Range of a function


\color{blue}\bf{Question} : Find the range of the function f given by f(x)=\sqrt{x}+\sqrt{1-x}.

Solution: The domain of the given function is closed interval [0,1]. Since this function  f is continuous on the  closed interval [0,1] , therefore it attains its bound (the greatest and least value), somewhere in the closed interval.

By Fermat’s theorem,  if a function attains its minimum or maximum at an interior point then the derivative at that point should either vanish or should be non-existant.

Now f'(x)=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{1-x}}.  Equating f'(x) to zero, we get x=\frac12.
Thus f(0)=1,f(1)=1,f(\frac12)=\frac{1}{\sqrt2} , and therefore the range of the function is closed interval [1,\sqrt2].

Leave a Reply

Your email address will not be published. Required fields are marked *