Jika fungsi f(x) = (1-x^5)^1/3 + 7 maka nilai f invers (6) = .... Question from @galangellwid

September 2018 2 13 Report

Tve F = ( 1 - x^5 )^ 1/3 + 7
maka :
y = ( 1 - x^5 )^1/3 + 7
( 1 - x^5 ) ^ -1/3 = 7 - y
( 1 - x^5 ) ^ 1/3 = -7 + y
1 - x^5 = ( -7 + y )^3
1- x^5 = y^3 - 21y^2 + 147y + 343
x^5 = y^3 - 21y^2 + 147y + 342
x = ( y^3 - 21y^2 + 147y + 342 )^1/5
ini dah betul maaf klo ada yang salah :D

0 votes Thanks 1

galangellwid itu ada pangkat 5 nya suhu hehe

Takamori37 $y=(1-x^5)^{1/3}+7 \\ (1-x^5)^{1/3}=y-7 \\ 1-x^5=(y-7)^3 \\ x^5=1-(y-7)^3 \\ x=\sqrt[5]{1-(y-7)^3} \\ $Maka:$ \\ f^{-1}(x)=\sqrt[5]{1-(x-7)^3} \\ $Sehingga:$ \\ f^{-1}(6)=\sqrt[5]{1-(6-7)^3} \\ f^{-1}(6)=\sqrt[5]{1-(-1)^3} \\ f^{-1}(6)=\sqrt[5]{1-(-1)} \\ f^{-1}(6)=\sqrt[5]{1+1} \\ \boxed{f^{-1}(6)=\sqrt[5]{2}}$

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