Question-172912

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

Answered by FongXD last updated on 03/Jul/22

★ f (x) = f (x + 4 n)

\Rightarrow f^{'} (x) = f^{'} (x + 4 n)

★ f (x) = f (x + 1) + f (3 x + 2)

\Rightarrow f^{'} (x) = f^{'} (x + 1) + {3 f}^{'} (3 x + 2)

∙ x = 1, f^{'} (1) = f^{'} (2) + {3 f}^{'} (5) = f^{'} (2) + {3 f}^{'} (1)

\Rightarrow f^{'} (2) = - {2 f}^{'} (1)

∙ x = 0, f^{'} (0) = f^{'} (1) + {3 f}^{'} (2) = f^{'} (1) - {6 f}^{'} (1)

\Rightarrow f^{'} (0) = - {5 f}^{'} (1)

∙ x = 2, f^{'} (2) = f^{'} (3) + {3 f}^{'} (8) = - 18 + {3 f}^{'} (0)

\Leftrightarrow - {2 f}^{'} (1) = - 18 - {15 f}^{'} (1)

\Rightarrow f^{'} (1) = - \frac{18}{13}

Commented by mnjuly1970 last updated on 03/Jul/22

thanks alot