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Question Number 38533 by behi83417@gmail.com last updated on 26/Jun/18

Commented by math khazana by abdo last updated on 03/Aug/18

$$1)f\left(a\right)=f\left(b\right)\Rightarrow \frac{pa+\sqrt{pa}}{\sqrt{pa+1}}=\frac{pb+\sqrt{pb}}{\sqrt{pb}+1}\Rightarrow$$$$2)f\left(p\right)=p-1\Rightarrow \frac{p^2+p}{\sqrt{p^2+1}}=p-1\Rightarrow$$$$p^2+p=(p-1)\sqrt{p^2+1}\Rightarrow {(p^2+p)}^2=(p^2+1){(p-1)}^2$$$$\Rightarrow p^4+2p^3+p^2=(p^2+1)(p^2-2p+1)\Rightarrow$$$$p^4+2p^3+p^2=p^4-2p^3+p^2+p^2-2p+1\Rightarrow$$$$2p^3+p^2=-2p^3+2p^2-2p+1\Rightarrow$$$$4p^3-p^2+2p-1=0therootsof$$$$p\left(x\right)=4x^3-x^2+2x-1are$$$$x_1\sim 0,432\left(real\right)$$$$x_2\sim -0,091+0,7552i\left(complex\right)$$$$x_3\sim -0,091-0,7552i\left(complex\right)forthatwe$$$$takep\sim 0,432$$

Commented by math khazana by abdo last updated on 03/Aug/18

$$p=0,432isnotsolutionbecausep-1<0?and$$$$\frac{p^2+p}{\sqrt{p^2+1}}>0sotheequationisimpossible.$$

Answered by behi83417@gmail.com last updated on 27/Jun/18

$$\frac{p^2+p}{\sqrt{p^2+1}}=p-1\Rightarrow p(p+1)=(p-1)\sqrt{p^2+1}$$$$\Rightarrow p^2(p^2+2p+1)=(p^2-2p+1)(p^2+1)$$$$\Rightarrow 4{\mathit{p}}^3-{\mathit{p}}^2+2\mathit{p}-1=0\Rightarrow \mathit{p}=0.43$$

Commented by MrW3 last updated on 27/Jun/18

$$\frac{0.{43}^2+0.43}{\sqrt{0.{43}^2+1}}=0.57!$$$$0.43-1=-0.57!$$$$i.e.p=0.43isnotasolutionofthe$$$$originalequation!$$$$Ithinkyouknowwheretheerror$$$$lies,seefollowingexample:$$$$\sqrt{x+1}=-2\left(ithasnosolution\right),but$$$${\left(\sqrt{x+1}\right)}^2={(-2)}^2\Rightarrow x+1=4\Rightarrow x=3$$

Commented by behi83417@gmail.com last updated on 27/Jun/18

$$thankyousomuchdearmaster.$$$$youareright.itshouldbe:p>1.$$

Answered by MrW3 last updated on 27/Jun/18

$$f^{\prime }\left(x\right)=(px+\sqrt{px})(-\frac{p}{2(px+1)\sqrt{px+1}})+\frac{1}{\sqrt{px+1}}(p+\frac{p}{2\sqrt{px}})$$$$f^{\prime }\left(x\right)=\frac{p}{2}[-\frac{\sqrt{px}}{px+1}+\frac{1}{\sqrt{px}}]$$$$f^{\prime }\left(x\right)=\frac{p}{2(px+1)\sqrt{px}}$$$$f^{\prime }\left(x\right)>0ifp>0$$$$f^{\prime }\left(x\right)<0ifp<0$$$$i.e.f\left(x\right)isalwaysincreasingor$$$$alwaysdecreasing.$$$$i.e.iff\left(a\right)=f\left(b\right)\Rightarrow a=b\Rightarrow \frac{a}{b}=1$$$$$$$$f\left(p\right)=\frac{p^2+p}{\sqrt{p^2+1}}=\frac{p(p+1)}{\sqrt{p^2+1}}>p$$$$butp-1<p$$$$\Rightarrow f\left(p\right)\ne p-1$$$$\Rightarrow nosolutionforf\left(p\right)=p-1$$

Commented by behi83417@gmail.com last updated on 27/Jun/18

$$dearmaster!greatjobdonebyyou.$$$$thanksinadvance.$$$$isthereanychancetofind{\mathit{f}}^{-1}\left(\mathit{x}\right)?$$

Commented by MrW3 last updated on 27/Jun/18

$$Iseenochancetofind{f}^{-1}\left(x\right).$$

Commented by ajfour last updated on 27/Jun/18

$$Canwefind\theta if$$$$\sin \theta (1+\tan \theta )=c?$$

Commented by behi83417@gmail.com last updated on 27/Jun/18

$$no,ithink.$$$$byputting:\mathit{p}\mathit{x}={\mathit{t}\mathit{g}}^2\mathit{\theta },wehavethe$$$$equation:$$$$\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }(1+\mathit{t}\mathit{g}\mathit{\theta })=\mathit{y}$$$$\mathit{a}\mathit{n}\mathit{d}\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{i}\mathit{n}\mathit{g}\mathit{t}\mathit{h}\mathit{i}\mathit{s},noteasy.$$

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