Algebra of Functions

Algebra of functions. Let $f$ and $g$ be functions with domains $A$ and $B$ respectively. Then the four combined functions $f+g$, $f-g$, $fg$, $\dfrac fg$ are all defined on the intersection $A\cap B$ (the set of inputs valid for both $f$ and $g$ at once), as follows: $(f+g)(x)=f(x)+g(x)$; $(f-g)(x)=f(x)-g(x)$; $(f\cdot g)(x)=f(x)\cdot g(x)$; $\left(\dfrac fg\right)(x)=\dfrac{f(x)}{g(x)}$ where $g(x)\ne0$ (this last one needs the further restriction that $g(x)$ isn't zero).

Ex. 1: If $f(x)=x^2+2$ and $g(x)=5x-8$, find (i) $(f+g)(1)$, (ii) $(f-g)(-2)$, (iii) $(fg)(3m)$, (iv) $\dfrac fg(0)$.

Solution: (i) $(f+g)(1)=f(1)+g(1)=[(1)^2+2]+[5(1)-8]=3+(-3)=0$.

(ii) $(f-g)(-2)=f(-2)-g(-2)=[(-2)^2+2]-[5(-2)-8]=[4+2]-[-10-8]=6+18=24$.

(iii) $(fg)(3m)=f(3m)\cdot g(3m)=[(3m)^2+2][5(3m)-8]=[9m^2+2][15m-8]=135m^3-72m^2+30m-16$.

(iv) $\dfrac fg(0)=\dfrac{f(0)}{g(0)}=\dfrac{0^2+2}{5(0)-8}=\dfrac{2}{-8}=-\dfrac14$.

Ex. 2: Given $f(x)=5x^2$ and $g(x)=\sqrt{4-x}$, find the domain of (i) $(f+g)(x)$, (ii) $(f\circ g)(x)$ [as printed — really the product/combination], (iii) $\dfrac fg(x)$.

Solution: (i) Domain of $f(x)=5x^2$ is $(-\infty,\infty)$. To find the domain of $g(x)=\sqrt{4-x}$: $4-x\ge0 \implies x\le4$, so domain is $(-\infty,4]$. Therefore, domain of $(f+g)(x)$ is $(-\infty,\infty)\cap(-\infty,4]=(-\infty,4]$. …