f(x)=2sin(2x+Ï/3)ï¼è¥å½x∊[Ï/12ï¼7Ï/12]æ¶å
¶åå½æ°ä¸ºf⁻¹(x)ï¼æ±f⁻¹(x)ã
解ï¼Ï/12â¦xâ¦7Ï/12ï¼Ï/6â¦2xâ¦7Ï/6ï¼Ï/2â¦2x+Ï/3â¦3Ï/2ï¼æ
å½x∊[Ï/12ï¼7Ï/12]æ¶f(x)ç¡®
æåå½æ°ã
y=2sin(2x+Ï/3)ï¼å®ä¹åï¼x∊[Ï/12ï¼7Ï/12]ï¼å¼åï¼[-2ï¼2]ã
sin(2x+Ï/3)=y/2ï¼2x+Ï/3=arcsin(y/2)ï¼2x=arcsin(y/2)-Ï/3ï¼
x=(1/2)arcsin(y/2)-Ï/6ï¼äº¤æ¢xï¼yï¼ä¾¿å¾åå½æ°f⁻¹(x)=(1/2)arcsin(x/2)-Ï/6ï¼(-2â¦xâ¦2)ï¼
ãé¢ç®æéï¼åªè½æ±åå½æ°ï¼ä¸è½æ±åå½æ°çå¼ï¼å ä¸ºä½ æ²¡æç»å®xï¼æ
åå½æ°çå¼æ¯ä¸çï¼ã
ãå¦æf⁻¹(x)=(1/2)arcsin(x/2)-Ï/6=Ï/4ï¼å(1/2)arcsin(x/2)=Ï/4+Ï/6=5Ï/12ï¼arcsin(x/2)=5Ï/6ï¼
x/2=sin(5Ï/6)=sin(Ï-Ï/6)=sin(Ï/6)=1/2ï¼æ
å¾x=1ãã
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