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 Take the intersection (i.e. common points) of Steps 2 and 3. <math>(  \infty, 1) \cup (2, \infty)</math>   Take the intersection (i.e. common points) of Steps 2 and 3. <math>(  \infty, 1) \cup (2, \infty)</math> 
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Revision as of 13:57, 23 February 2015
2. Find the domain of the following function. Your answer should use interval notation.
f(x) = $\displaystyle {\frac {1}{\sqrt {x^{2}x2}}}$
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 $x^{2}x2$

$x^{2}x2=(x+1)(x2)$

So we can rewrite f(x) as $f(x)=\displaystyle {\frac {1}{\sqrt {(x+1)(x2)}}}$

Step 2:

When does the denominator of f(x) = 0?

${\sqrt {(x+1)(x2)}}=0$

(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 $h(x)={\sqrt {(x+1)(x2)}}$

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) = $\displaystyle {\frac {1}{\sqrt {x^{2}x2}}}$
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 $x^{2}x2$

$x^{2}x2=(x+1)(x2)$

So we can rewrite f(x) as $f(x)=\displaystyle {\frac {1}{\sqrt {(x+1)(x2)}}}$

Step 2:

When does the denominator of f(x) = 0?

${\sqrt {(x+1)(x2)}}=0$

(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 $h(x)={\sqrt {(x+1)(x2)}}$

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.
$f(x)=\displaystyle {\frac {1}{\sqrt {x^{2}x2}}}$
Hint 1

Which xvalues lead to division by 0 or square rooting a negative number

Hint 2

Use a sign chart to determine for which xvalues $x^{2}x2>0$

Solution:
Solution

Since the domain is the collection of xvalues for which we don't divide by zero or square root a negative number we want to solve the inequality $x^{2}x2>0$

$(x2)(x+1)>0$

Now we use a sign chart with test numbers 2, 0, and 3

$x=2:(22)(2+1)=(4)(1)=4>0$

$x=0:(02)(0+1)=(2)(1)=2<0$

$x=3:(32)(3+1)=(1)(4)=4>0$

So the solution is $(\infty ,1)\cup (2,\infty )$

2. Find the domain of the following function. Your answer should use interval notation.
$f(x)=\displaystyle {\frac {1}{\sqrt {x^{2}x2}}}$
Hint 1

Which xvalues lead to division by 0 or square rooting a negative number

Hint 2

Use a sign chart to determine for which xvalues $x^{2}x2>0$

Solution:
Solution

Since the domain is the collection of xvalues for which we don't divide by zero or square root a negative number we want to solve the inequality $x^{2}x2>0$

$(x2)(x+1)>0$

Now we use a sign chart with test numbers 2, 0, and 3

$x=2:(22)(2+1)=(4)(1)=4>0$

$x=0:(02)(0+1)=(2)(1)=2<0$

$x=3:(32)(3+1)=(1)(4)=4>0$

So the solution is $(\infty ,1)\cup (2,\infty )$
