Question

Diketahui fungsi f (x) = x^2 - 2x + 1 ,g(x) = 2x - 1 dan (f ¤ g)(t) =0 nilai t yang memenuhi

Answer 1

(fog)(x)
= f(g(x))
= (2x-1)^2 - 2.(2x-1) + 1
= (4x^2 - 4x + 1) - 4x + 2 + 1
= 4x^2 - 8x + 4

(fog)(t) = 0
4t^2 - 8t + 4 = 0
4t^2 - 4t - 4t + 4 = 0
4t(t-1) - 4(t-1) = 0
(4t-4)(t-1) = 0
t = 1

Answer 2

F(x)=x²-2x+1
g(x)= 2x-1

f0g(x)= (2x-1)²-2(2x-1)+1
f0g(x)= 4x²-4x+1-4x+2+1
f0g(x)=4x²-8x+4

f0g(t)= 4t²-8t+4
0=4t²-8t+4                  :4
0=t²-2t+1
(t-1)(t-1)=0
t=1
Semoga membantu