If f (x) = 1-2x, G [f (x)] = 1-x & # 178; (x is not equal to 0), then G (1 / 2) = g (1 / 2) brings 1 / 2 directly into G [f (x)] = 1-x & # 178/ If the function f (x) = 1-2x, G [f (x)] = 1-x & # 178 / / X & # 178; (x is not equal to 0), then G (1 / 2)= G (1 / 2) directly takes 1 / 2 into G [f (x)] = 1-x & # 178 / / X & # 178; why not take it into f (x) and then calculate x into another Isn't this the wrong way to get x

If f (x) = 1-2x, G [f (x)] = 1-x & # 178; (x is not equal to 0), then G (1 / 2) = g (1 / 2) brings 1 / 2 directly into G [f (x)] = 1-x & # 178/ If the function f (x) = 1-2x, G [f (x)] = 1-x & # 178 / / X & # 178; (x is not equal to 0), then G (1 / 2)= G (1 / 2) directly takes 1 / 2 into G [f (x)] = 1-x & # 178 / / X & # 178; why not take it into f (x) and then calculate x into another Isn't this the wrong way to get x

Yes, G [f (x)] = 1-x & sup2 / / X & sup2;
Instead, G [x] = 1-x & sup2 / X & sup2;
So g (1 / 2) and G [f (x)] = 1-x & sup2 / X & sup2; are compared
F (x) = 1 / 2, not x = 1 / 2
So we can't substitute x = 1 / 2 into 1-x & sup2 / X & sup2 directly;

If f (x) = 1-2x, G [f (x)] = 1 − x2x2 (x ≠ 0), then the value of G (12) is ()
A. 1B. 3C. 15D. 30

Let f (x) = 1-2x = 12, then we can get x = 14. G [f (x)] = 1 − x2x2 (x ≠ 0), х g (12) = 1 − (14) 2 (14) 2 = 15, so we choose C

If G (x) = 2x-1, f [g (x)] = the square of X / the square of 1-x (x is not equal to plus or minus 1), then f (- 1 / 2) is equal to___ A.1 B.1/3 C.1/15 D.1/30
The answer is C

f(-1/2)=f[g(x)]
g(x)=-1/2
x=1/4
Take x into the square of F [g (x)] = x / 1-x (x is not equal to plus or minus 1),
So f [g (x)] = 1 / 15

Given that G (x) = 1-2x, f [g (x)] = 1-x ^ 2 / x ^ 2 (x is not equal to 0), then f (1 / 2) = 15, why is it equal to 15

g(x)=1-2x=t
x=1/2(1-t)
f[g(x)]=f(t)=(1-x^2)/x^2=1/x^2-1=4/(1-t)^2-1
f(x)=4/(1-x)^2-1
So f (1 / 2) = 15