Samuel Dominic Chukwuemeka (SamDom For Peace)

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Solved Examples on the Arithmetic Operations on Functions

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
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(1.) $ f(x) = x + 2 \\[3ex] g(x) = x^2 - x \\[3ex] $ Calculate $(f + g)(-5)$ using at least two methods.


First Method (long way)
$ (f + g)(x) \\[3ex] = f(x) + g(x) \\[3ex] = (x + 2) + (x^2 - x) \\[3ex] = x + 2 + x^2 - x \\[3ex] = x^2 + 2 \\[3ex] (f + g)(x) = x^2 + 2 \\[3ex] (f + g)(-5) \\[3ex] = ({-5})^2 + 2 \\[3ex] = 25 + 2 \\[3ex] = 27 \\[3ex] $ Second Method (short way)
$ (f + g)(-5) \\[3ex] = f(-5) + g(-5) \\[3ex] f(x) = x + 2 \\[3ex] f(-5) = -5 + 2 \\[3ex] = -3 \\[3ex] g(x) = x^2 - x \\[3ex] g(-5) = ({-5})^2 - (-5) \\[3ex] = 25 + 5 \\[3ex] = 30 \\[3ex] f(-5) + g(-5) \\[3ex] = -3 + 30 \\[3ex] = 27 $
(2.) $ f(x) = 10x^2 - 2x \\[3ex] g(x) = 2x \\[3ex] $ Calculate:
(a.) $(f + g)(x)$ and the domain of the sum
(b.) $(f - g)(x)$ and the domain of the difference
(c.) $(fg)(x)$ and the domain of the product
(d.) $\left(\dfrac{f}{g}\right)(x)$ and the domain of the quotient


$ (f + g)(x) = f(x) + g(x) \\[3ex] = (10x^2 - 2x) + 2x \\[3ex] = 10x^2 - 2x + 2x \\[3ex] = 10x^2 \\[3ex] D = (-\infty, \infty) \\[7ex] (f - g)(x) = f(x) - g(x) \\[3ex] = (10x^2 - 2x) - 2x \\[3ex] = 10x^2 - 2x - 2x \\[3ex] = 10x^2 - 4x \\[3ex] D = (-\infty, \infty) \\[7ex] (f * g)(x) = f(x) * g(x) \\[3ex] = (10x^2 - 2x) * 2x \\[3ex] = 2x(10x^2 - 2x) \\[3ex] = 20x^3 - 4x \\[3ex] D = (-\infty, \infty) \\[7ex] \left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)} \\[5ex] = (10x^2 - 2x) \div 2x \\[3ex] = \dfrac{10x^2 - 2x}{2x} \\[5ex] Factor\:\: by\:\: GCF \\[3ex] GCF = 2x \\[2ex] = \dfrac{2x(5x - 1)}{2x} \\[5ex] = 5x - 1 \\[3ex] D = (-\infty, \infty) $
(3.) $ f(x) = x^2 - 1 \\[3ex] g(x) = 18 - x \\[3ex] $ Calculate $(f - g)(-4)$ using at least two methods.


First Method (long way)
$ (f - g)(x) \\[3ex] = f(x) - g(x) \\[3ex] = (x^2 - 1) - (18 - x) \\[3ex] = x^2 - 1 - 18 + x \\[3ex] = x^2 + x - 19 \\[3ex] (f - g)(x) = x^2 - x + 17 \\[3ex] (f - g)(-4) \\[3ex] = (-4)^2 + (-4) - 19 \\[3ex] = 16 - 4 - 19 \\[3ex] = -7 \\[3ex] $ Second Method (short way)
$ (f - g)(-4) \\[3ex] = f(-4) - g(-4) \\[3ex] f(x) = x^2 - 1 \\[3ex] f(-4) = {-4}^2 - 1 \\[3ex] = 16 - 1 \\[3ex] = 15 \\[3ex] g(x) = 18 - x \\[3ex] g(-4) = 18 - (-4) \\[3ex] = 18 + 4 \\[3ex] = 22 \\[3ex] f(-4) - g(-4) \\[3ex] = 15 - 22 \\[3ex] = -7 $
(4.) $ f(x) = \dfrac{4}{x - 3} \\[5ex] g(x) = \sqrt{x - 2} \\[3ex] $ Calculate:
(a.) $(f + g)(x)$ and the domain of the sum
(b.) $(f - g)(x)$ and the domain of the difference
(c.) $(fg)(x)$ and the domain of the product
(d.) $\left(\dfrac{f}{g}\right)(x)$ and the domain of the quotient


$ (f + g)(x) = f(x) + g(x) \\[3ex] = \dfrac{4}{x - 3} + \sqrt{x - 2} \\[5ex] For\:\: \dfrac{4}{x - 3}; D = (-\infty, 3) \cup (3, \infty) \\[5ex] For\:\: \sqrt{x - 2}; D = [2, \infty) \\[3ex] $ Find the domain that is common to the sum of both functions
Domain of the sum, $D = [2, 3) \cup (3, \infty)$

$ (f - g)(x) = f(x) - g(x) \\[3ex] = \dfrac{4}{x - 3} - \sqrt{x - 2} \\[5ex] For\:\: \dfrac{4}{x - 3}; D = (-\infty, 3) \cup (3, \infty) \\[5ex] For\:\: \sqrt{x - 2}; D = [2, \infty) \\[3ex] $ Find the domain that is common to the difference of both functions
Domain of the difference, $D = [2, 3) \cup (3, \infty)$

$ (f * g)(x) = f(x) * g(x) \\[3ex] = \dfrac{4}{x - 3} * \sqrt{x - 2} \\[3ex] = \dfrac{4\sqrt{x - 2}}{x - 3} \\[3ex] For\:\: the\:\: denominator; D = (-\infty, 3) \cup (3, \infty) \\[3ex] For\:\: the\:\: numerator; D = [2, \infty) \\[3ex] $ Find the intersection of both domains
Domain of the product, $D = [2, 3) \cup (3, \infty)$

$ \left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)} \\[5ex] = \dfrac{4}{x - 3} \div \sqrt{x - 2} \\[5ex] = \dfrac{4}{x - 3} * \dfrac{1}{\sqrt{x - 2}} \\[5ex] = \dfrac{4}{(x - 3)\sqrt{x - 2}} \\[5ex] = \dfrac{4}{(x - 3)\sqrt{x - 2}} * \dfrac{\sqrt{x - 2}}{\sqrt{x - 2}} \\[5ex] = \dfrac{4\sqrt{x - 2}}{(x - 3)(x - 2)} \\[5ex] For\:\: the\:\: denominator; D = (-\infty, 2) \cup (2, 3) \cup (3, \infty) \\[3ex] For\:\: the\:\: numerator; D = [2, \infty) \\[3ex] $ Find the intersection of both domains
Domain of the quotient, $D = (2, 3) \cup (3, \infty)$