Let f(x)=x1/3 ⇒f′(x)=31x−2/3
Now f(x+Δx)−f(x)=f′(x)⋅Δx=3(x2/3)Δx
We may write, 0.007=0.008−0.001,
taking. x=0.008 and dx=−0.001
we have f(0.007)−f(0.008)=−3(0.008)2/30.001 ⇒f(0.007)−(0.008)1/3=−3(0.2)20.001 ⇒f(0.007)=0.2−3(0.04)0.001 =0.2−1201=12023
Hence (0.007)1/3=12023