Applications Of Derivatives Question 28
Question: The approximate value of $ {{(0.007)}^{1/3}} $
Options:
A) $ \frac{23}{120} $
B) $ \frac{27}{120} $
C) $ \frac{19}{120} $
D) $ \frac{17}{120} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ f(x)={x^{1/3}}\Rightarrow f’(x)=\frac{1}{3}{x^{-2/3}} $ Now $ f(x+\Delta x)-f(x)=f’(x)\cdot \Delta x=\frac{\Delta x}{3({x^{2/3}})} $ 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)=-\frac{0.001}{3{{(0.008)}^{2/3}}} $
$ \Rightarrow f(0.007)-{{(0.008)}^{1/3}}=-\frac{0.001}{3{{(0.2)}^{2}}} $
$ \Rightarrow f(0.007)=0.2-\frac{0.001}{3(0.04)}=0.2-\frac{1}{120}=\frac{23}{120} $
Hence $ {{(0.007)}^{1/3}}=\frac{23}{120} $
BETA
