If f(x) = x + 3 and g(x) = 2x - 7, what is (f@g)(x)?

Oct 17, 2014

$\left(f \circ g\right) \left(x\right)$ is equivalent to $f \left(g \left(x\right)\right)$. We solve this problem just as we solve $f \left(x\right)$. But since it asks us to find out $f \left(g \left(x\right)\right)$, in $f \left(x\right)$, each time we encounter x, we replace it with $g \left(x\right)$.
In the above problem, $f \left(x\right) = x + 3$.
Therefore, $f \left(g \left(x\right)\right) = g \left(x\right) + 3$.
$\implies \left(f \circ g\right) \left(x\right) = 2 x - 7 + 3$
$\implies \left(f \circ g\right) \left(x\right) = 2 x - 4$

Basically, write the $g \left(x\right)$ equation where you see the x in the $f \left(x\right)$ equation.

$f \circ g \left(x\right) = \left(g \left(x\right)\right) + 3$ Replace $g \left(x\right)$ with the equation
$f \circ g \left(x\right) = \left(2 x - 7\right) + 3$
$f \circ g \left(x\right) = 2 x - 7 + 3$ we just took away the parentheses
$f \circ g \left(x\right) = 2 x - 4$ Because the $- 7 + 3 = 4$
This is it
$g \circ f \left(x\right)$ would be the other way around

$g \circ f \left(x\right) = 2 \left(x + 3\right) - 7$

now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them.
$g \circ f \left(x\right) = 2 x + 6 - 7$

Next, $+ 6 - 7 = - 1$

$g \circ f \left(x\right) = 2 x - 1$