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Question Number 180391 by Shrinava last updated on 11/Nov/22

y = x − (2/x)  Find the set of values of the function.

$$\mathrm{y}\:=\:\mathrm{x}\:−\:\frac{\mathrm{2}}{\mathrm{x}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}. \\ $$

Commented by Shrinava last updated on 11/Nov/22

Answer:  (−∞ ; +∞)

$$\mathrm{Answer}:\:\:\left(−\infty\:;\:+\infty\right) \\ $$

Answered by aleks041103 last updated on 12/Nov/22

x∈R⇒∀y∈I,∃x∈R,y=x−(2/x)  ⇒x^2 −yx−2=0  for this to have a soln in the R:  D=y^2 −4(1)(−2)=y^2 +8>0⇔y∈R  ⇒ans. I=(−∞,+∞)≡R

$${x}\in\mathbb{R}\Rightarrow\forall{y}\in{I},\exists{x}\in\mathbb{R},{y}={x}−\frac{\mathrm{2}}{{x}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{yx}−\mathrm{2}=\mathrm{0} \\ $$$${for}\:{this}\:{to}\:{have}\:{a}\:{soln}\:{in}\:{the}\:\mathbb{R}: \\ $$$${D}={y}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{2}\right)={y}^{\mathrm{2}} +\mathrm{8}>\mathrm{0}\Leftrightarrow{y}\in\mathbb{R} \\ $$$$\Rightarrow{ans}.\:{I}=\left(−\infty,+\infty\right)\equiv\mathbb{R} \\ $$$$ \\ $$

Commented by Shrinava last updated on 12/Nov/22

thank you dear professor

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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