find-the-range-f-x-2-6-x-2-

Question Number 73649 by aliesam last updated on 14/Nov/19

$${find}\:{the}\:{range} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{6}−\sqrt{{x}+\mathrm{2}}} \\ $$

Commented by mathmax by abdo last updated on 14/Nov/19

x ∈D_f ⇔ x≥−2 and (√(x+2))≠6 ⇔x≥−2 and x+2 ≠36 ⇔x≥−2 and x≠34 ⇒D_f =[−2,34[∪]34,+∞[ f^′ (x)=2(−((−(1/(2(√(x+2)))))/((6−(√(x+2)))^2 ))) =(1/( (√(x+2))(6−(√(x+2)))^2 ))>0 ⇒f is increazing on D_(f ) we have f(−2)=(1/3) and lim_(x→34^− ) f(x) =+∞ lim_(x→34^+ ) f(x)=−∞ ⇒f([−2,34[)=[(1/3),+∞[ and f(]34,+∞[) =]−∞,0[ variation of f x −2 34 +∞ f^′ + ∣∣ + f (1/3) inc +∞∣∣−∞ inc 0

$${x}\:\in{D}_{{f}} \Leftrightarrow\:{x}\geqslant−\mathrm{2}\:\:{and}\:\sqrt{{x}+\mathrm{2}}\neq\mathrm{6}\:\Leftrightarrow{x}\geqslant−\mathrm{2}\:{and}\:{x}+\mathrm{2}\:\neq\mathrm{36}\:\Leftrightarrow{x}\geqslant−\mathrm{2} \\ $$$${and}\:{x}\neq\mathrm{34}\:\:\Rightarrow{D}_{{f}} =\left[−\mathrm{2},\mathrm{34}\left[\cup\right]\mathrm{34},+\infty\left[\right.\right. \\ $$$${f}^{'} \left({x}\right)=\mathrm{2}\left(−\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}}{\left(\mathrm{6}−\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} }\right)\:=\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{2}}\left(\mathrm{6}−\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} }>\mathrm{0}\:\Rightarrow{f}\:{is}\:{increazing} \\ $$$${on}\:{D}_{{f}\:\:} \:\:{we}\:{have}\:{f}\left(−\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{34}^{−} } {f}\left({x}\right)\:\:\:=+\infty \\ $$$${lim}_{{x}\rightarrow\mathrm{34}^{+} } \:\:{f}\left({x}\right)=−\infty\:\Rightarrow{f}\left(\left[−\mathrm{2},\mathrm{34}\left[\right)=\left[\frac{\mathrm{1}}{\mathrm{3}},+\infty\left[\:{and}\right.\right.\right.\right. \\ $$$$\left.{f}\left(\right]\mathrm{34},+\infty\left[\right)\:=\right]−\infty,\mathrm{0}\left[\:\:{variation}\:{of}\:{f}\right. \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{34}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\:\:{inc}\:+\infty\mid\mid−\infty\:\:\:\:\:\:{inc}\:\:\:\:\:\:\:\:\mathrm{0} \\ $$

Commented by malwaan last updated on 14/Nov/19

x≥−2 and x≠ 34 ⇒−2≤ x < 34 ∨ x > 34 −2≤ x < 34⇒0≤x+2<36 ⇒0≤(√(x+2)) <6⇒0≥−(√(x+2)) >−6 ⇒6≥6−(√(x+2)) >0 ⇒(1/3)≤ (2/(6−(√(x+2)))) < ∞ ⇒(1/3) ≤ y < ∞ ....(1) x > 34⇒34<x<∞ ⇒36< x+2 < ∞ ⇒6<(√(x+2)) <∞ ⇒−6>−(√(x+2)) >−∞ ⇒0>6−(√(x+2)) >−∞ ⇒−∞<(2/(6−(√(x+2)))) < 0 ⇒−∞<y<0 ....(2) from(1);(2) ⇒the range = [1 ╱ 3 ; ∞[ ∪ ]−∞ ; 0 [

$$\boldsymbol{{x}}\geqslant−\mathrm{2}\:\boldsymbol{{and}}\:\boldsymbol{{x}}\neq\:\mathrm{34} \\ $$$$\Rightarrow−\mathrm{2}\leqslant\:\boldsymbol{{x}}\:<\:\mathrm{34}\:\vee\:\boldsymbol{{x}}\:>\:\mathrm{34} \\ $$$$−\mathrm{2}\leqslant\:\boldsymbol{{x}}\:<\:\mathrm{34}\Rightarrow\mathrm{0}\leqslant\boldsymbol{{x}}+\mathrm{2}<\mathrm{36} \\ $$$$\Rightarrow\mathrm{0}\leqslant\sqrt{\boldsymbol{{x}}+\mathrm{2}}\:<\mathrm{6}\Rightarrow\mathrm{0}\geqslant−\sqrt{\boldsymbol{{x}}+\mathrm{2}}\:>−\mathrm{6} \\ $$$$\Rightarrow\mathrm{6}\geqslant\mathrm{6}−\sqrt{\boldsymbol{{x}}+\mathrm{2}}\:>\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant\:\frac{\mathrm{2}}{\mathrm{6}−\sqrt{\boldsymbol{{x}}+\mathrm{2}}}\:<\:\infty \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\:\leqslant\:\boldsymbol{{y}}\:<\:\infty\:….\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{x}}\:>\:\mathrm{34}\Rightarrow\mathrm{34}<\boldsymbol{{x}}<\infty \\ $$$$\Rightarrow\mathrm{36}<\:\boldsymbol{{x}}+\mathrm{2}\:<\:\infty \\ $$$$\Rightarrow\mathrm{6}<\sqrt{\boldsymbol{{x}}+\mathrm{2}}\:<\infty \\ $$$$\Rightarrow−\mathrm{6}>−\sqrt{\boldsymbol{{x}}+\mathrm{2}}\:>−\infty \\ $$$$\Rightarrow\mathrm{0}>\mathrm{6}−\sqrt{\boldsymbol{{x}}+\mathrm{2}}\:>−\infty \\ $$$$\Rightarrow−\infty<\frac{\mathrm{2}}{\mathrm{6}−\sqrt{\boldsymbol{{x}}+\mathrm{2}}}\:<\:\mathrm{0} \\ $$$$\Rightarrow−\infty<\boldsymbol{{y}}<\mathrm{0}\:….\left(\mathrm{2}\right) \\ $$$$\boldsymbol{{from}}\left(\mathrm{1}\right);\left(\mathrm{2}\right) \\ $$$$\Rightarrow\boldsymbol{{the}}\:\boldsymbol{{range}}\:= \\ $$$$\left[\mathrm{1}\:\diagup\:\mathrm{3}\:;\:\infty\left[\:\cup\:\right]−\infty\:;\:\mathrm{0}\:\left[\right.\right. \\ $$

Answered by MJS last updated on 14/Nov/19

defined for x≥−2∧x≠34 f(−2)=(1/3) lim_(x→34^− ) f(x)=+∞ lim_(x→34^+ ) f(x)=−∞ lim_(x→+∞) f(x)=0 ⇒ −∞<y≤0∨(1/3)≤y<+∞

$$\mathrm{defined}\:\mathrm{for}\:{x}\geqslant−\mathrm{2}\wedge{x}\neq\mathrm{34} \\ $$$${f}\left(−\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\underset{{x}\rightarrow\mathrm{34}^{−} } {\mathrm{lim}}{f}\left({x}\right)=+\infty \\ $$$$\underset{{x}\rightarrow\mathrm{34}^{+} } {\mathrm{lim}}{f}\left({x}\right)=−\infty \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow\:−\infty<{y}\leqslant\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{3}}\leqslant{y}<+\infty \\ $$

Commented by aliesam last updated on 14/Nov/19