Prototype questions

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 we 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.e. common points) of Steps 2 and 3. $(-\infty ,-1)\cup (2,\infty )$

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 we 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 undefinend 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.e. common points) of Steps 2 and 3. $(-\infty ,-1)\cup (2,\infty )$

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

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 $x^{2}-x-2>0$

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 $x^{2}-x-2>0$
$(x-2)(x+1)>0$
Now we use a sign chart with test numbers -2, 0, and 3
$x=-2:(-2-2)(-2+1)=(-4)(-1)=4>0$
$x=0:(0-2)(0+1)=(-2)(1)=-2<0$
$x=3:(3-2)(3+1)=(1)(4)=4>0$
So the solution is $(-\infty ,-1)\cup (2,\infty )$

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

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 $x^{2}-x-2>0$

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 $x^{2}-x-2>0$
$(x-2)(x+1)>0$
Now we use a sign chart with test numbers -2, 0, and 3
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = -2: (-2 - 2)(-2 + 1) = (-4)(-1) = 4 > 0}
$x=0:(0-2)(0+1)=(-2)(1)=-2<0$
$x=3:(3-2)(3+1)=(1)(4)=4>0$
So the solution is $(-\infty ,-1)\cup (2,\infty )$