Tolong di jawab ya pake caranya :))))))))))
tolong banget !!!!
invers fungsi
adalah.....
thanks :)
tolong banget !!!!
invers fungsi
thanks :)
Brillianttt
![h(x) = x^{3} - 27 \\ \\ misalkan~ h(x) = y~maka \\ ~~~y = x^{3} - 27 \\ x^{3} = y+27 \\ ~~~x= \sqrt[3]{y+27} \\ \\ Jadi~h^{-1}(x)=\sqrt[3]{x+27}~atau~(x+27)^{ \frac{1}{3} } \\ Semoga~membantu... ](https://tex.z-dn.net/?f=h%28x%29+%3D+x%5E%7B3%7D+-+27++%5C%5C++%5C%5C+misalkan%7E+h%28x%29+%3D+y%7Emaka+%5C%5C+%0A%7E%7E%7Ey+%3D++x%5E%7B3%7D+-+27+%5C%5C+%0A+x%5E%7B3%7D+%3D+y%2B27+%5C%5C+%0A%7E%7E%7Ex%3D+%5Csqrt%5B3%5D%7By%2B27%7D++%5C%5C++%5C%5C+%0AJadi%7Eh%5E%7B-1%7D%28x%29%3D%5Csqrt%5B3%5D%7Bx%2B27%7D%7Eatau%7E%28x%2B27%29%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D++%5C%5C+Semoga%7Emembantu...%0A+%0A "h(x) = x^{3} - 27 \\ \\ misalkan~ h(x) = y~maka \\ ~~~y = x^{3} - 27 \\ x^{3} = y+27 \\ ~~~x= \sqrt[3]{y+27} \\ \\ Jadi~h^{-1}(x)=\sqrt[3]{x+27}~atau~(x+27)^{ \frac{1}{3} } \\ Semoga~membantu... ")
1 votes
Thanks 2
AnggitaSeveners
![h(x) = x^3 - 27 \\ \\ y = x^3 - 27 \\ \\ x^3 = y + 27 \\ \\ x = \sqrt[3]{y+27} \\ \\ h^-^1(x) = \sqrt[3]{y+27} = (y+27)^ \frac{1}{3}](https://tex.z-dn.net/?f=h%28x%29+%3D+x%5E3+-+27+%5C%5C++%5C%5C+y+%3D+x%5E3+-+27+%5C%5C++%5C%5C+x%5E3+%3D+y+%2B+27+%5C%5C++%5C%5C+x+%3D++%5Csqrt%5B3%5D%7By%2B27%7D++%5C%5C++%5C%5C+h%5E-%5E1%28x%29+%3D++%5Csqrt%5B3%5D%7By%2B27%7D++%3D+%28y%2B27%29%5E+%5Cfrac%7B1%7D%7B3%7D+ "h(x) = x^3 - 27 \\ \\ y = x^3 - 27 \\ \\ x^3 = y + 27 \\ \\ x = \sqrt[3]{y+27} \\ \\ h^-^1(x) = \sqrt[3]{y+27} = (y+27)^ \frac{1}{3}")
0 votes
Thanks 1
Brillianttt
jawabannya kurang tepat... 27 nya nggak bisa disederhanaain...
AnggitaSeveners
serius nggax bisa
Brillianttt
iya....
Brillianttt
kecuali itu dikali... kalau operasi jumlah dan kurang tak bisa....
AnggitaSeveners
oh, ya trima kasih
tyashigihara
jawaban yang bener yang mana?
