Function Arithmetic

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 \( g(x) = 3x+6?\)

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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

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

what is the value \(h(3)?\)

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We know that

\[ h(3)=(f\cdot g)(3)=f(3)\cdot g(3).\]

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

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

Given

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

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

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The expression can be rewritten as

\[\begin{align} \big((f\cdot h)-g\big)(3)&=(f\cdot h)(3)-g(3)\\ &=f(3)\cdot h(3)-g(3). \end{align}\]

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:

1. \(\ f + g\)
2. \(\ f - g\)
3. \(\ 2f + 4g\)
4. \(\ f + 4\)
5. \(\ fg\)
6. \(\ \frac{f}{g}\)
7. \(\ |f|\)
8. \(\ \sqrt{f}\)
9. \(\ f^2\)
10. \(\ f^3\)

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

1. \(f + g = \big\{(4,5 - 4),(6,-4 + 5)\big\} = \big\{(4,1), (6,1)\big\}\)
2. \(f - g = \big\{(4,5 + 4), (6, -4 - 5)\big\} = \big\{(4,9),(6,-9)\big\}\)
3. 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 \(f + g = {4,6}:\) \[2f + 4g = \big\{(4,10 - 16), (6, -8 + 20)\big\} = \big\{(4,-6),(6,12)\big\}.\]
4. \(f + 4 = \big\{(4, 5+4),(5,6+4),(6,-4+4)\big\} = \big\{(4,9),(5,10),(6,0)\big\}\)
5. 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: \[ f \cdot g = \left\{\big(4, 5 \times (-4)\big),\big(6,(-4) \times (-5)\big)\right\} = \big\{(4,-20),(6,20)\big\}.\]
6. 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: \[\frac{f}{g} = \left\{\left(4,\dfrac{-5}{4} \right), \left(6,\dfrac{-4}{5} \right)\right\}.\]
7. \(|f| = \left\{\big(4,|5|\big),\big(5,|6|\big),\big(6,|-4|\big)\right\} = \big\{ (4,5),(5,6),(6,4) \big\}\)
8. Here too the domain of \(\sqrt{f}\) is \(\{4,6\}\) as \(\sqrt{-4}\) is not defined: \[\sqrt{f} = \Big\{\big(4,\sqrt{5}\big),\big(5,\sqrt{6}\big)\Big\}.\]
9. \(f^2 = \Big\{\big(4,5^2\big),\big(5,6^2\big),\big(6,(-4)^2\big)\Big\} = \big\{(4,25),(5,36),(6,16)\big\}\)
10. \(f^3 = \Big\{\big(4,5^3\big),\big(5,6^3\big),\big(6,(-4)^3\big)\Big\} = \big\{(4,125),(5,216),(6,-64)\big\}\) \(_\square\)