When we work with functions, we have a specific notation that applies. Generally, instead of y, we will now see f(x), g(x), or h(x). When we have a function, we can use the notation to ask for the function's value, given a certain input. If we see f(2), this means replace the independent variable with 2 and evaluate.

Test Objectives
• Demonstrate the ability to write a function using function notation
• Demonstrate an understanding of f(a), where a is a real number
• Demonstrate an understanding of f(x + a), where a is a real number
Function Notation Practice Test:

#1:

Instructions: for the given function, find f(1), f(9), and f(2).

a)

#2:

Instructions: for the given function, find f(-3), f(1), and f(5).

a)

#3:

Instructions: for the given function, find f(-2) and f(2).

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

Instructions: for the given function, find f(1) and f(6).

$$b)\hspace{.2em}f(x)=\frac{1}{x^2 - 5x - 6}$$

#4:

Instructions: for the given function, find f(10) and f(-2).

$$a)\hspace{.2em}f(x)=\sqrt{3x + 6}$$

Instructions: for the given function, find f(a) and f(a + 1).

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

#5:

Instructions: for the given function, find f(z - 1) and f((z2)).

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

Instructions: for the given function, find f(q - 3) and f((q3)).

$$b)\hspace{.2em}f(x)=\sqrt[3]{x}+ 7$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}f(1)=3, f(9)=19, f(2)=-4$$

#2:

Solutions:

$$a)\hspace{.2em}f(-3)=3, f(1)=-1, f(5)=3$$

#3:

Solutions:

$$a)\hspace{.2em}f(-2)=5, f(2)=-7$$

$$b)\hspace{.2em}f(1)=-\frac{1}{10}, f(6) \hspace{.2em}is \hspace{.2em}undefined$$

#4:

Solutions:

$$a)\hspace{.2em}f(10)=6, f(-2)=0$$

$$b)\hspace{.2em}f(a)=a^2 - 2a - 1$$ $$f(a + 1)=a^2 - 2$$

#5:

Solutions:

$$a)\hspace{.2em}f(z - 1)=z^3 - 3z^2 + 3z + 1$$ $$f(z^2)=z^6 + 2$$

$$b)\hspace{.2em}f(q - 3)=\sqrt[3]{q - 3}+ 7$$ $$f(q^3)=q + 7$$