反函數微分律的問題!!

Question

題目:
f(X)=X^3+2x+3
請問f(6)的微分的反函數是多少?
就是在問 f'^-1(6)=?
下禮拜期中考要考!請幫幫我!謝!

Answer 1

f(f-1(x))=xf'(f-1(x))(f-1)'(x)=1(f-1)'(x)=1/f'(f-1(x))(f-1)'(6)=1/f'(f-1(6))( 以下求f-1(6) )x³+2x-3=0(x-1)(x²+x+3)=0x²+x+3=0　沒有實數根x-1=0, x=1故知f-1(6)=1( 以下求f'(1) )f(x)=x³+2x+3f'(x)=3x²+2f'(1)=3(1)²+2=3+2=5故(f-1)'(6)=1/f'(f-1(6))=1/(f'(1)=1/5