Q3 of 168 Page 216

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

Let x = 5 and Δx = 0.001. Then, we get,

f(5.001) = f(x + Δx) = (x +Δx)3 - 7 ( x + Δx)2 + 15


Now, Δy = f (x + Δx) – f(x)


⇒ f (x + Δx) = f(x) + Δy


≈ f(x) + f’(x).Δx (as dx = Δx)


⇒ f(5.001) ≈ (x3 – 7x2 + 15) + (3x2 – 14x) Δx


= [(5)2 – 7(5)2 + 15] +[3(5)2 -14(5)] (0.001)


= (125 – 175 + 15) + (75 - 70)(0.001)


= -35 + 0.005


= -34.995


Therefore, the approximate value of f (5.001) is -34.995.


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