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Question Number 121956 by bemath last updated on 13/Nov/20
Find concave up and concove down  of function f(x) = x^2  ∣x−3∣
$${Find}\:{concave}\:{up}\:{and}\:{concove}\:{down} \\ $$$${of}\:{function}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:\mid{x}−\mathrm{3}\mid\: \\ $$
Answered by MJS_new last updated on 13/Nov/20
f(x)= { ((3x^2 −x^3 ; x<3)),((x^3 −3x^2 ; x≥3)) :}  f′′(x)= { ((6−6x; x<3)),((6x−6; x≥3)) :}  ⇒  f(x) is concave  { ((up; x<1)),((down; 1<x<3)),((up; 3<x)) :}
$${f}\left({x}\right)=\begin{cases}{\mathrm{3}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} ;\:{x}<\mathrm{3}}\\{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} ;\:{x}\geqslant\mathrm{3}}\end{cases} \\ $$$${f}''\left({x}\right)=\begin{cases}{\mathrm{6}−\mathrm{6}{x};\:{x}<\mathrm{3}}\\{\mathrm{6}{x}−\mathrm{6};\:{x}\geqslant\mathrm{3}}\end{cases} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{concave}\:\begin{cases}{\mathrm{up};\:{x}<\mathrm{1}}\\{\mathrm{down};\:\mathrm{1}<{x}<\mathrm{3}}\\{\mathrm{up};\:\mathrm{3}<{x}}\end{cases} \\ $$

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