# Given F(x) = 3x+1, G(x) = 2x, H(x) = x^2, how do you find the rule for ((F @ G) @ H) (x)?

Apr 27, 2017

$\left(\left(F o G\right) o H\right) \left(x\right) = 6 {x}^{2} + 1.$

#### Explanation:

$F \left(x\right) = 3 x + 1 , G \left(x\right) = 2 x , \mathmr{and} , H \left(x\right) = {x}^{2.}$

$\therefore \left(F o G\right) \left(x\right) = F \left(G \left(x\right)\right) ,$

$= 3 G \left(x\right) + 1 , \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left[\because , F \left(x\right) = 3 x + 1 ,\right] ,$

$= 3 \left(2 x\right) + 1 , \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left[\because , G \left(x\right) = 2 x ,\right] ,$

$\therefore \left(F o G\right) \left(x\right) = 6 x + 1.$

Now, $\left(\left(F o G\right) o H\right) \left(x\right) = \left(\left(F o G\right)\right) \left(H \left(x\right)\right)$

$= 6 H \left(x\right) + 1 , \ldots \ldots \ldots \ldots . . \left[\because , \left(F o G\right) \left(x\right) = 6 x + 1 ,\right] ,$

$= 6 \left({x}^{2}\right) + 1 , \ldots \ldots \ldots \ldots \ldots . \left[\because , H \left(x\right) = {x}^{2} ,\right] ,$

$\Rightarrow \left(\left(F o G\right) o H\right) \left(x\right) = 6 {x}^{2} + 1.$