Question
Find the approximate value of f (5.001), where f (x) = x3 –7x2 + 15.
Solution
f (x) = x3 – 7 x2 + 15
Let x = 5, ∆x = 0.001
Now ∆y = f (x + ∆y) – f (x)
∴ f (x + ∆x) = f (x) + ∆x
⇒ f (x + ∆x) = f (x) + f ' (x) ∆x [∵ dx = ∆x]
∴ f (x + ∆x) = f (x)+ (3 x2 – 14 x) ∆y
∴ f (5.001) = f (5) + {3 (5)2 – 14 (5)} (0.001)
= [(5)3 – 7(5)2 +15] + {3 (25) – 70} (0.001)
= (125 – 175 + 15)+ (75 – 70) (0.001)
= (– 35) + 5 (0.001) = – 35 + 0.005
∴ f (5.001) = – 34. 995
