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Question Number 74138 by MASANJAJ last updated on 19/Nov/19

Commented by mathmax by abdo last updated on 19/Nov/19

$$f\left(x\right)=x^2-x-2=x^2-2\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}-2={(x-\frac{1}{2})}^2-\frac{9}{4}$$$$\Rightarrow minf\left(x\right)=-\frac{9}{4}andnomaximum$$$$isupoosethattheturningpointisfixedpoint\Rightarrow f\left(x\right)=x\Rightarrow$$$$x^2-x-2=x\Rightarrow x^2-2x-2=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1+2=3\Rightarrow x_1=1+\sqrt{3}$$$$and{x}_2=1-\sqrt{3}$$$$theaxisofsymetrieisthelinex=\frac{1}{2}$$$$fisdefinedonRandthevaristionofis$$$$x-\mathrm{\infty }\frac{1}{2}+\mathrm{\infty }$$$$f^{{}^{\prime }}\left(x\right)-0+$$$$f\left(x\right)+\mathrm{\infty }decr-\frac{9}{4}incr+\mathrm{\infty }$$$$\Rightarrow f\left(\right]-\mathrm{\infty },\frac{1}{2}]=[-\frac{9}{4},+\mathrm{\infty }[andf[\frac{1}{2},+\mathrm{\infty }\left[\right)=[-\frac{9}{4},+\mathrm{\infty }[$$

Commented by MJS last updated on 20/Nov/19

$$\mathrm{turning}\mathrm{point}=\mathrm{inflection}\mathrm{point}?$$$$f^{″}\left(x\right)=0$$$$f^{″}\left(x\right)=2\ne 0\Rightarrow \mathrm{no}\mathrm{turning}\mathrm{point}$$

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