When we work with functions, we use a very specific notation to ask for the function's value given a certain input for the independent variable. Additionally, we will practice creating a new function by adding, subtracting, multiplying, and dividing functions. We will need to be able to determine the domain for the newly created function.

Test Objectives
• Demonstrate an understanding of function notation
• Demonstrate the ability to add or subtract functions
• Demonstrate the ability to multiply or divide functions
• Demonstrate the ability to find the domain for a function
Operations on Functions Practice Test:

#1:

Instructions: find (f + g)(x), (f - g)(x), and state the domain.

$$a)\hspace{.2em}$$$$f(x)=x^2 + 5x - 1$$$$g(x)=-x - 3$$

$$b)\hspace{.2em}$$$$f(x)= 7 - x^2$$$$g(x)= x^2 - 4$$

#2:

Instructions: find (fg)(x), (f/g)(x), and state the domain.

$$a)\hspace{.2em}$$$$f(x)=-4x - 1$$$$g(x)=2x^2 - x$$

$$b)\hspace{.2em}$$$$f(x)=\sqrt{x - 2}$$$$g(x)=\frac{x}{x + 11}$$

#3:

Instructions: find (4f - 5g)(x) and state the domain.

$$a)\hspace{.2em}$$ $$f(x)=9x - 1$$ $$g(x)=5x^2 - x - 2$$

$$b)\hspace{.2em}$$ $$f(x)=\frac{3x - 2}{2}$$ $$g(x)=x - 1$$

#4:

Instructions: find (f + g)(x), (fg)(x), and state the domain. Find (f + g)(5) and (fg)(-1).

$$a)\hspace{.2em}$$ $$f(x)=\sqrt{7x + 1}$$ $$g(x)=3x + 5$$

$$b)\hspace{.2em}$$ $$f(x)=\frac{\sqrt{x + 1}}{2}$$ $$g(x)=\frac{9x - 5}{x}$$

#5:

Instructions: find (f - g)(x), (f/g)(x), and state the domain. Find (f - g)(1) and (f/g)(2).

$$a)\hspace{.2em}$$ $$f(x)=\sqrt{x - 1}$$ $$g(x)=\sqrt{5 - x^2}$$

$$b)\hspace{.2em}$$ $$f(x)=\sqrt[3]{x - 8}$$ $$g(x)=\sqrt[3]{8 - x}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}$$ $$(f + g)(x)=x^2 + 4x - 4$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$ $$(f - g)(x) = x^2 + 6x + 2$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$

$$b)\hspace{.2em}$$ $$(f + g)(x) = 3$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$ $$(f - g)(x) = -2x^2 + 11$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$

#2:

Solutions:

$$a)\hspace{.2em}$$ $$(fg)(x)=-8x^3 + 2x^2 + x$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$ $$(f/g)(x)=-\frac{4x + 1}{x(2x - 1)}$$ $$\text{Domain}: \left\{x | x ≠ 0, \frac{1}{2}\right\}$$

$$b)\hspace{.2em}$$ $$(fg)(x)= \frac{x\sqrt{x - 2}}{x + 11}$$ $$\text{Domain}: \{x | x ≥ 2\}$$ $$(f/g)(x)=\frac{(x+11)\sqrt{x - 2}}{x}$$ $$\text{Domain}: \left\{x | x ≥ 2\right\}$$

#3:

Solutions:

$$a)\hspace{.2em}$$ $$(4f - 5g)(x) = -25x^2 + 41x + 6$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$

$$b)\hspace{.2em}$$ $$(4f - 5g)(x) = x + 1$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$

#4:

Solutions:

$$a)\hspace{.2em}$$ $$(f + g)(x) = \sqrt{7x + 1} + 3x + 5$$ $$\text{Domain}: \left\{x | x ≥ -\frac{1}{7}\right\}$$ $$(fg)(x)=(3x + 5)\sqrt{7x + 1}$$ $$\text{Domain}: \left\{x | x ≥ -\frac{1}{7}\right\}$$ $$(f + g)(5)=26$$ $$(fg)(-1) \hspace{.2em}\text{is} \hspace{.2em}\text{undefined}$$

$$b)\hspace{.2em}$$ $$(f + g)(x) = \frac{x\sqrt{x + 1} + 18x - 10}{2x}$$ $$\text{Domain}: \left\{x | x ≥ -1, x ≠ 0\right\}$$ $$(fg)(x)=\frac{(9x - 5)\sqrt{x + 1}}{2x}$$ $$\text{Domain}: \left\{x | x ≥ -1, x ≠ 0\right\}$$ $$(f + g)(5)=\frac{\sqrt{6} + 16}{2}$$ $$(fg)(-1) = 0$$

#5:

Solutions:

$$a)\hspace{.2em}$$ $$(f - g)(x) = \sqrt{x - 1} - \sqrt{5 - x^2}$$ $$\text{Domain}: \{x | 1 ≤ x ≤ \sqrt{5}\}$$ $$\left(\frac{f}{g}\right)(x)=\frac{\sqrt{x - 1}}{\sqrt{5 - x^2}}$$ Note: You may want to rationalize the denominator. The analysis of the domain should be done before you take this step. $$\left(\frac{f}{g}\right)(x)=\frac{\sqrt{(x - 1)(5 - x^2)}}{5 - x^2}$$ $$\text{Domain}: \{x | 1 ≤ x < \sqrt{5}\}$$ $$(f - g)(1) = -2$$ $$\left(\frac{f}{g}\right)(2)=1$$

$$b)\hspace{.2em}$$ $$(f - g)(x) = 2\sqrt[3]{x - 8}$$ $$\text{Domain}: \{x | x ∈ \mathbb{R}\}$$ $$\left(\frac{f}{g}\right)(x)= -1$$ $$\text{Domain}: \{x | x ≠ 8\}$$ $$(f - g)(1) = -2\sqrt[3]{7}$$ $$\left(\frac{f}{g}\right)(2)= -1$$