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Determine-the-values-of-m-R-for-which-the-function-f-x-1-2x-2-mx-m-is-the-set-of-real-numbers-




Question Number 74265 by Maclaurin Stickker last updated on 21/Nov/19
Determine the values of m∈R for   which the function f(x)=(1/( (√(2x^2 −mx+m))))  is the set of real numbers.
$${Determine}\:{the}\:{values}\:{of}\:{m}\in\mathbb{R}\:{for}\: \\ $$$${which}\:{the}\:{function}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −{mx}+{m}}} \\ $$$${is}\:{the}\:{set}\:{of}\:{real}\:{numbers}. \\ $$
Answered by MJS last updated on 21/Nov/19
2x^2 −mx+m=0  x=((m±(√(m(m−8))))/4)  to get no zeros:  m(m−8)<0  ⇒ 0<m<8  for 0<m<8: ∀x∈R: 2x^2 −mx+m>0  in this case f(x) is defined for x∈R but the  range is 0<f(x)≤((2(√2))/( (√(8m−m^2 ))))
$$\mathrm{2}{x}^{\mathrm{2}} −{mx}+{m}=\mathrm{0} \\ $$$${x}=\frac{{m}\pm\sqrt{{m}\left({m}−\mathrm{8}\right)}}{\mathrm{4}} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{no}\:\mathrm{zeros}: \\ $$$${m}\left({m}−\mathrm{8}\right)<\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{0}<{m}<\mathrm{8} \\ $$$$\mathrm{for}\:\mathrm{0}<{m}<\mathrm{8}:\:\forall{x}\in\mathbb{R}:\:\mathrm{2}{x}^{\mathrm{2}} −{mx}+{m}>\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{range}\:\mathrm{is}\:\mathrm{0}<{f}\left({x}\right)\leqslant\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{8}{m}−{m}^{\mathrm{2}} }} \\ $$

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