Smallest Maximum Using For Loop Only Part2 From Question Please 2 (16.57 KiB) Viewed 11 times
Given the function f(n) defined as log ((n − 650)¹4 + sin¹6 (505)) 16 f(n)=8- 909 log (n) +9 -here n is a positive integer. . (2 Marks) Verify that f(5) = -0.53669 then find y₁ = f(70). - (2 Marks) Find the smallest integer ng such that n3 21 and f(n3) ≥ 4. Y1 = n3 = Mark
log ((n - 399)4+ sin² (579)) f(n) = 2- log (n) + 5. where n is a positive integer. • (2 Marks) Verify that f(6)= -1.5183 then find y₁ = f (60). (2 Marks) Find the maximum value of f(n) in the interval n € [1,1000].
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