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Question Number 59788 by aliesam last updated on 14/May/19
find the local minimum and maximum value a^2 y=x^2 (a−x) f(x)=(4/(2−x))+(9/(x−3))
$${find}\:{the}\:{local}\:{minimum}\:{and}\:{maximum}\:{value} \\ $$$$ \\ $$$${a}^{\mathrm{2}} {y}={x}^{\mathrm{2}} \left({a}−{x}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{4}}{\mathrm{2}−{x}}+\frac{\mathrm{9}}{{x}−\mathrm{3}} \\ $$
Commented by kaivan.ahmadi last updated on 14/May/19
y=((x^2 (a−x))/a^2 ) and let a>0 y′=((2ax−3x^2 )/a^2 )=0⇒x(2a−3x)=0⇒ { ((x=0)),((x=((2a)/3))) :} { ((x<0⇒f′(x)<0)),((0<x<((2a)/3)⇒f′(x)>0)),((x>((2a)/3)⇒f′(x)<0)) :} ⇒x=0 is local min and x=((2a)/3) is local max if a<0 then x=0 is local max and x=((2a)/3) is local min
$${y}=\frac{{x}^{\mathrm{2}} \left({a}−{x}\right)}{{a}^{\mathrm{2}} }\:{and}\:{let}\:{a}>\mathrm{0} \\ $$$${y}'=\frac{\mathrm{2}{ax}−\mathrm{3}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{0}\Rightarrow{x}\left(\mathrm{2}{a}−\mathrm{3}{x}\right)=\mathrm{0}\Rightarrow\begin{cases}{{x}=\mathrm{0}}\\{{x}=\frac{\mathrm{2}{a}}{\mathrm{3}}}\end{cases} \\ $$$$\begin{cases}{{x}<\mathrm{0}\Rightarrow{f}'\left({x}\right)<\mathrm{0}}\\{\mathrm{0}<{x}<\frac{\mathrm{2}{a}}{\mathrm{3}}\Rightarrow{f}'\left({x}\right)>\mathrm{0}}\\{{x}>\frac{\mathrm{2}{a}}{\mathrm{3}}\Rightarrow{f}'\left({x}\right)<\mathrm{0}}\end{cases} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{is}\:{local}\:{min} \\ $$$${and} \\ $$$${x}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:{is}\:{local}\:{max} \\ $$$${if}\:{a}<\mathrm{0}\:{then} \\ $$$${x}=\mathrm{0}\:{is}\:{local}\:{max}\:{and}\:{x}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:{is}\:{local}\:{min} \\ $$
Commented by aliesam last updated on 14/May/19
thank you sir i am waiting number to solution
$${thank}\:{you}\:{sir}\:{i}\:{am}\:{waiting}\:{number}\:{to}\:{solution} \\ $$
Commented by kaivan.ahmadi last updated on 14/May/19
f(x)=(4/(2−x))+(9/(x−3))⇒ D_f =R−{2,3} f′(x)=(4/((2−x)^2 ))−(9/((x−3)^2 ))=0⇒ 4(x−3)^2 =9(2−x)^2 ⇒ 4x^2 −24x+36=9x^2 −36x+36⇒ 5x^2 −12x=0⇒x(5x−12)=0⇒ { ((x=0)),((x=((12)/5))) :} x<0⇒f′(x)<0 0<x<((12)/5)⇒f′(x)>0 x>((12)/5)⇒f′(x)<0 ⇒x=0 min and x=((12)/5)max
$${f}\left({x}\right)=\frac{\mathrm{4}}{\mathrm{2}−{x}}+\frac{\mathrm{9}}{{x}−\mathrm{3}}\Rightarrow \\ $$$${D}_{{f}} =\mathbb{R}−\left\{\mathrm{2},\mathrm{3}\right\} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{4}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} }−\frac{\mathrm{9}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{0}\Rightarrow \\ $$$$\mathrm{4}\left({x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{9}\left(\mathrm{2}−{x}\right)^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{36}=\mathrm{9}{x}^{\mathrm{2}} −\mathrm{36}{x}+\mathrm{36}\Rightarrow \\ $$$$\mathrm{5}{x}^{\mathrm{2}} −\mathrm{12}{x}=\mathrm{0}\Rightarrow{x}\left(\mathrm{5}{x}−\mathrm{12}\right)=\mathrm{0}\Rightarrow\begin{cases}{{x}=\mathrm{0}}\\{{x}=\frac{\mathrm{12}}{\mathrm{5}}}\end{cases} \\ $$$${x}<\mathrm{0}\Rightarrow{f}'\left({x}\right)<\mathrm{0} \\ $$$$\mathrm{0}<{x}<\frac{\mathrm{12}}{\mathrm{5}}\Rightarrow{f}'\left({x}\right)>\mathrm{0} \\ $$$${x}>\frac{\mathrm{12}}{\mathrm{5}}\Rightarrow{f}'\left({x}\right)<\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{min}\:{and}\:{x}=\frac{\mathrm{12}}{\mathrm{5}}{max} \\ $$
Commented by aliesam last updated on 14/May/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by meme last updated on 14/May/19
y=((x^2 (a−x))/a^2 ). a≠0 y^′ =((2x(a−x)−x^2 )/a^2 )=0 2x(a−x)−x^2 =0 2(a−x)=x 2a=3x x=((2a)/3) if x∈]−∞;((2a)/3)]y^′ ≤0.y has minimal if x∈[((2a)/3);+∞[y^′ ≥0.y has maximal
$${y}=\frac{{x}^{\mathrm{2}} \left({a}−{x}\right)}{{a}^{\mathrm{2}} }.\:{a}\neq\mathrm{0} \\ $$$${y}^{'} =\frac{\mathrm{2}{x}\left({a}−{x}\right)−{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{2}{x}\left({a}−{x}\right)−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}\left({a}−{x}\right)={x} \\ $$$$\mathrm{2}{a}=\mathrm{3}{x} \\ $$$${x}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\left.{i}\left.{f}\:{x}\in\right]−\infty;\frac{\mathrm{2}{a}}{\mathrm{3}}\right]{y}^{'} \leqslant\mathrm{0}.{y}\:{has}\:{minimal} \\ $$$${if}\:{x}\in\left[\frac{\mathrm{2}{a}}{\mathrm{3}};+\infty\left[{y}^{'} \geqslant\mathrm{0}.{y}\:{has}\:{maximal}\right.\right. \\ $$
Answered by MJS last updated on 14/May/19
y=−(1/a^2 )x^3 +(1/a)x^2 y′=−(3/a^2 )x^2 +(2/a)x=0 ⇒ x=0∨x=((2a)/3) y′′=−(6/a^2 )x+(2/a) x=0 ⇒ y′′=(2/a) ⇒ max with a<0; min with a>0 x=((2a)/3) ⇒ y′′=−(2/a) ⇒ min with a<0; max with a>0 f(x)=(4/(2−x))+(9/(x−3))=((5x−6)/((x−3)(x−2))) f′(x)=−((x(5x−12))/((x−3)^2 (x−2)^2 ))=0 ⇒ x=0∨x=((12)/5) f′′(x)=((2(5x^3 −18x^2 +36))/((x−3)^3 (x−2)^3 )) f′′(0)=(1/3)>0 ⇒ min at x=0 f′′(((12)/5))=−((625)/3) ⇒ max at x=((12)/5)
$${y}=−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}}{x}^{\mathrm{2}} \\ $$$${y}'=−\frac{\mathrm{3}}{{a}^{\mathrm{2}} }{x}^{\mathrm{2}} +\frac{\mathrm{2}}{{a}}{x}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{0}\vee{x}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$${y}''=−\frac{\mathrm{6}}{{a}^{\mathrm{2}} }{x}+\frac{\mathrm{2}}{{a}} \\ $$$$\:\:\:\:\:{x}=\mathrm{0}\:\Rightarrow\:{y}''=\frac{\mathrm{2}}{{a}}\:\Rightarrow\:\mathrm{max}\:\mathrm{with}\:{a}<\mathrm{0};\:\mathrm{min}\:\mathrm{with}\:{a}>\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:\Rightarrow\:{y}''=−\frac{\mathrm{2}}{{a}}\:\Rightarrow\:\mathrm{min}\:\mathrm{with}\:{a}<\mathrm{0};\:\mathrm{max}\:\mathrm{with}\:{a}>\mathrm{0} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{4}}{\mathrm{2}−{x}}+\frac{\mathrm{9}}{{x}−\mathrm{3}}=\frac{\mathrm{5}{x}−\mathrm{6}}{\left({x}−\mathrm{3}\right)\left({x}−\mathrm{2}\right)} \\ $$$${f}'\left({x}\right)=−\frac{{x}\left(\mathrm{5}{x}−\mathrm{12}\right)}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{0}\vee{x}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$${f}''\left({x}\right)=\frac{\mathrm{2}\left(\mathrm{5}{x}^{\mathrm{3}} −\mathrm{18}{x}^{\mathrm{2}} +\mathrm{36}\right)}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$${f}''\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{3}}>\mathrm{0}\:\Rightarrow\:\mathrm{min}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$${f}''\left(\frac{\mathrm{12}}{\mathrm{5}}\right)=−\frac{\mathrm{625}}{\mathrm{3}}\:\Rightarrow\:\mathrm{max}\:\mathrm{at}\:{x}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$

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