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Question Number 85176 by jagoll last updated on 19/Mar/20

$$\mathrm{what}\mathrm{is}\mathrm{range}$$$$\mathrm{function}\mathrm{y}=\sqrt{\mathrm{x}-1}+\sqrt{5-\mathrm{x}}$$

Answered by john santu last updated on 19/Mar/20

$$\mathrm{Domain}:\mathrm{x}⩾1\wedge \mathrm{x}⩽5\Rightarrow 1⩽\mathrm{x}⩽5$$$$\mathrm{Range}:\mathrm{y}{}^{\prime }=\frac{1}{2\sqrt{\mathrm{x}-1}}-\frac{1}{2\sqrt{5-\mathrm{x}}}=0$$$$\sqrt{\mathrm{x}-1}=\sqrt{5-\mathrm{x}}\Rightarrow \mathrm{x}-1=5-\mathrm{x}$$$$2\mathrm{x}=3\Rightarrow \mathrm{x}=3.$$$${\mathrm{y}}_{\max }=\sqrt{2}+\sqrt{2}=2\sqrt{2}$$$$\mathrm{for}\mathrm{x}=1\Rightarrow \mathrm{y}=\sqrt{5-1}=2$$$$\mathrm{for}\mathrm{x}=5\Rightarrow \mathrm{y}=\sqrt{5-1}=2$$$$\mathrm{range}:2⩽\mathrm{y}⩽2\sqrt{2}$$

Commented by jagoll last updated on 19/Mar/20

$$\mathrm{thanks}\mathrm{you}$$

Answered by mr W last updated on 19/Mar/20

$$y=\sqrt{x-1}+\sqrt{5-x}⩾0$$$$y^2=4+2\sqrt{(x-1)(5-x})⩾4$$$$\Rightarrow y⩾2...\left(i\right)$$$$\frac{y^2}{2}-2=\sqrt{(x-1)(5-x)}$$$${(\frac{y^2}{2}-2)}^2=(x-1)(5-x)$$$$x^2+6x+5+{(\frac{y^2}{2}-2)}^2=0$$$$D=6^2-4(5+{(\frac{y^2}{2}-2)}^2)⩾0$$$$4-{(\frac{y^2}{2}-2)}^2⩾0$$$$-2⩽\frac{y^2}{2}-2⩽2$$$$0⩽\frac{y^2}{2}⩽4$$$$\Rightarrow y⩽2\sqrt{2}...\left(ii\right)$$$$$$$$from\left(i\right)and\left(ii\right)wegetrange:$$$$2⩽y⩽2\sqrt{2}$$

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