Answer
It is not true in general that: $f\circ (g+h) = f\circ g+f\circ h$
Work Step by Step
We can find a counterexample to show that the given statement is not true in general.
Suppose $f(x) = x^2$
Let $g(x) = 1$ and let $h(x) = x$
We can find $f \circ (g+h):$
$f \circ (g+h)$
$= f(x+1)$
$= (x+1)^2$
$= x^2+2x+1$
We can find $f\circ g + f \circ h$:
$f\circ g + f \circ h$
$= f(1) + f(x)$
$= (1)^2 + x^2$
$= x^2+1$
We can see that: $f\circ (g+h) \neq f\circ g+f\circ h$
