Diketahui f(x) = ax^7+bx^3+cx-5. Jika f(-7) = 7 tentukan nilai f(7) ?
rirismaulidina
![f(x)~~=ax^7+bx^3+cx-5 \\ f(-7)=a(-7)^7+b(-7)^5+c(-7)-5 \\ ~~~~7~~~~=-[a(7)^7+b(7)^5+c(7)+5] \\ ~~~~7~~~~=-[a(7)^7+b(7)^5+c(7)]-5 \\ a(7)^7+b(7)^5+c(7)=-5-7 \\ a(7)^7+b(7)^5+c(7)=-12 \\ \\ f(7)=a(7)^7+b(7)^5+c(7)-5 \\ f(7)=-12-5 \\ f(7)=-17](https://tex.z-dn.net/?f=f%28x%29%7E%7E%3Dax%5E7%2Bbx%5E3%2Bcx-5+%5C%5C+f%28-7%29%3Da%28-7%29%5E7%2Bb%28-7%29%5E5%2Bc%28-7%29-5+%5C%5C+%7E%7E%7E%7E7%7E%7E%7E%7E%3D-%5Ba%287%29%5E7%2Bb%287%29%5E5%2Bc%287%29%2B5%5D+%5C%5C+%7E%7E%7E%7E7%7E%7E%7E%7E%3D-%5Ba%287%29%5E7%2Bb%287%29%5E5%2Bc%287%29%5D-5+%5C%5C+a%287%29%5E7%2Bb%287%29%5E5%2Bc%287%29%3D-5-7+%5C%5C+a%287%29%5E7%2Bb%287%29%5E5%2Bc%287%29%3D-12+%5C%5C++%5C%5C+f%287%29%3Da%287%29%5E7%2Bb%287%29%5E5%2Bc%287%29-5+%5C%5C+f%287%29%3D-12-5+%5C%5C+f%287%29%3D-17 "f(x)~~=ax^7+bx^3+cx-5 \\ f(-7)=a(-7)^7+b(-7)^5+c(-7)-5 \\ ~~~~7~~~~=-[a(7)^7+b(7)^5+c(7)+5] \\ ~~~~7~~~~=-[a(7)^7+b(7)^5+c(7)]-5 \\ a(7)^7+b(7)^5+c(7)=-5-7 \\ a(7)^7+b(7)^5+c(7)=-12 \\ \\ f(7)=a(7)^7+b(7)^5+c(7)-5 \\ f(7)=-12-5 \\ f(7)=-17")
2 votes
Thanks 12
AlbertHM
woo
rirismaulidina
ada yang salah ya? maaf suka gak teliti
AlbertHM
bener kok! *hebat!
rirismaulidina
makasih :D
MathSolver74
Good :D
nadhifas1
Kak itu yang bx^3 kok jadi b(-7)^5 ga pangkat 3kak?
rirismaulidina
hahhaha iya salah ketik, makasih ya koreksinya
kalo di buat b(-7)^3 jawabannya juga tetap -12
sayang banget ga bisa di edit -_-
nadhifas1
ya :( ya uda ga apa Ka :) :D
shintiamustika
Makasihh. :D