Q. 115.0( 1 Vote )

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

Answer :

Given f(x) = x3 – 7x2 + 15


Let x = 5 so that x + Δx = 5.001


5 + Δx = 5.001


Δx = 0.001


On differentiating f(x) with respect to x, we get





We know and derivative of a constant is 0.




When x = 5, we have




Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as



Here, and Δx = 0.001


Δf = (5)(0.001)


Δf = 0.005


Now, we have f(5.001) = f(5) + Δf


f(5.001) = 53 – 7(5)2 + 15 + 0.005


f(5.001) = 125 – 175 + 15 + 0.005


f(5.001) = –35 + 0.005


f(5.001) = –34.995


Thus, f(5.001) = –34.995


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