Dwayne Fritz

Individual Assessment Coversheet To be attached to the front of the assessment. Campus: EAST LONDON Faculty: INFORMATION TECHNOLOGY Group: Intake 1 Lecturer’s Name: Mr. Nnachi Student Full Name: Dwayne Fritz Student Number: Indicate PE. 2023.Z7Y6M4 Yes No Plagiarism report attached Declaration: I declare that this assessment is my own original work except for source material explicitly acknowledged. I also declare that this assessment or any other of my original work related to it has not been previously, or is not being simultaneously, submitted for this or any other course. I also acknowledge that I am aware of the Institution’s policy and regulations on honesty in academic work as set out in the Eduvos Conditions of Enrolment, and of the disciplinary guidelines applicable to breaches of such policy and regulations. Signature D.F Date 2023/11/11 Lecturer’s Comments: Marks Awarded: Signature % Date Eduvos (Pty) Ltd. (formerly Pearson Institute of Higher Education) is registered with the Department of Higher Education and Training as a private higher education institution under the Higher Education Act, 101, of 1997. Registration Certificate number: 2001/HE07/008 Contents Question 1 ................................................................................................................................................................... 3 Question 2 ................................................................................................................................................................... 4 Question 3 ................................................................................................................................................................... 6 Question 4 ................................................................................................................................................................... 6 Question 5 ................................................................................................................................................................... 8 Question 1 a. 2(𝑘 2 −1)+𝑘 𝑘+(𝑘 2 −1) 2𝑘 2 − 2 + 𝑘 𝑘(𝑘 2 − 1) 2𝑘 2 + 𝑘 − 2 = 𝑘3 − 𝑘 b. 2(𝑘 2 −1)−𝑘 𝑘−(𝑘 2 −1) 2𝑘 2 − 2 − 𝑘 𝑘(𝑘 2 − 1) 2𝑘 2 − 𝑘 − 2 = 𝑘3 − 𝑘 c. 2 𝑥(𝑥 2 −1) 2 =𝑥 3 −𝑥 d. 2(𝑥 2 −1) 𝑥 = 2𝑥 2 −2 𝑥 Question 2 a. 𝑓 (0) = 0.5 (0,0.5) b. lim 𝑓(𝑥) = 0.5 𝑥→−2 c. 𝑓(−2) = 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 d. lim 𝑓(𝑥 ) = 4 𝑥→0 e. 𝑓(2) = 1 (2,1) f. lim 𝑓(𝑥 ) = 0.5 𝑥→2 g. 𝑓(4) = 2 (4,2) h. lim 𝑓(𝑥 ) = 2 𝑥→4 2.1 lim √2+𝑥−√2 𝑥 𝑥→0 𝑑 (√2+𝑥−√2 𝑑𝑥 𝑑 𝑥→0 (𝑥) 𝑑𝑥 = lim = lim ( 1 2√2+𝑥 1 𝑥→0 =lim (2 𝑥→0 =2 1 √2+𝑥 ) ) 1 √2+0 √2 =4 2.3 𝑓 (𝑥 ) = 1 5𝑥 2 −3 𝑑 1 =𝑓 ′ (𝑥 ) = 𝑑𝑥 (5𝑥 2 −3) =𝑓 ′ (𝑥 ) = − 𝑑 (5𝑥 2 −3) 𝑑𝑥 (5𝑥 2 −3)2 =𝑓 ′ (𝑥 ) = − 𝑑 𝑑 (5𝑥 2 )− (3) 𝑑𝑥 𝑑𝑥 2 (5𝑥 −3)2 5×2𝑥−0 =𝑓 ′ (𝑥 ) = − (5𝑥 2 −3)2 10𝑥 =𝑓 ′ (𝑥 ) = − (5𝑥 2 −3)2 Question 3 𝐷 = 𝑅4.10 + 𝑅0,12 = = 𝑅4,23 1500𝑙𝑖𝑡𝑟𝑒𝑠 1,2𝐿𝑖𝑡𝑟𝑒𝑠 = 1250𝑙𝑖𝑡𝑟𝑒𝑠 𝐷 = 𝑅4.22 + 𝑅0,12 = 𝑅4.34 = 1250𝑙𝑖𝑡𝑟𝑒𝑠 1,2𝑙𝑖𝑡𝑟𝑒𝑠 Approx. litres sold per day= 1042𝑙𝑖𝑡𝑟𝑒𝑠 Question 4 a. 𝑔(𝑥 ) = 3𝑥−1 √𝑥 3𝑥−1 =𝑔′ (𝑥 ) = ( ′( =𝑔 𝑥 ) = √𝑥 ) (3𝑥−1)×√𝑥−(3𝑥−1)×(√𝑥 √𝑥 1 3 =𝑔′ (𝑥 ) = 2 √𝑥−(3𝑥−1)× 2 √𝑥 √𝑥 3𝑥+1 =𝑔′ (𝑥 ) = 2𝑥 √𝑥 2 2 b. 𝑓 (𝑥 ) = 𝑥 4 × (1 − 𝑥+1) 2 =𝑓 ′ (𝑥 ) = (𝑥 4 × (1 − 𝑥+1)) =𝑓 ′ (𝑥 ) = (𝑥 4 × 𝑥+1−2 𝑥+1 ) 𝑥−1 =𝑓 ′ (𝑥 ) = (𝑥 4 × 𝑥+1) 𝑥 4 ×(𝑥−1) ′( =𝑓 𝑥 ) = ( 𝑥 5 −𝑥 4 ′( =𝑓 𝑥 ) = ( ′( =𝑓 𝑥 ) = =𝑓 ′ (𝑥 ) = ) 𝑥+1 𝑥+1 ) (5𝑥 4 −4𝑥 3 )×(𝑥+1)−(𝑥 5 −𝑥 4 )×1 (𝑥+1)2 4𝑥 5 +2𝑥 4 −4𝑥 3 (𝑥+1)2 3𝑥 2 −2 −2 c. ℎ(𝑥 ) = ( 2𝑥+3 ) ′( 3𝑥 2 −2 −2 =ℎ 𝑥 ) = (( 2𝑥+3 ) ) ′( 3𝑥 2 −2 =ℎ 𝑥 ) = (𝑔−2 ) × ( 2𝑥+3 ) =ℎ′ (𝑥 ) = −2𝑔−3 × ′( 3×2𝑥×(2𝑥+3)−(3𝑥 2 −2)×2 (2𝑥+3)2 3𝑥 2 −2 =ℎ 𝑥 ) = −2 × ( 2𝑥+3 ) ′( =ℎ 𝑥 ) = −3 × 3×2𝑥×(2𝑥+3)−(3𝑥 2 −2)×2 (2𝑥+3)2 24𝑥 3 +108𝑥 2 +124𝑥+24 − (3𝑥 2 −2)3 Question 5 a. 𝐶 (𝑥 ) = 𝑥(𝑅6,67) = 𝐶 ′ (𝑥) = 1200 × (𝑅6,67) = 𝐶 ′ (𝑥) = 𝑅8000 b. 𝑅9000 𝑅6,67 estimated televisions sold = 1349