Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 172912 by mnjuly1970 last updated on 03/Jul/22

Answered by FongXD last updated on 03/Jul/22

★ f(x)=f(x+4n) ⇒ f′(x)=f′(x+4n) ★ f(x)=f(x+1)+f(3x+2) ⇒ f′(x)=f′(x+1)+3f′(3x+2) • x=1, f′(1)=f′(2)+3f′(5)=f′(2)+3f′(1) ⇒ f′(2)=−2f′(1) • x=0, f′(0)=f′(1)+3f′(2)=f′(1)−6f′(1) ⇒ f′(0)=−5f′(1) • x=2, f′(2)=f′(3)+3f′(8)=−18+3f′(0) ⇔ −2f′(1)=−18−15f′(1) ⇒ f′(1)=−((18)/(13))

$$\bigstar\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}+\mathrm{4n}\right) \\ $$$$\Rightarrow\:\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{f}'\left(\mathrm{x}+\mathrm{4n}\right) \\ $$$$\bigstar\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{f}\left(\mathrm{3x}+\mathrm{2}\right) \\ $$$$\Rightarrow\:\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{f}'\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{3f}'\left(\mathrm{3x}+\mathrm{2}\right) \\ $$$$\bullet\:\mathrm{x}=\mathrm{1},\:\mathrm{f}'\left(\mathrm{1}\right)=\mathrm{f}'\left(\mathrm{2}\right)+\mathrm{3f}'\left(\mathrm{5}\right)=\mathrm{f}'\left(\mathrm{2}\right)+\mathrm{3f}'\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{f}'\left(\mathrm{2}\right)=−\mathrm{2f}'\left(\mathrm{1}\right) \\ $$$$\bullet\:\mathrm{x}=\mathrm{0},\:\mathrm{f}'\left(\mathrm{0}\right)=\mathrm{f}'\left(\mathrm{1}\right)+\mathrm{3f}'\left(\mathrm{2}\right)=\mathrm{f}'\left(\mathrm{1}\right)−\mathrm{6f}'\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{f}'\left(\mathrm{0}\right)=−\mathrm{5f}'\left(\mathrm{1}\right) \\ $$$$\bullet\:\mathrm{x}=\mathrm{2},\:\mathrm{f}'\left(\mathrm{2}\right)=\mathrm{f}'\left(\mathrm{3}\right)+\mathrm{3f}'\left(\mathrm{8}\right)=−\mathrm{18}+\mathrm{3f}'\left(\mathrm{0}\right) \\ $$$$\Leftrightarrow\:−\mathrm{2f}'\left(\mathrm{1}\right)=−\mathrm{18}−\mathrm{15f}'\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{f}'\left(\mathrm{1}\right)=−\frac{\mathrm{18}}{\mathrm{13}} \\ $$

Commented by mnjuly1970 last updated on 03/Jul/22

thanks alot

$$\mathrm{thanks}\:\mathrm{alot} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com