Q. 9 W5.0( 1 Vote )

# Using diffe

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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