# Difference between revisions of "Prototype questions"

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

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

Solution:

Step 1:
Factor ${\displaystyle x^{2}-x-2}$
${\displaystyle x^{2}-x-2=(x+1)(x-2)}$
So we can rewrite f(x) as ${\displaystyle f(x)=\displaystyle {\frac {1}{\sqrt {(x+1)(x-2)}}}}$
Step 2:
When does the denominator of f(x) = 0?
${\displaystyle {\sqrt {(x+1)(x-2)}}=0}$
(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 ${\displaystyle \neq }$ -1 and x ${\displaystyle \neq }$ 2
Step 3:
What is the domain of ${\displaystyle h(x)={\sqrt {(x+1)(x-2)}}}$
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 ${\displaystyle (-\infty ,-1]\cup [2,\infty )}$
Step 4:
Take the intersection (i.e. common points) of Steps 2 and 3. ${\displaystyle (-\infty ,-1)\cup (2,\infty )}$

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

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

Solution:

Step 1:
Factor ${\displaystyle x^{2}-x-2}$
${\displaystyle x^{2}-x-2=(x+1)(x-2)}$
So we can rewrite f(x) as ${\displaystyle f(x)=\displaystyle {\frac {1}{\sqrt {(x+1)(x-2)}}}}$
Step 2:
When does the denominator of f(x) = 0?
${\displaystyle {\sqrt {(x+1)(x-2)}}=0}$
(x + 1)(x - 2) = 0
(x + 1) = 0 or (x - 2) = 0
x = -1 or x = 2
So, since the function is undefinend when the denominator is zero, x ${\displaystyle \neq }$ -1 and x ${\displaystyle \neq }$ 2
Step 3:
What is the domain of ${\displaystyle h(x)={\sqrt {(x+1)(x-2)}}}$
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 ${\displaystyle (-\infty ,-1]\cup [2,\infty )}$
Step 4:
Take the intersection (i.e. common points) of Steps 2 and 3. ${\displaystyle (-\infty ,-1)\cup (2,\infty )}$

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

Hint 1
Which x-values lead to division by 0 or square rooting a negative number
Hint 2
Use a sign chart to determine for which x-values ${\displaystyle x^{2}-x-2>0}$

Solution:

Solution
Since the domain is the collection of x-values for which we don't divide by zero or square root a negative number we want to solve the inequality ${\displaystyle x^{2}-x-2>0}$
${\displaystyle (x-2)(x+1)>0}$
Now we use a sign chart with test numbers -2, 0, and 3
${\displaystyle x=-2:(-2-2)(-2+1)=(-4)(-1)=4>0}$
${\displaystyle x=0:(0-2)(0+1)=(-2)(1)=-2<0}$
${\displaystyle x=3:(3-2)(3+1)=(1)(4)=4>0}$
So the solution is ${\displaystyle (-\infty ,-1)\cup (2,\infty )}$

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

Hint 1
Which x-values lead to division by 0 or square rooting a negative number
Hint 2
Use a sign chart to determine for which x-values ${\displaystyle x^{2}-x-2>0}$

Solution:

Solution
Since the domain is the collection of x-values for which we don't divide by zero or square root a negative number we want to solve the inequality ${\displaystyle x^{2}-x-2>0}$
${\displaystyle (x-2)(x+1)>0}$
Now we use a sign chart with test numbers -2, 0, and 3
${\displaystyle x=-2:(-2-2)(-2+1)=(-4)(-1)=4>0}$
${\displaystyle x=0:(0-2)(0+1)=(-2)(1)=-2<0}$
${\displaystyle x=3:(3-2)(3+1)=(1)(4)=4>0}$
So the solution is ${\displaystyle (-\infty ,-1)\cup (2,\infty )}$