Operation with functions

To continue, we first need to quickly have a look at operations between functions. Consider two functions $f$ and $g$ , and a constant $c$ . It is possible to form a new function $h$ in several ways based on $f$ , $g$ and $c$ :

Definition 1

We can create the following new functions

Equation 1

c\cdot f,\, f+g,\, f-g,\, f\cdot g,\, \frac{f}{g},\, f\circ g

which are defined as follows:

$(c\cdot f)(x)=c\cdot f(x)$ (multiplication of a function with a constant)
$(f+g)(x)=f(x)+g(x),\,(f-g)(x)=f(x)+g(x)$ (addition or subtraction of two functions)
$(f\cdot g)(x)=f(x)\cdot g(x)\, \frac{f}{g}(x)=\frac{f(x)}{g(x)}$ (multiplication or division of two functions)
$(f\circ g)(x)=f(g(x))$ (composition of two functions)

for all input values $x$ .

Note 1

The composition of two functions $f(g(x))$ is a chain, where the output of $g(x)$ is the input of $f$ :

\begin{array}{cll} x & \\ \large\downarrow & \\ \large\boxed{g} & \\ \large\downarrow & \\ y=g(x) & \\ \large\downarrow & \\ \large\boxed{f} & \\ \large\downarrow & \\ y=f(g(x)) \end{array}

Example 1

Consider the functions $f(x)=3x$ and $g(x)=x^2+1$ , and $c=4$ .

$(4\cdot f)(2) = 4\cdot f(2) = 4\cdot 6= 24$
$(f+g)(2)=f(2)+g(2)=6+5=11$
$(f-g)(2)=f(2)-g(2)=6-5=1$
$(f\cdot g)(2)=f(2)\cdot g(2)=6\cdot 5=30$
$\frac{f}{g}(2)=\frac{f(2)}{g(2)}=\frac{6}{5}=1.2$
$(f\circ g)(2)=f(g(2))=f(5)=15$

Exercise 1

Consider the functions $f(x)=x^2$ , $g(x)=0.5x$ , and the constant $c=2$ . Find the function equation of $h$ , where

$h=c\cdot f$
$h=f \pm g$
$h=f\cdot g$
$h=\frac{f}{g}$
$h = f \circ g$

Simplify the expression for $h$ as much as possible.

Solution

$h(x)=c\cdot f(x)= 2\cdot x^2 =2x^2$
$h(x)=f(x)+g(x)=x^2+0.5x=x(x+0.5)$ $h(x)=f(x)-g(x)=x^2-0.5x=x(x-0.5)$
$h(x)=f(x)\cdot g(x)=x^2\cdot 0.5x=0.5x^3$
$h(x)=\frac{f(x)}{g(x)}=\frac{x^2}{0.5x}=2x$
$h(x)=f(g(x))=(0.5x)^2=0.25x^2$

Note that in general it is

u\circ v \neq v\circ u

See the following exercise ... .

Exercise 2

Determine the function equation of $u=f\circ g$ and $v=g\circ f$ . Simplify the function equation of $u$ and $v$ as much as possible.

$f(x)=\sqrt{x}\, \text{ and }\, g(x)=x^3$
$f(x)=x^2+2x+2\, \text{ and }\, g(x)=3x$
$f(x)=x^2+2x+2\, \text{ and }\, g(x)=x+1$
$f(x)=\frac{2}{x}\, \text{ and }\, g(x)=2 x^2$
$f(x)=(2x+1)^2\, \text{ and }\, g(x)=x-1$

Solution

$u(x)=f(g(x))=\sqrt{x^3} =x^{3/2}$ , $v(x)=g(f(x))=(\sqrt{x})^3 = x^{3/2}$
$u(x)=f(g(x))=(3x)^2+2(3x)+2=9x^2+6x+2$ , $v(x)=g(f(x))=3(x^2+2x+2)=3x^2+6x+6$
$u(x)=f(g(x))=(x+1)^2+2(x+1)+2=x^2+4x+5$ , $v(x)=g(f(x))=x^2+2x+2+1=x^2+2x+3$
$u(x)=f(g(x))=\frac{2}{2x^2}=\frac{1}{x^2}$ , $v(x)=g(f(x))=2 \left(\frac{2}{x}\right)^2=\frac{8}{x^2}$
$u(x)=f(g(x))=(2(x-1)+1)^2=(2x-1)^2=4x^2-4x+1$ , $v(x)=g(f(x))=(2x+1)^2-1=4x^2+4x =4x(x+1)$

Exercise 3

Consider two functions $f$ and $g$ with $f(2)=7, f(3)=8.5, g(2)=3$ . Determine $h(2)$ for

$h=4\cdot f$
$h=f+g$
$h=f\cdot g$
$h=\frac{f}{g}$
$h=f\circ g$

Solution

$h(2)=4\cdot f(2)= 4\cdot 7=28$
$h(2)=f(2)+g(2)=7+3=10$
$h(2)=f(2)\cdot g(2)=7\cdot 3=21$
$h(2)=\frac{f(2)}{g(2)}=\frac{7}{3}$
$h(2)=f(g(2))=f(3)=8.5$

Exercise 4

Consider the graph of the following functions $f$ and $g$ shown below. Construct the graph of the function $u=f+g$ and $v=f\circ g$ without determining the function equations of $f$ and $g$ .

Solution