DSC1520 ASSIGNMENT 2
SEMESTER 2 2025
ANSWER
𝑑𝑦
𝑆𝑙𝑜𝑝𝑒 = 𝑑𝑥
𝑑
𝑑𝑥
(3𝑥 3 + 4𝑥 2 − 7𝑥 + 9)
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑟𝑢𝑙𝑒: (𝑓 ± 𝑔)′ = 𝑓′ ± 𝑔′
𝑑 𝑑 𝑑 𝑑
𝑑𝑥
(3𝑥 3 ) + 𝑑𝑥 (4𝑥 2 ) − 𝑑𝑥 (7𝑥 ) + 𝑑𝑥 (9)
𝑑
𝑑𝑥
(3𝑥 2 ) = 9𝑥 2
𝑑
𝑑𝑥
(4𝑥 2 ) = 8𝑥
𝑑
𝑑𝑥
(7𝑥 ) = 7
𝑑
𝑑𝑥
(9) = 0
9𝑥 2 + 8𝑥 − 7 + 0
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦
9𝑥2 + 8𝑥 − 7
𝐹𝑜𝑟 9𝑥2 + 8𝑥 − 7 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 𝑤𝑖𝑡ℎ 9
= 9 × 92 + 8 × 9 − 7
,= 9 × 92 + 8 × 9 − 7 = 794
= 794
∴ 𝑂𝑃𝑇𝐼𝑂𝑁 𝐴
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ANSWER
𝑑
𝑑𝑥
(ξ 𝑟 3 + 9𝑡 2 + 16)
1 𝑑
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑐ℎ𝑎𝑖𝑛 𝑟𝑢𝑙𝑒: 2ξ 𝑡 2 +9𝑡 2 +16 𝑑𝑡
(𝑡 3 + 9𝑡 2 + 16)
1 𝑑
= (𝑡 3 + 9𝑡 2 + 16)
2ξ 𝑡 3 +9𝑡 2 +16 𝑑𝑡
𝑑
𝑑𝑡
(𝑡 3 + 9𝑡 2 + 18𝑡
1
= 2ξ 𝑡 3 +9𝑡 2 +16 (3𝑡 2 + 18𝑡)
1 3𝑡 2 +18 𝑡
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦 (3𝑡 2 + 18𝑡) :
2ξ 𝑡 3 +9𝑡 2 +16 2ξ 𝑡 3 +9𝑡 2 +16
3𝑡 2 +18 𝑡
= 2ξ 𝑡 3 +9𝑡 2 +16
∴ 𝑂𝑃𝑇𝐼𝑂𝑁 𝐴
, ANSWER
𝑀𝑜𝑛𝑜𝑡𝑜𝑛𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛
𝐼𝑓 𝑓 ′ (𝑥) > 0 𝑡ℎ𝑒𝑛 𝑓(𝑥 ) 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝐼𝑓 𝑓 ′ (𝑥) < 0 𝑡ℎ𝑒𝑛 𝑓(𝑥 ) 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝑓 ′ (𝑝) = −3𝑝2 + 18𝑝 − 24
𝐹𝑖𝑛𝑑 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠: 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: − ∞ < 𝑝 < 2, 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 2 < 𝑝 < 4, 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 4 < 𝑝 < ∞
𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: −∞ < 𝑝 < 2, 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 2 < 𝑝 < 4, 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 4 < 𝑝 < ∞
∴ 𝑂𝑃𝑇𝐼𝑂𝑁 𝐷
SEMESTER 2 2025
ANSWER
𝑑𝑦
𝑆𝑙𝑜𝑝𝑒 = 𝑑𝑥
𝑑
𝑑𝑥
(3𝑥 3 + 4𝑥 2 − 7𝑥 + 9)
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑟𝑢𝑙𝑒: (𝑓 ± 𝑔)′ = 𝑓′ ± 𝑔′
𝑑 𝑑 𝑑 𝑑
𝑑𝑥
(3𝑥 3 ) + 𝑑𝑥 (4𝑥 2 ) − 𝑑𝑥 (7𝑥 ) + 𝑑𝑥 (9)
𝑑
𝑑𝑥
(3𝑥 2 ) = 9𝑥 2
𝑑
𝑑𝑥
(4𝑥 2 ) = 8𝑥
𝑑
𝑑𝑥
(7𝑥 ) = 7
𝑑
𝑑𝑥
(9) = 0
9𝑥 2 + 8𝑥 − 7 + 0
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦
9𝑥2 + 8𝑥 − 7
𝐹𝑜𝑟 9𝑥2 + 8𝑥 − 7 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 𝑤𝑖𝑡ℎ 9
= 9 × 92 + 8 × 9 − 7
,= 9 × 92 + 8 × 9 − 7 = 794
= 794
∴ 𝑂𝑃𝑇𝐼𝑂𝑁 𝐴
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ANSWER
𝑑
𝑑𝑥
(ξ 𝑟 3 + 9𝑡 2 + 16)
1 𝑑
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑐ℎ𝑎𝑖𝑛 𝑟𝑢𝑙𝑒: 2ξ 𝑡 2 +9𝑡 2 +16 𝑑𝑡
(𝑡 3 + 9𝑡 2 + 16)
1 𝑑
= (𝑡 3 + 9𝑡 2 + 16)
2ξ 𝑡 3 +9𝑡 2 +16 𝑑𝑡
𝑑
𝑑𝑡
(𝑡 3 + 9𝑡 2 + 18𝑡
1
= 2ξ 𝑡 3 +9𝑡 2 +16 (3𝑡 2 + 18𝑡)
1 3𝑡 2 +18 𝑡
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦 (3𝑡 2 + 18𝑡) :
2ξ 𝑡 3 +9𝑡 2 +16 2ξ 𝑡 3 +9𝑡 2 +16
3𝑡 2 +18 𝑡
= 2ξ 𝑡 3 +9𝑡 2 +16
∴ 𝑂𝑃𝑇𝐼𝑂𝑁 𝐴
, ANSWER
𝑀𝑜𝑛𝑜𝑡𝑜𝑛𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛
𝐼𝑓 𝑓 ′ (𝑥) > 0 𝑡ℎ𝑒𝑛 𝑓(𝑥 ) 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝐼𝑓 𝑓 ′ (𝑥) < 0 𝑡ℎ𝑒𝑛 𝑓(𝑥 ) 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝑓 ′ (𝑝) = −3𝑝2 + 18𝑝 − 24
𝐹𝑖𝑛𝑑 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠: 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: − ∞ < 𝑝 < 2, 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 2 < 𝑝 < 4, 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 4 < 𝑝 < ∞
𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: −∞ < 𝑝 < 2, 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 2 < 𝑝 < 4, 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔: 4 < 𝑝 < ∞
∴ 𝑂𝑃𝑇𝐼𝑂𝑁 𝐷