Question

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

Collected in the board: Function

Steven Zheng posted 2 weeks ago

Answer

y=\sqrt{1-x}+\sqrt{x-\dfrac{1}{2} }
(1)

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
(2)

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}

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

Steven Zheng posted 2 weeks ago

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