Invers dari f(x)=(1-x^3)^1/5 +2 adalah?
rirismaulidina
![y=f(x) = (1-x^3)^ \frac{1}{5}+2 \\ y= (1-x^3)^ \frac{1}{5}+2 \\ y-2 = (1-x^3)^ \frac{1}{5} \\ --------kedua~ruas~dipangkatkan~5 \\ (y-2)^5 = (1-x^3) \\ x^3=1- (y-2)^5 \\ x = \sqrt[3]{1- (y-2)^5} \\ x=( 1- (y-2)^5)^ \frac{1}{3} \\ f'(x) = ( 1- (x-2)^5)^ \frac{1}{3}](https://tex.z-dn.net/?f=y%3Df%28x%29+%3D+%281-x%5E3%29%5E+%5Cfrac%7B1%7D%7B5%7D%2B2+%5C%5C++y%3D+%281-x%5E3%29%5E+%5Cfrac%7B1%7D%7B5%7D%2B2+%5C%5C+y-2+%3D+%281-x%5E3%29%5E+%5Cfrac%7B1%7D%7B5%7D+%5C%5C+--------kedua%7Eruas%7Edipangkatkan%7E5+%5C%5C+%28y-2%29%5E5+%3D+%281-x%5E3%29+%5C%5C+x%5E3%3D1-+%28y-2%29%5E5+%5C%5C+x+%3D++%5Csqrt%5B3%5D%7B1-+%28y-2%29%5E5%7D++%5C%5C+x%3D%28+1-+%28y-2%29%5E5%29%5E+%5Cfrac%7B1%7D%7B3%7D++%5C%5C+f%27%28x%29+%3D+%28+1-+%28x-2%29%5E5%29%5E+%5Cfrac%7B1%7D%7B3%7D "y=f(x) = (1-x^3)^ \frac{1}{5}+2 \\ y= (1-x^3)^ \frac{1}{5}+2 \\ y-2 = (1-x^3)^ \frac{1}{5} \\ --------kedua~ruas~dipangkatkan~5 \\ (y-2)^5 = (1-x^3) \\ x^3=1- (y-2)^5 \\ x = \sqrt[3]{1- (y-2)^5} \\ x=( 1- (y-2)^5)^ \frac{1}{3} \\ f'(x) = ( 1- (x-2)^5)^ \frac{1}{3}")
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