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

what is range function y = (√(x−1)) + (√(5−x))

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

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

Domain : x ≥ 1 ∧ x ≤ 5 ⇒ 1 ≤x≤5 Range : y ′ = (1/(2(√(x−1)))) − (1/(2(√(5−x)))) = 0 (√(x−1)) = (√(5−x)) ⇒ x−1 = 5−x 2x = 3 ⇒ x = 3 . y_(max ) = (√2) + (√2) = 2(√2) for x = 1 ⇒ y = (√(5−1)) = 2 for x = 5 ⇒ y =(√(5−1)) = 2 range : 2 ≤ y ≤ 2(√2)

$$\mathrm{Domain}\::\:\mathrm{x}\:\geqslant\:\mathrm{1}\:\wedge\:\mathrm{x}\:\leqslant\:\mathrm{5}\:\Rightarrow\:\mathrm{1}\:\leqslant\mathrm{x}\leqslant\mathrm{5} \\ $$$$\mathrm{Range}\::\:\mathrm{y}\:'\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}\:−\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}−\mathrm{x}}}\:=\:\mathrm{0} \\ $$$$\sqrt{\mathrm{x}−\mathrm{1}}\:=\:\sqrt{\mathrm{5}−\mathrm{x}}\:\Rightarrow\:\mathrm{x}−\mathrm{1}\:=\:\mathrm{5}−\mathrm{x} \\ $$$$\mathrm{2x}\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{x}\:=\:\mathrm{3}\:.\: \\ $$$$\mathrm{y}_{\mathrm{max}\:} \:=\:\sqrt{\mathrm{2}}\:+\:\sqrt{\mathrm{2}}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$$\mathrm{for}\:\mathrm{x}\:=\:\mathrm{1}\:\Rightarrow\:\mathrm{y}\:=\:\sqrt{\mathrm{5}−\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\mathrm{for}\:\mathrm{x}\:=\:\mathrm{5}\:\Rightarrow\:\mathrm{y}\:=\sqrt{\mathrm{5}−\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\mathrm{range}\::\:\mathrm{2}\:\leqslant\:\mathrm{y}\:\leqslant\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$

Commented by jagoll last updated on 19/Mar/20

thanks you

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

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

y=(√(x−1))+(√(5−x))≥0 y^2 =4+2(√((x−1)(5−x)))≥4 ⇒y≥2 ...(i) (y^2 /2)−2=(√((x−1)(5−x))) ((y^2 /2)−2)^2 =(x−1)(5−x) x^2 +6x+5+((y^2 /2)−2)^2 =0 D=6^2 −4(5+((y^2 /2)−2)^2 )≥0 4−((y^2 /2)−2)^2 ≥0 −2≤(y^2 /2)−2≤2 0≤(y^2 /2)≤4 ⇒y≤2(√2) ...(ii) from (i) and (ii) we get range: 2≤y≤2(√2)

$${y}=\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{5}−{x}}\geqslant\mathrm{0} \\ $$$$\left.{y}^{\mathrm{2}} =\mathrm{4}+\mathrm{2}\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{5}−{x}\right.}\right)\geqslant\mathrm{4} \\ $$$$\Rightarrow{y}\geqslant\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}=\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{5}−{x}\right)} \\ $$$$\left(\frac{{y}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} =\left({x}−\mathrm{1}\right)\left(\mathrm{5}−{x}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{5}+\left(\frac{{y}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${D}=\mathrm{6}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{5}+\left(\frac{{y}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\mathrm{4}−\left(\frac{{y}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$−\mathrm{2}\leqslant\frac{{y}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\leqslant\mathrm{2} \\ $$$$\mathrm{0}\leqslant\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\leqslant\mathrm{4} \\ $$$$\Rightarrow{y}\leqslant\mathrm{2}\sqrt{\mathrm{2}}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}\:{range}: \\ $$$$\mathrm{2}\leqslant{y}\leqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$

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