Question Number 56179 by problem solverd last updated on 11/Mar/19
$$\mathrm{draw}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of} \\ $$$$\mathrm{f}\left({x}\right)=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{for}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1} \\ $$
Commented by mr W last updated on 12/Mar/19
$${y}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$${the}\:{graph}\:{is}\:{just}\:{a}\:{part}\:{of}\:{a}\:{circle}\:{with} \\ $$$${radius}\:\mathrm{1}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
![with the use of graphical app you can see the graph.. trying to solve... f(0)=1 f(1)=0 (df/dx)=(1/(2(√(1−x^2 ))))×−2x=((−x)/(√(1−x^2 ))) (d^2 f/dx^2 )=−1×{(((√(1−x^2 )) ×(dx/dx)−x×(d/dx)((√(1−x^2 )) ) )/(1−x^2 ))} (d^2 f/dx^2 )=−1×{(((√(1−x^2 )) −x×(1/(2(√(1−x^2 )) ))×−2x)/(1−x^2 ))} =−1×{(((√(1−x^2 )) +(x^2 /(√(1−x^2 ))))/((1−x^2 )))} =((−1)/((1−x^2 )^(3/2) )) (d^2 f/dx^2 )<0 in x [0,1] now if (d^2 f/dx^2 )>0 hold water that means concave up if (d^2 f/dx^2 )<0 spill water that means cocave down that means CONVEX so in x[0,1] shape of f(x) is convex ..](Q56194.png)
$${with}\:{the}\:{use}\:{of}\:{graphical}\:{app}\:{you}\:{can}\:{see}\:{the} \\ $$$${graph}.. \\ $$$${trying}\:{to}\:{solve}... \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{{df}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}×−\mathrm{2}{x}=\frac{−{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }=−\mathrm{1}×\left\{\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:×\frac{{dx}}{{dx}}−{x}×\frac{{d}}{{dx}}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\right)\:}{\mathrm{1}−{x}^{\mathrm{2}} }\right\} \\ $$$$\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }=−\mathrm{1}×\left\{\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:−{x}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:}×−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right\} \\ $$$$=−\mathrm{1}×\left\{\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\right\} \\ $$$$=\frac{−\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }<\mathrm{0}\:\:{in}\:{x}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$${now}\:\:\:{if}\:\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }>\mathrm{0}\:\:{hold}\:{water}\:{that}\:{means}\:{concave}\:{up} \\ $$$${if}\:\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }<\mathrm{0}\:\:{spill}\:{water}\:{that}\:{means}\:{cocave}\:{down}\:\:{that}\:{means}\:\boldsymbol{{C}}{ONVEX} \\ $$$${so}\:{in}\:{x}\left[\mathrm{0},\mathrm{1}\right]\:\:{shape}\:{of}\:{f}\left({x}\right)\:{is}\:{convex} \\ $$$$.. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
