Domain and Range of Functions 3

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Question 1 of 4

1. Question
Graph and find the domain and range

$y=sqrt(x-5)$
- 1. Domain: All real numbers; Range: All real numbers
- 2. Domain: All real numbers $≠5$ ; Range: All real numbers $≠0$
- 3. Domain: $x>5$ ; Range: $y>0$
- 4. Domain: $x≥5$ ; Range: $y≥0$
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The domain and range is the set of $x$ and $y$ values of a function

First, use a table of values and test several values of $x$ to get the value of $y$

$x$ $5$ $6$ $7$ $9$ $10$

$y$

Substitute the values of $x$ to the function to get their $y$ values

$x=5$

$f(x)$ $=$ $sqrt(x-5)$

$f(5)$ $=$ $sqrt(5-5)$ Substitute $x=5$

$=$ $sqrt0$

$=$ $0$

$x=6$

$f(x)$ $=$ $sqrt(x-5)$

$f(6)$ $=$ $sqrt(6-5)$ Substitute $x=6$

$=$ $sqrt1$

$=$ $1$

$x=7$

$f(x)$ $=$ $sqrt(x-5)$

$f(7)$ $=$ $sqrt(7-5)$ Substitute $x=7$

$=$ $sqrt2$

$x=9$

$f(x)$ $=$ $sqrt(x-5)$

$f(9)$ $=$ $sqrt(9-5)$ Substitute $x=9$

$=$ $sqrt4$

$=$ $2$

$x=10$

$f(x)$ $=$ $sqrt(x-5)$

$f(10)$ $=$ $sqrt(10-5)$ Substitute $x=10$

$=$ $sqrt5$

$x$ $5$ $6$ $7$ $9$ $10$

$y$ $0$ $1$ $sqrt2$ $2$ $sqrt5$

Next, plot the points to the graph and connect them to form the curve.

Notice that the curve Starts at the point $(5,0)$ and extends infinitely to the right side

This means that the curves will cover all real values of $x$ greater than or equal to $5$ and all real values of $y$ greater than or equal to $0$

$\text(Domain:) x≥5$

$\text(Range:) y≥0$
Question 2 of 4

2. Question
Find the domain

$y=3/(2x-1)$
- 1. All real numbers $≠1/2$
- 2. All whole numbers
- 3. All real numbers greater than $>1/2$
- 4. All real numbers less than $<1/2$
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The domain and range is the set of $x$ and $y$ values of a function

For finding the domain, focus on the value of the denominator

Recall that fractions cannot have $0$ as the denominator or the value will be undefined

$2x-1$ $≠$ $0$

$2x-1$ $+1$ $≠$ $0$ $+1$ Add $1$ to both sides

$2x$ $divide2$ $≠$ $1$ $divide2$ Divide both sides by $2$

$x$ $≠$ $1/2$

Since $x=1/2$ makes the denominator equal to $0$ , $x≠\frac{1}{2}$

Therefore, the domain is all real numbers $≠\frac{1}{2}$

$\text(Domain: All real numbers)≠1/2$
Question 3 of 4

3. Question
Find the domain

$y=1/(sqrt(2x-3))$
- 1. Domain: $x≥3/2$
- 2. Domain: $x>3/2$
- 3. Domain: $x<3/2$
- 4. Domain: $x≠3/2$
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The domain and range is the set of $x$ and $y$ values of a function

For finding the domain, focus on the value of the denominator

Recall that fractions cannot have $0$ as the denominator or the value will be undefined. Also, since the value of the denominator is a square root, it cannot have a negative value.

$2x-3$ $>$ $0$

$2x-1$ $+3$ $>$ $0$ $+3$ Add $3$ to both sides

$2x$ $divide2$ $>$ $3$ $divide2$ Divide both sides by $2$

$x$ $>$ $3/2$

Therefore, the domain is all real numbers greater than $3/2$

$\text(Domain):x>3/2$
Question 4 of 4

4. Question
Find the domain

$y=3/(sqrt(3-4x))$
- 1. Domain: $x<3/4$
- 2. Domain: $x>3/4$
- 3. Domain: $x≥3/4$
- 4. Domain: $x≠3/4$
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The domain and range is the set of $x$ and $y$ values of a function

For finding the domain, focus on the value of the denominator

Recall that fractions cannot have $0$ as the denominator or the value will be undefined. Also, since the value of the denominator is a square root, it cannot have a negative value.

$3-4x$ $>$ $0$

$3-4x$ $-3$ $>$ $0$ $-3$ Subtract $3$ from both sides

$-4x$ $divide(-4)$ $>$ $-3$ $divide(-4)$ Divide both sides by $-4$

$x$ $<$ $3/4$ Flip the inequality

Therefore, the domain is all real numbers less than $3/4$

$\text(Domain):x<3/4$

Domain and Range of Functions 3

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Quizzes

$f(x)$	$=$	$sqrt(x-5)$
$f(5)$	$=$	$sqrt(5-5)$	Substitute $x=5$
	$=$	$sqrt0$
	$=$	$0$

$f(x)$	$=$	$sqrt(x-5)$
$f(6)$	$=$	$sqrt(6-5)$	Substitute $x=6$
	$=$	$sqrt1$
	$=$	$1$

$f(x)$	$=$	$sqrt(x-5)$
$f(7)$	$=$	$sqrt(7-5)$	Substitute $x=7$
	$=$	$sqrt2$

$f(x)$	$=$	$sqrt(x-5)$
$f(9)$	$=$	$sqrt(9-5)$	Substitute $x=9$
	$=$	$sqrt4$
	$=$	$2$

$f(x)$	$=$	$sqrt(x-5)$
$f(10)$	$=$	$sqrt(10-5)$	Substitute $x=10$
	$=$	$sqrt5$

$2x-1$	$≠$	$0$
$2x-1$ $+1$	$≠$	$0$ $+1$	Add $1$ to both sides
$2x$ $divide2$	$≠$	$1$ $divide2$	Divide both sides by $2$

$x$	$≠$	$1/2$

$2x-3$	$>$	$0$
$2x-1$ $+3$	$>$	$0$ $+3$	Add $3$ to both sides
$2x$ $divide2$	$>$	$3$ $divide2$	Divide both sides by $2$

$x$	$>$	$3/2$

$3-4x$	$>$	$0$
$3-4x$ $-3$	$>$	$0$ $-3$	Subtract $3$ from both sides
$-4x$ $divide(-4)$	$>$	$-3$ $divide(-4)$	Divide both sides by $-4$

$x$	$<$	$3/4$	Flip the inequality

$x$	$5$	$6$	$7$	$9$	$10$
$y$

$x$	$5$	$6$	$7$	$9$	$10$
$y$	$0$	$1$	$sqrt2$	$2$	$sqrt5$