Question
If f (x) = 3x2 + 15x + 5, then find the approximate value of f (3.02) is
-
47.66
-
57.66
-
67.66
-
77.66
Solution
D.
77.66
f (x) = 3 x2 + 15 x + 5
Let x = 3 and ∆x = 0.02
∆y = f (x + ∆x) – f (x)
∴ f (x + ∆x) = f (x) + ∆y = f (x) + f ' (x) ∆x [∵ dx = ∆x]
⇒ f (x + ∆x) = f (x) + (6 x + 15) ∆x
∴ f (3.02) = f (3)+ {6 (3) + 15} (0.02)
= [3 (3)2 + 15 (3) + 5]+ (18 + 15) (0.02)1111
= (27+ 45 + 5) + 33 (0.02)
= 77 + 0.66 = 77.66
∴ approximate value of f (3.02) is 77.66.
