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Q. The approximate value of $ {{(0.009)}^{1/3}} $ is

J & K CETJ & K CET 2013Application of Derivatives

Solution:

Use the relation; $ f(x+\Delta x)=f(x)+\Delta xf'(x); $
to determine the approximate value. Consider,
$ f(x)={{(x)}^{1/3}}\,\,\,\Rightarrow \,\,\,\,f'(x)=\frac{1}{3}{{x}^{-2/3}} $
Let $ x=\frac{8}{1000} $ and $ \Delta x=\frac{1}{1000} $
Now, $ f'(x+\Delta x)=f(x)+\Delta xf'(x) $
$ \Rightarrow $ $ {{(x+\Delta x)}^{1/3}}={{x}^{1/3}}+\frac{1}{3{{x}^{2/3}}}\times \Delta x $
$ \Rightarrow $ $ {{\left( \frac{8}{1000}+\frac{1}{1000} \right)}^{1/3}}={{\left( \frac{8}{1000} \right)}^{1/3}}+\frac{\frac{1}{1000}}{3{{\left( \frac{8}{1000} \right)}^{1/3}}} $
$ \Rightarrow $ $ {{\left( \frac{9}{1000} \right)}^{1/3}}=\frac{2}{10}+\frac{1}{1000\times 3\times \frac{2}{10}} $
$ =\frac{1}{5}+\frac{1}{600}=\frac{121}{600} $
$ \therefore $ $ {{(0.009)}^{1/3}}=0.20166 $