Find the max and min values of the function $$y=\sqrt{1-x}+\sqrt{x-\dfrac{1}{2} } $$

Question

Find the max and min values of the function $$y=\sqrt{1-x}+\sqrt{x-\dfrac{1}{2}  }    $$

Uniteasy Admin · Accepted Answer

$$y=\sqrt{1-x}+\sqrt{x-\dfrac{1}{2} }  $$nc
The radicant is no less than 0 implies
$$1-x \geq 0$$ and $$x-\dfrac{1}{2} \geq 0 $$, that is
$$\dfrac{1}{2} \leq x \leq 1 $$nc
Square the equation (1)
 
$$\begin{aligned} y^2 &= \dfrac{1}{2}+2\sqrt{-x^2+\dfrac{3}{2}x -\dfrac{1}{2}  } \  &= \dfrac{1}{2}+2\sqrt{-(x-\dfrac{3}{4} )^2+\dfrac{1}{16} }  \end{aligned}$$c
 Therefore, $$y_{max} = 1$$, when $$x = \dfrac{3}{4} $$
Since $$\dfrac{1}{2}$$ and $$1 $$ are symmetric about $$\dfrac{3}{4}$$, therefore,  $$y_{min} = \dfrac{\sqrt{2} }{2} $$, when $$x = \dfrac{1}{2}$$ or $$1 $$

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