Q.If $f,g,h$ are real valued functions defined on $R$, then prove that $(f+g)\circ h=f\circ h+g\circ h$. What can you say about $f\circ(g+h)$? Justify your answer.

Step 1 (Proof that $(f+g)\circ h=f\circ h+g\circ h$). For every $x\in R$:

This chain of equalities holds for EVERY $x$ and for ANY functions $f,g,h$, so $(f+g)\circ h=f\circ h+g\circ h$ always. $\blacksquare$ (Composing on the right by $h$ distributes over addition.)

Step 2 (Investigating $f\circ(g+h)$). Compare $\big[f\circ(g+h)\big](x)=f(g(x)+h(x))$ with $\big[f\circ g+f\circ h\big](x)=f(g(x))+f(h(x))$. These agree for every $g,h$ ONLY if $f$ itself satisfies $f(a+b)=f(a)+f(b)$ for all $a,b$ (i.e. $f$ is "additive", such as $f(x)=cx$). For a general $f$ (e.g. anything nonlinear), the two sides differ. …

Prove the right-composition identity directly from definitions; test the left-composition versio …

• Assuming composition distributes over addition symmetrically on both sides just because it does on one side -- composing on the LEFT by a nonlinear …