) = 𝑝(𝑥 2 ) . Then 3𝑥 1 − 2 = 3𝑥 2 − 2 ⇒ 3𝑥 1 = 3𝑥 2 ⇒ 𝑥 1 = 𝑥 2 is the only solution. Hence, 𝑝(𝑥) is one-to-one and so its inverse exists. Let 𝑦 = 3𝑥 − 2 . Then 𝑦 + 2 = 3𝑥 ⇒ 𝑥 = 𝑦+2 3 ∴ 𝑝 −1 (𝑥) = 𝑥+2 3 For every 𝑥 ∈ 𝐷𝑜𝑚(𝑝) = ℝ , 3𝑥 − 2 can be any real number, i.e. 𝑅𝑎𝑛(𝑝) = ℝ ∴ 𝐷𝑜𝑚(𝑝 −1 ) = 𝑅𝑎𝑛(𝑝) = ℝ (b) 𝑞(𝑥) = 𝑥 2 − 1 , 𝑥 ≥ 0 Let 𝑞(𝑥 1 ) = 𝑞(𝑥 2 ) Then 𝑥 1 2 − 1 = 𝑥 2 2 − 1 ⇒ 𝑥 1 2 = 𝑥 2 2 ⇒ 𝑥 1 = 𝑥 2 or 𝑥 1 = −𝑥 2 (rejected since 𝑥 ≥ 0 ) ⇒ 𝑥 1 = 𝑥 2 is the only solution. Hence, 𝑞(𝑥) is one-to-one and so its inverse exists. Let 𝑦 = 𝑥 2 − 1 , 𝑥 ≥ 0 Then 𝑦 + 1 = 𝑥 2 ⇒ 𝑥 = √𝑦 + 1 or 𝑥 = −√𝑦 + 1 (rejected since 𝑥 ≥ 0 ) ∴ 𝑞 −1 (𝑥) = √𝑥 + 1 Dr. Emily Chan Page 6 For every 𝑥 ∈ 𝐷𝑜𝑚(𝑞) = [0, ∞) , 𝑥 2 ≥ 0 ⇒ 𝑥 2 − 1 ≥ −1 , i.e. 𝑅𝑎𝑛(𝑞) = [−1, ∞) ∴ 𝐷𝑜𝑚(𝑞 −1 ) = 𝑅𝑎𝑛(𝑞) = [−1, ∞) - End - GE1359