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Q.
The greatest value of $ f(x)={{(x+1)}^{1/3}}-{{(x-1)}^{1/3}} $ on [0, 1] is:
JamiaJamia 2005
Solution:
We have $ f(x)={{(x+1)}^{1/3}}-{{(x-1)}^{1/3}} $ $ \therefore $ $ f(x)=\frac{1}{3}\left[ \frac{1}{{{(x+1)}^{2/3}}}-\frac{1}{{{(x-1)}^{2/3}}} \right] $ $ =\frac{{{(x-1)}^{2/3}}-{{(x+1)}^{2/3}}}{3{{({{x}^{2}}-1)}^{2/3}}} $ Clearly, $ f(x) $ does not exists at $ x=\pm 1 $ Now, $ f(x)\ne 0 $ then $ {{(x-1)}^{2/3}}={{(x+1)}^{2/3}} $ $ \Rightarrow $ $ x=0 $ Clearly $ f(x)\ne 0 $ for any other value of $ x\in [0,1] $ . The value of $ f(x) $ at $ x=0 $ is 2. Hence, the greatest value of $ f(x) $ is 2