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Question Number 27847 by NECx last updated on 15/Jan/18

$$Findtherangeoff\left(x\right)=\frac{x+5}{x^2-4}$$

Commented by NECx last updated on 16/Jan/18

$$pleasehelp$$

Answered by prakash jain last updated on 17/Jan/18

$$y=\frac{x+5}{x^2-4}$$$$x^{2}y-x-(4y+5)=0$$$$x\in \mathrm{R}\Rightarrow \mathrm{\Delta }⩾0$$$${(-1)}^2-4\left(y\right)(-)(4y+5)⩾0$$$$1+16y^2+20y⩾0$$$$16y^2+20y+1⩾0$$$$(y-(-\frac{5}{8}+\frac{\sqrt{21}}{8}))(y-(-\frac{5}{8}-\frac{\sqrt{21}}{8}))⩾0$$$$\because \mathrm{A}\mathrm{quadratic}\mathrm{expression}\mathrm{has}\mathrm{sine}$$$$\mathrm{same}\mathrm{as}\mathrm{coefficient}\mathrm{of}{x}^2\mathrm{except}\mathrm{in}$$$$\mathrm{between}\mathrm{roots}.$$$$y\in (-\mathrm{\infty },-\frac{5}{8}-\frac{\sqrt{21}}{8}]\cup [-\frac{5}{8}+\frac{\sqrt{21}}{8},\mathrm{\infty })$$

Commented by Rasheed.Sindhi last updated on 16/Jan/18

$$\mathrm{Pl}\mathrm{insert}\mathrm{some}\mathrm{more}\mathrm{steps}\mathrm{between}$$$$\mathrm{last}\mathrm{two}\mathrm{steps}.$$

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