Find the approximate value of f (5.001), where f (x) = x3 –7

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389.

Find the approximate value of f (5.001), where f (x) = x3 –7x2 + 15.


 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

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