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Q. What is approximate value of $\sqrt[5]{242.999} ?$

Gujarat CETGujarat CET 2018

Solution:

$\sqrt[5]{242.999}$
Let $f (x)=x^{1 / 5}$
$f ( a + h )= f ( a )+ hf ^{\prime}( a ) f ^{\prime}( x )=\frac{1}{5} x ^{-\frac{4}{5}}$
$a =243, h =0.001$
$f (243-0.001)=(243)^{\frac{1}{5}}-(0.001) \frac{1}{5}(243)^{-\frac{4}{5}}$
$f (242.999)=3-\frac{1}{1000} \times \frac{1}{5} \times \frac{1}{81}$
$=3-\frac{1}{405000}=\frac{1214999}{405000}$