Function Arithmetic | Brilliant Math & Science Wiki

When performing arithmetic on functions, it is necessary to understand the rules that relate functions to each other. For the following examples, we'll use these functions:

$a(x) = x^2$
$b(x) = x+3$
$c(x) = 4x$ .

Addition: $(f+g)(x) = f(x) + g(x)$ , so for example,

$(b+c)(x) = b(x) + c(x) = (x+3) + (4x) = 5x + 3 .$

Subtraction: $(f-g)(x) = f(x) - g(x)$ , so for example,

$(b-c)(x) = b(x) - c(x) = (x+3) - (4x) = -3x + 3.$

Multiplication: $( f\cdot g)(x) = f(x) \cdot g(x)$ , so for example,

$(a\cdot b)(x) = (x^2) \cdot (x+3) = x^3 + 3x^2.$

Division: $\frac{f}{g}(x)=\frac{f(x)}{g(x)}$ , so for example,

$\frac{b}{c}(x)=\frac{b(x)}{c(x)}=\frac{x+3}{4x}.$

Exponents: $f^n(x) = \big(f(x)\big)^n$ where $n \in \mathbb N$ , so for example,

$a^5(x) = \big(a(x)\big)^5 = \big(x^2\big)^5 = x^{10}.$

Square Root: It is a part of "exponents," but here $n$ is not an integer. $\sqrt{f} (x) = \sqrt{f(x)}$ , so for example,

$\sqrt{b} (x) = \sqrt{b(x)} = \sqrt{x + 3}.$

Note: The domain of all these combinations is the intersection of the domains of $f$ and $g$ . That is, both functions must be defined at a point of the combination to be defined. Another requirement for division is that the denominator must be non-zero: in this case $g(x)\neq0$ .

What is $(f-g)(3)$ if $f(x) = x^2 + 1$ and $g(x) = 2x-2?$

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Since $f(3) = 3^2 + 1 = 10$ and $g(3) = 2(3) - 2 = 4$ , we can see that

$(f-g)(3) = f(3) - g(3) =10 - 4 = 6. \ _\square$

What is $(f+g)(4)$ if $f(x) = 5x^{2} + 4$ and $g(x) = 3x+6?$

Since $f(4) = 5\big(4^{2}\big)+4 = 84$ and $g(4) = 3(4) +6 = 18$ , we can see that

$(f+g)(4) = f(4) + g(3) =84 +18 = 102. \ _\square$

Given

$\begin{aligned} f(x)&=4{ x }^{ 3 }-5\\ g(x)&=6{ x }^{ 2 }+9\\ h(x)&=(f\cdot g)(x), \end{aligned}$

what is the value $h(3)?$

We know that

$h(3)=(f\cdot g)(3)=f(3)\cdot g(3).$

Since $f(3)=4(3)^{3}-5=103$ and $g(3)=6(3)^{2}+9=63,$

$\begin{aligned} h(3)&=f(3)\cdot g(3)\\ &=103\times 63\\ &=6489.\ _\square \end{aligned}$

Given

$\begin{aligned} f(x)&={ x }^{ 2 }+1\\ g(x)&=6x\\ h(x)&=3x+4, \end{aligned}$

what is the value of $\big((f\cdot h)-g\big)(3)?$

The expression can be rewritten as

$\begin{aligned} \big((f\cdot h)-g\big)(3)&=(f\cdot h)(3)-g(3)\\ &=f(3)\cdot h(3)-g(3). \end{aligned}$

Since $f(3)=3^{2}+1=10$ , $g(3)=6(3)=18$ , and $h(3)=3(3)+4=13,$ we have

$f(3)\cdot h(3)-g(3)=10\cdot 13-18=112.\ _\square$

In some questions, instead of giving the general form of the function along with its domain and co-domain, the function will be directly given mentioning for what value what output will come. In such a way, the value and its output will be enclosed and expressed as one pair in a pair of parenthesis " $( )$ ". So, every first term from each pair is taken as domain and the second term from each pair is taken as co-domain. The rules and algebra of functions are applicable here, too, but while applying any arithmetic operation between two functions, remember that the domain will not change and only co-domain will change. For example, if it is given that $f = {(x,a), (y,b), (z,c)},$ it means that if we give input as $x$ the output will be $a$ , if we give input as $y$ the output will be $b,$ and if we give input as $z$ the output will be $c,$ so if the domain is ${x,y,z}$ its co=domain is ${a,b,c}$ . The below-given example gives the complete picture on how to do problems related to this way of expressing:

If $f = \big\{(4,5),(5,6),(6,-4) \big\}$ and $g = \big\{ (4,-4),(6,5),(8,5) \big\},$ then find the values of the following:

$\ f + g$

$\ f - g$

$\ 2f + 4g$

$\ f + 4$

$\ fg$

$\ \frac{f}{g}$

$\ |f|$

$\ \sqrt{f}$

$\ f^2$

$\ f^3$

The domain of $f = A = \{4,5,6\}$ and the domain of $g = B = \{4,6,8\},$ and hence the domain of $f \pm g = A \cap B = {4,6}$ .

$f + g = \big\{(4,5 - 4),(6,-4 + 5)\big\} = \big\{(4,1), (6,1)\big\}$

$f - g = \big\{(4,5 + 4), (6, -4 - 5)\big\} = \big\{(4,9),(6,-9)\big\}$

The domain of $2f$ and $4g$ will not change but the co-domain will change. As we need $2f,$ multiply the second term in each pair by two and do the same for $4g:$ $2f = \big\{(4,10),(5,12),(6,-8)\big\},\ 4g = \big\{(4,-16),(6,20),(8,20)\big\}.$ The domain of $2f + 4g$ is the same as $f + g = {4,6}:$ $2f + 4g = \big\{(4,10 - 16), (6, -8 + 20)\big\} = \big\{(4,-6),(6,12)\big\}.$

$f + 4 = \big\{(4, 5+4),(5,6+4),(6,-4+4)\big\} = \big\{(4,9),(5,10),(6,0)\big\}$

The domain of $f \cdot g = A \cap B = {4,6}$ . So, we need only the pairs whose first term is 4 or 6 and only need to change their respective co-domains: $f \cdot g = \left\{\big(4, 5 \times (-4)\big),\big(6,(-4) \times (-5)\big)\right\} = \big\{(4,-20),(6,20)\big\}.$

Here too the domain of $\frac{f}{g} = {4,6}$ . So, we need to change only those two terms from the domain of each function: $\frac{f}{g} = \left\{\left(4,\dfrac{-5}{4} \right), \left(6,\dfrac{-4}{5} \right)\right\}.$

$|f| = \left\{\big(4,|5|\big),\big(5,|6|\big),\big(6,|-4|\big)\right\} = \big\{ (4,5),(5,6),(6,4) \big\}$

Here too the domain of $\sqrt{f}$ is $\{4,6\}$ as $\sqrt{-4}$ is not defined: $\sqrt{f} = \Big\{\big(4,\sqrt{5}\big),\big(5,\sqrt{6}\big)\Big\}.$

$f^2 = \Big\{\big(4,5^2\big),\big(5,6^2\big),\big(6,(-4)^2\big)\Big\} = \big\{(4,25),(5,36),(6,16)\big\}$

$f^3 = \Big\{\big(4,5^3\big),\big(5,6^3\big),\big(6,(-4)^3\big)\Big\} = \big\{(4,125),(5,216),(6,-64)\big\}$ $_\square$