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Precalculus

Presented below is the template for one of the sample questions Parker presented during 302.

2. Question Statement

Foundations
The foundations:
Provide an short explanation about the prerequisite material required to complete this problem.

Solution:

Step 1:
Provide as many steps as necessary to complete the problem.
The steps should split the solution based on the foundation topics

Step 2:
Additional step provided to make the template longer

Example

2. Find the domain of the following function. Your answer should use interval notation. f(x) = $\displaystyle {\frac {1}{\sqrt {x^{2}-x-2}}}$

Foundations
The foundations:
What is the domain of g(x) = ${\frac {1}{x}}$?
The function is undefined if the denominator is zero, so x $\neq$0.
Rewriting"x $\neq$0" in interval notation( $-\infty$, 0) $\cup$(0, $\infty$)
What is the domain of h(x) = ${\sqrt {x}}$?
The function is undefined if wwe have a negative number inside the square root, so x $\geq$ 0

Solution:

Step 1:
Factor $x^{2}-x-2$
$x^{2}-x-2=(x+1)(x-2)$
So we can rewrite f(x) as f(x) = $\displaystyle {\frac {1}{\sqrt {(x+1)(x-2)}}}$

Step 2:
When does the denominator of f(x) = 0?
$sqrt{(x+1)(x-2)}=0$
(x + 1)(x - 2) = 0
(x + 1) = 0 or (x - 2) = 0
x = -1 or x = 2
So, since the function is undefiend when the denominator is zero, x $\neq$ -1 and x $\neq$ 2

Step 3:
What is the domain of h(x) = ${\sqrt {(x+1)(x-2)}}$
critical points: x = -1, x = 2
Test points:
x = -2: (-2 + 1)(-2 - 2): (-1)(-4) = 4 > 0
x = 0: (0 + 1)(0 - 2) = -2 < 0
x = 3: (3 + 1)(3 - 2): 4*1 = 4 > 0
So the domain of h(x) is ($-\infty$, -1] $\cup$ [2, $\infty$)

Step 4:
Take the intersection (i.3. common points) of Steps 2 and 3. ( $-\infty$, -1) $\cup$ (2, $\infty$)