Answer:
Step-by-step explanation:
g(t)=t^2 - t
f(x) = (1 + x)
g(f(x)) = f(x)^2 - f(x)
g(f(x)) = (x + 1)^2 - x - 1
g(f(0)) = (0 +1)^2 - x - 1
g(f(0)) = 1 - 1 - 1 = -1
================================
f(x) = 1 + x
f(g(t)) = 1 + g(t)
f(g(t)) = 1 + t^2 - t
f(g(0)) = 1 + 0 - 0
f(g(0)) = 1
The answer I'm getting is 0.
I dont have paper with me but just solve 2(-2)^3+2 for the first one. The second one is -28 though
We know that x > 2 ( or : x ≠ +/- 2 )
We have to factorize the numerator and the denominator:
x² - x - 6 = x² - 3 x + 2 x - 6 = x ( x - 3 ) + 2 ( x - 3 ) = ( x - 3 ) ( x + 2 )
x ² - 4 = ( x - 2 ) ( x + 2 )
Answer:
(x+2)(x+1)=0
so x = -2,-1
Step-by-step explanation:
I’m pretty sure you just start solving for x like
x2 + 4 = x - 1
so on and so on