Answer :

Let us assume that


Also, let x = 32 so that x + Δx = 33


32 + Δx = 33


Δx = 1


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



We know





When x = 32, 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 = 1


Δf = (0.0125)(1)


Δf = 0.0125


Now, we have f(33) = f(32) + Δf




f(33) = 2 + 0.0125


f(33) = 2.0125


Thus, (33)1/5 ≈ 2.0125


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