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Question Number 187651 by Mastermind last updated on 19/Feb/23
Find the range of this function  x^2  −13x + 36 = 0      Help!
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{this}\:\mathrm{function} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\mathrm{13x}\:+\:\mathrm{36}\:=\:\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by Ar Brandon last updated on 19/Feb/23
Let y=x^2 −13x+36  ⇒x^2 −13x+36−y=0  ⇒x=((13±(√(169−4(36−y))))/2)           =((13±(√(25+4y)))/2)  25+4y ≥0 ⇒y≥−((25)/4)  Hence range f(x)≥−((25)/4)
$$\mathrm{Let}\:\mathrm{y}={x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{36} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{36}−\mathrm{y}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{13}\pm\sqrt{\mathrm{169}−\mathrm{4}\left(\mathrm{36}−\mathrm{y}\right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{13}\pm\sqrt{\mathrm{25}+\mathrm{4y}}}{\mathrm{2}} \\ $$$$\mathrm{25}+\mathrm{4y}\:\geqslant\mathrm{0}\:\Rightarrow\mathrm{y}\geqslant−\frac{\mathrm{25}}{\mathrm{4}} \\ $$$$\mathrm{Hence}\:\mathrm{range}\:{f}\left({x}\right)\geqslant−\frac{\mathrm{25}}{\mathrm{4}} \\ $$
Commented by Mastermind last updated on 20/Feb/23
Thank you BOSS, that′s what i also  got but with diff. method.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{BOSS},\:\mathrm{that}'\mathrm{s}\:\mathrm{what}\:\mathrm{i}\:\mathrm{also} \\ $$$$\mathrm{got}\:\mathrm{but}\:\mathrm{with}\:\mathrm{diff}.\:\mathrm{method}. \\ $$
Answered by mr W last updated on 20/Feb/23
x^2 −13x+36  =(x−((13)/2))^2 +36−(((13)/2))^2   ≥36−(((13)/2))^2 =−((25)/4)  ⇒range is [−((25)/4),+∞)
$${x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{36} \\ $$$$=\left({x}−\frac{\mathrm{13}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{36}−\left(\frac{\mathrm{13}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\geqslant\mathrm{36}−\left(\frac{\mathrm{13}}{\mathrm{2}}\right)^{\mathrm{2}} =−\frac{\mathrm{25}}{\mathrm{4}} \\ $$$$\Rightarrow{range}\:{is}\:\left[−\frac{\mathrm{25}}{\mathrm{4}},+\infty\right) \\ $$

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