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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(x)=x^2 −x−2 =x^2 −2(1/2)x +(1/4) −(1/4)−2 =(x−(1/2))^2 −(9/4)  ⇒minf(x)=−(9/4)    and no maximum  i supoose that the turning point is fixed point ⇒f(x)=x ⇒  x^2 −x−2=x ⇒x^2 −2x−2=0 →Δ^′ =1+2=3 ⇒x_1 =1+(√3)  and x_2 =1−(√3)  the axis of symetrie is the line x=(1/2)  f is defined on R  and the varistion of is  x          −∞                    (1/2)               +∞  f^′ (x)                    −        0          +  f(x)     +∞    decr  −(9/4)    incr  +∞  ⇒f(]−∞ ,(1/2)] =[−(9/4),+∞[   and  f[(1/2),+∞[) =[−(9/4),+∞[

$${f}\left({x}\right)={x}^{\mathrm{2}} −{x}−\mathrm{2}\:={x}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}\:=\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow{minf}\left({x}\right)=−\frac{\mathrm{9}}{\mathrm{4}}\:\:\:\:{and}\:{no}\:{maximum} \\ $$$${i}\:{supoose}\:{that}\:{the}\:{turning}\:{point}\:{is}\:{fixed}\:{point}\:\Rightarrow{f}\left({x}\right)={x}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{2}={x}\:\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{2}=\mathrm{0}\:\rightarrow\Delta^{'} =\mathrm{1}+\mathrm{2}=\mathrm{3}\:\Rightarrow{x}_{\mathrm{1}} =\mathrm{1}+\sqrt{\mathrm{3}} \\ $$$${and}\:{x}_{\mathrm{2}} =\mathrm{1}−\sqrt{\mathrm{3}} \\ $$$${the}\:{axis}\:{of}\:{symetrie}\:{is}\:{the}\:{line}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\:{is}\:{defined}\:{on}\:{R}\:\:{and}\:{the}\:{varistion}\:{of}\:{is} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:+ \\ $$$${f}\left({x}\right)\:\:\:\:\:+\infty\:\:\:\:{decr}\:\:−\frac{\mathrm{9}}{\mathrm{4}}\:\:\:\:{incr}\:\:+\infty \\ $$$$\left.\Rightarrow{f}\left(\right]−\infty\:,\frac{\mathrm{1}}{\mathrm{2}}\right]\:=\left[−\frac{\mathrm{9}}{\mathrm{4}},+\infty\left[\:\:\:{and}\:\:{f}\left[\frac{\mathrm{1}}{\mathrm{2}},+\infty\left[\right)\:=\left[−\frac{\mathrm{9}}{\mathrm{4}},+\infty\left[\right.\right.\right.\right.\right. \\ $$

Commented by MJS last updated on 20/Nov/19

turning point = inflection point?  f′′(x)=0  f′′(x)=2 ≠0 ⇒ no turning point

$$\mathrm{turning}\:\mathrm{point}\:=\:\mathrm{inflection}\:\mathrm{point}? \\ $$$${f}''\left({x}\right)=\mathrm{0} \\ $$$${f}''\left({x}\right)=\mathrm{2}\:\neq\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{turning}\:\mathrm{point} \\ $$

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