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Question-132295




Question Number 132295 by shaker last updated on 13/Feb/21
Commented by mathmax by abdo last updated on 13/Feb/21
f(x)=logx−(√(5x−5))   f defined on [1,+∞[  f^′ (x)=(1/x)−(5/(2(√(5x−5)))) =(1/x)−((√5)/(2(√(x−1)))) =((2(√(x−1))−(√5)x)/(2x(√(x−1))))  =(((2(√(x−1))−(√5)x)(2(√(x−1))+(√5)x))/(2x(√(x−1))(2(√(x−1))+(√5)x))) =((4(x−1)−5x^2 )/((...)))=((−5x^2 +4x−4)/((....)))  Δ^′  =4−20<0 ⇒f^′ (x)<0 ⇒f is decreazing on[1,+∞[  f(1)=0  and lim_(x→+∞) f(x)=lim_(x→+∞) (√(5x)){((logx)/( (√(5x))))−(√(1−(1/x)))}  =lim_(x→+∞) (√(5x)){2 ((log((√x)))/( (√(5x))))−(√(1−(1/x)))}=−∞  ⇒∃!x_0 ∈[1,+∞[ /  f(x_0 )=0   and we see that x_0 =1
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}−\sqrt{\mathrm{5x}−\mathrm{5}}\:\:\:\mathrm{f}\:\mathrm{defined}\:\mathrm{on}\:\left[\mathrm{1},+\infty\left[\right.\right. \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5x}−\mathrm{5}}}\:=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}\:=\frac{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}} \\ $$$$=\frac{\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}\right)\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}\:=\frac{\mathrm{4}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{5x}^{\mathrm{2}} }{\left(…\right)}=\frac{−\mathrm{5x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{4}}{\left(….\right)} \\ $$$$\Delta^{'} \:=\mathrm{4}−\mathrm{20}<\mathrm{0}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)<\mathrm{0}\:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{decreazing}\:\mathrm{on}\left[\mathrm{1},+\infty\left[\right.\right. \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{0}\:\:\mathrm{and}\:\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \sqrt{\mathrm{5x}}\left\{\frac{\mathrm{logx}}{\:\sqrt{\mathrm{5x}}}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}}\right\} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \sqrt{\mathrm{5x}}\left\{\mathrm{2}\:\frac{\mathrm{log}\left(\sqrt{\mathrm{x}}\right)}{\:\sqrt{\mathrm{5x}}}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}}\right\}=−\infty\:\:\Rightarrow\exists!\mathrm{x}_{\mathrm{0}} \in\left[\mathrm{1},+\infty\left[\:/\right.\right. \\ $$$$\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{0}\:\:\:\mathrm{and}\:\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{x}_{\mathrm{0}} =\mathrm{1} \\ $$
Commented by liki last updated on 13/Feb/21
Commented by liki last updated on 13/Feb/21
Help me please
$$\mathrm{Help}\:\mathrm{me}\:\mathrm{please} \\ $$
Commented by mr W last updated on 13/Feb/21
x=1
$${x}=\mathrm{1} \\ $$

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