150. Diketahui fungsi π(π₯) = β2π₯ + 2 , π(π₯) = βπ₯^2 , dan β(π₯) = π₯^2 + 2π₯ + 1 . Tentukan (π β β β π)(β2) !
a. 22
b. 32
c. 18
d. 17
e. 49
Jawaban :
π(π₯) = β2π₯ + 2
π(π₯) = βπ₯^2
β(π₯) = π₯^2 + 2π₯ + 1
(π β β β π)(β2)
= g (h (f (-2 )))
maka perlu dihitung f(-2) terlebih dahulu
π(-2) = β2(-2) + 2
f(-2) = 4 + 2
f (-2) = 6
selanjutnya,
(π β β β π)(β2)
= g (h (f (-2 )))
= g(h (6))
sehingga, perlu dihitung h (6)
β(π₯) = π₯^2 + 2π₯ + 1
h(6) = 6^2 + 2(6) + 1
h(6) = 36 + 12 + 1
h(6) = 49
selanjutnya,
(π β β β π)(β2)
= g (h (f (-2 )))
= g(h (6))
= g(49)
π(π₯) = βπ₯^2
π(49) = β(49)^2
g(49) = 49
Jawaban : E
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