Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 32283 by abdo imad last updated on 22/Mar/18

$$letgivef\left(x\right)=x+2-\sqrt{x+1}$$$$1)find{f}^{-1}\left(x\right)inverseoff\left(x\right)$$$$2)calculate{\left(f^{-1}\right)}^{{}^{\prime }}\left(x\right).$$

Commented by abdo imad last updated on 24/Mar/18

$$D_f=[-1,+\mathrm{\infty }[f\left(x\right)=y\iff y=x+2-\sqrt{x+1}\iff$$$${(x+2-y)}^2=x+1\Rightarrow {(x+2)}^2-2y(x+2)+y^2-x-1=0$$$$\Rightarrow x^2+4x+4-2yx-4y+y^2-x-1=0\Rightarrow$$$$x^2+(3-2y)x+y^2-4y+3=0$$$$\mathrm{\Delta }={(3-2y)}^2-4(y^2-4y+3)=$$$$=9-12y+4y^2-4y^2+16y-12=4y-3$$$$x_1=\frac{-3+2y+\sqrt{4y-3}}{2}and{x}_2=\frac{-3+2y-\sqrt{4h-3}}{2}$$$$wemusthavex+2-y⩾0letverify$$$$x_2+2-y=\frac{-3+2y-\sqrt{4y-3}}{2}+2-y$$$$=\frac{-3+2y-\sqrt{4y-3}+4-2y}{2}=\frac{1-y-\sqrt{4y-3}}{2}<0sothe$$$$soutionis{x}_{1}and{f}^{-1}\left(x\right)=\frac{2x+\sqrt{4x-3}-3}{2}$$$$f^{-1}\left(x\right)=x+\frac{1}{2}\sqrt{4x-3}-\frac{3}{2}\Rightarrow$$$${\left(f^{-1}\right)}^{{}^{\prime }}\left(x\right)=1+\frac{1}{2}\frac{4}{2\sqrt{4x-3}}=1+\frac{1}{\sqrt{4x-3}}.withx>\frac{3}{4}.$$

Answered by MJS last updated on 22/Mar/18

$$x=y+2-\sqrt{y+1}$$$$\sqrt{y+1}=-x+y+2$$$$y+1=y^2+2(2-x)y+{(2-x)}^2$$$$y^2+(3-2x)y+(x^2-4x+3)=0$$$$p=3-2x;q=(x^2-4x+3)$$$$y=-\frac{p}{2}\pm \frac{\sqrt{p^2-4q}}{2}$$$$y=x-\frac{3}{2}\pm \frac{\sqrt{4x-3}}{2}$$$$y^{\prime }=1\pm \frac{\frac{1}{2}{(4x-3)}^{-\frac{1}{2}}\times 4}{2}$$$$y^{\prime }=1\pm \frac{1}{\sqrt{4x-3}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com