Question Number 194085 by 073 last updated on 27/Jun/23
$$\mathrm{f}\left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{f}'\left(\mathrm{6}\right)=? \\ $$$$\mathrm{solution}?? \\ $$
Answered by qaz last updated on 27/Jun/23
$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${f}\left({x}−\mathrm{1}\right)={a}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{b}\left({x}−\mathrm{1}\right)+{c} \\ $$$${f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)={a}\left(\mathrm{2}{x}−\mathrm{1}\right)+{b}={x}+\mathrm{1} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:{b}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{x}+{c}\:\:\:\:\Rightarrow{f}\left(\mathrm{6}\right)'=\frac{\mathrm{15}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 27/Jun/23
$${a}_{{k}} −{a}_{{k}−\mathrm{1}} ={k}+\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({a}_{{k}} −{a}_{{k}−\mathrm{1}} \right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right) \\ $$$${a}_{{n}} −{a}_{\mathrm{0}} =\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{2}} \\ $$$${a}_{{n}} =\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{2}}+{a}_{\mathrm{0}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{3}\right)}{\mathrm{2}}+{f}\left(\mathrm{0}\right) \\ $$$$\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{f}'\left(\mathrm{6}\right)=\frac{\mathrm{15}}{\mathrm{2}} \\ $$