Question

precalculus chapter 1.1 - 1.2 and chapter 2 quiz review
directions: you may use a calculator for all problems. show all work for each problem.
chapter 1.1: determining functions

1. circle the graphs that are functions. put an x over graphs that are not functions.

chapter 1.1 evaluating and solving functions

1. using the function f(x) = √(x + 2), evaluate and solve for the following:

a) evaluate for f(2)
b) evaluate for f(98)
c) solve for f(x) = 5
d) solve for f(x) = 0
chapter 1.1. identifying domain and range

1. what is the domain of the following relation? {(3, 5), (4, 10), (5, 15), (6, 20)}
2. what is the range of the function from #3?
3. is the relation from #3 a function? explain your answer.

Explanation:

Step1: Use vertical - line test for graphs

If a vertical line intersects the graph at most once, it is a function. Graphs a and b pass the vertical - line test, so they are functions. Graph c fails as a vertical line can intersect it more than once.

Step2: Evaluate f(2)

Substitute x = 2 into f(x)=\sqrt{x + 2}.
$f(2)=\sqrt{2+2}=\sqrt{4}=2$

Step3: Evaluate f(98)

Substitute x = 98 into f(x)=\sqrt{x + 2}.
$f(98)=\sqrt{98 + 2}=\sqrt{100}=10$

Step4: Solve f(x)=5

Set \sqrt{x + 2}=5. Square both sides: $x + 2=25$, then $x=23$.

Step5: Solve f(x)=0

Set \sqrt{x + 2}=0. Square both sides: $x+2 = 0$, then $x=-2$.

Step6: Find domain of relation

The domain of a relation \{(x,y)\} is the set of all x - values. For \{(3,5),(4,10),(5,15),(6,20)\}, the domain is \{3,4,5,6\}.

Step7: Find range of relation

The range of a relation \{(x,y)\} is the set of all y - values. For \{(3,5),(4,10),(5,15),(6,20)\}, the range is \{5,10,15,20\}.

Step8: Determine if relation is a function

In the relation \{(3,5),(4,10),(5,15),(6,20)\}, each x - value has exactly one y - value. So it is a function.

Answer:

1. a and b are functions, c is not.
2. a) $f(2)=2$; b) $f(98)=10$; c) $x = 23$; d) $x=-2$
3. Domain: \{3,4,5,6\}
4. Range: \{5,10,15,20\}
5. Yes, because each x - value has exactly one y - value.