\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].