Question Number 25397 by naka3546 last updated on 09/Dec/17
$${Let}\:\:{y}\:\:=\:\:{f}\:\left({x}\right) \\ $$$$\:\:\:\:{f}\:\left(\mathrm{0}\right)\:\:=\:\:\mathrm{5} \\ $$$${f}\:'\left({x}\right)\:\:=\:\:−\mathrm{3}{x}\:+\:\mathrm{2}\:\underset{\mathrm{0}} {\int}\overset{\mathrm{2}} {\:}{f}\:\left({x}\right)\:{dx} \\ $$$${Find}\:\:{the}\:\:{maximum}\:\:{value}\:\:{of}\:\:\:\boldsymbol{{y}}\:\:=\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:. \\ $$$$ \\ $$
Commented by prakash jain last updated on 09/Dec/17
![c=2∫_0 ^2 f(x)dx f^′ (x)=−3x+c f(x)=−((3x^2 )/2)+cx+c_1 f(0)=5⇒c_1 =5 f(x)=−((3x^2 )/2)+cx+5 2∫_0 ^( 2) f(x)dx=2[−(x^3 /2)+((cx^2 )/2)+5x]_0 ^2 c=−8+4c+20 c=−4 f(x)=−((3x^2 )/2)−4x+5 quadratic with a<0 so maximum value at x=((−b)/(2a)) x=(4/(−3))=−(4/3) f_(max) (x)=−(3/2)((4/3))^2 +((16)/3)+5 =−(8/3)+((16)/3)+5=((23)/3)](https://www.tinkutara.com/question/Q25398.png)
$${c}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx} \\ $$$${f}^{'} \left({x}\right)=−\mathrm{3}{x}+{c} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+{cx}+{c}_{\mathrm{1}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{5}\Rightarrow{c}_{\mathrm{1}} =\mathrm{5} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+{cx}+\mathrm{5} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\:\:\mathrm{2}} {f}\left({x}\right){dx}=\mathrm{2}\left[−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{cx}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{5}{x}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$${c}=−\mathrm{8}+\mathrm{4}{c}+\mathrm{20} \\ $$$${c}=−\mathrm{4} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4}{x}+\mathrm{5} \\ $$$${quadratic}\:{with}\:{a}<\mathrm{0}\:{so}\:{maximum} \\ $$$${value}\:{at}\:{x}=\frac{−{b}}{\mathrm{2}{a}} \\ $$$${x}=\frac{\mathrm{4}}{−\mathrm{3}}=−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${f}_{{max}} \left({x}\right)=−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{3}}+\mathrm{5} \\ $$$$=−\frac{\mathrm{8}}{\mathrm{3}}+\frac{\mathrm{16}}{\mathrm{3}}+\mathrm{5}=\frac{\mathrm{23}}{\mathrm{3}} \\ $$
Commented by abwayh last updated on 09/Dec/17
![another way; put f(x)=Ax^2 +Bx+c f(0)=c=5 f ^ (x)=2Ax+B ∵ f ^ (x)=−3x+2∫_0 ^2 f(x)dx ∴2Ax+B=−3x+2∫_0 ^2 (Ax^2 +Bx+c)dx 2Ax+B=−3x+2[((Ax^3 )/3) +((Bx^2 )/2) +cx]_0 ^2 2Ax+B=−3x+((16A)/3) +4B +4c (by compairing) 2A=−3 ⇒ A=((−3)/2) B=((16A)/3) +4B+4c B=((16)/3)(((−3)/2))+4B+4(5) B=−8+4B+20 ⇒3B=−12 ⇒B=−4 ∴f(x)=((−3)/2) x^2 −4x+5 continue](https://www.tinkutara.com/question/Q25409.png)
$$\mathrm{another}\:\mathrm{way}; \\ $$$$\mathrm{put}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{Ax}^{\mathrm{2}} +\mathrm{Bx}+\mathrm{c} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{0}\right)=\mathrm{c}=\mathrm{5} \\ $$$$\mathrm{f}\acute {\:}\left(\mathrm{x}\right)=\mathrm{2Ax}+\mathrm{B} \\ $$$$\because\:\mathrm{f}\acute {\:}\left(\mathrm{x}\right)=−\mathrm{3x}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\therefore\mathrm{2Ax}+\mathrm{B}=−\mathrm{3x}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{Ax}^{\mathrm{2}} +\mathrm{Bx}+\mathrm{c}\right)\mathrm{dx} \\ $$$$\mathrm{2Ax}+\mathrm{B}=−\mathrm{3x}+\mathrm{2}\left[\frac{\mathrm{Ax}^{\mathrm{3}} }{\mathrm{3}}\:+\frac{\mathrm{Bx}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{cx}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\mathrm{2Ax}+\mathrm{B}=−\mathrm{3x}+\frac{\mathrm{16A}}{\mathrm{3}}\:+\mathrm{4B}\:+\mathrm{4c} \\ $$$$\left(\mathrm{by}\:\mathrm{compairing}\right) \\ $$$$\mathrm{2A}=−\mathrm{3}\:\Rightarrow\:\:\mathrm{A}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$\:\mathrm{B}=\frac{\mathrm{16A}}{\mathrm{3}}\:+\mathrm{4B}+\mathrm{4c} \\ $$$$\mathrm{B}=\frac{\mathrm{16}}{\mathrm{3}}\left(\frac{−\mathrm{3}}{\mathrm{2}}\right)+\mathrm{4B}+\mathrm{4}\left(\mathrm{5}\right) \\ $$$$\mathrm{B}=−\mathrm{8}+\mathrm{4B}+\mathrm{20}\:\:\Rightarrow\mathrm{3B}=−\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{B}=−\mathrm{4} \\ $$$$\therefore\mathrm{f}\left(\mathrm{x}\right)=\frac{−\mathrm{3}}{\mathrm{2}}\:\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{5} \\ $$$$\mathrm{continue} \\ $$$$ \\ $$