Question Number 87799 by M±th+et£s last updated on 06/Apr/20
![f(x)= { ((ax^2 +bx −1≤x≤0)),((cx^2 +d 0<x≤(1/2))),((bx+d (1/2)<x≤1)) :} f(x) is continuous on[−1,1] prove d=0 c=2b](https://www.tinkutara.com/question/Q87799.png)
$${f}\left({x}\right)=\begin{cases}{{ax}^{\mathrm{2}} +{bx}\:\:\:\:\:\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{0}}\\{{cx}^{\mathrm{2}} +{d}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}}\\{{bx}+{d}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}}\end{cases} \\ $$$${f}\left({x}\right)\:{is}\:{continuous}\:{on}\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${prove}\:{d}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}=\mathrm{2}{b} \\ $$
Commented by john santu last updated on 06/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\left(\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}\right)\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\mathrm{cx}^{\mathrm{2}} +\mathrm{d}\right) \\ $$$$\mathrm{0}\:=\:\mathrm{d}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}.\mathrm{5}^{−} } {\mathrm{lim}}\:\left(\mathrm{cx}^{\mathrm{2}} +\mathrm{d}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}.\mathrm{5}^{+} } {\mathrm{lim}}\:\left(\mathrm{bx}+\mathrm{d}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{b}\:\Rightarrow\mathrm{c}\:=\:\mathrm{2b} \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
$${god}\:{bless}\:{you}\:{sir} \\ $$$$ \\ $$