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Question Number 74265 by Maclaurin Stickker last updated on 21/Nov/19

$$Determinethevaluesofm\in \mathbb{R}for$$$$whichthefunctionf\left(x\right)=\frac{1}{\sqrt{2x^2-mx+m}}$$$$isthesetofrealnumbers.$$

Answered by MJS last updated on 21/Nov/19

$$2x^2-mx+m=0$$$$x=\frac{m\pm \sqrt{m(m-8)}}{4}$$$$\mathrm{to}\mathrm{get}\mathrm{no}\mathrm{zeros}:$$$$m(m-8)<0$$$$\Rightarrow 0<m<8$$$$\mathrm{for}0<m<8:\mathrm{\forall }x\in \mathbb{R}:2x^2-mx+m>0$$$$\mathrm{in}\mathrm{this}\mathrm{case}f\left(x\right)\mathrm{is}\mathrm{defined}\mathrm{for}x\in \mathbb{R}\mathrm{but}\mathrm{the}$$$$\mathrm{range}\mathrm{is}0<f\left(x\right)⩽\frac{2\sqrt{2}}{\sqrt{8m-m^2}}$$

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