# Thread:

1. ## Domain And Range

1) State the domain and range of each of the following functions:
a) y = 3x + 4
b) y = $3(x-4)^2 +5$
c) y = √ 2x-10
(the root is over all the numbers for c)
d) y = √ $-x^2 + 4$
(the square root sign is over all the numbers for d) as well
e) y = 2 |1− x |−17
f) $y = x/ (x^2-x-6)$

Can someone help me find these plus explain why for d) e) and f)

2. Originally Posted by foreverbrokenpromises
1) State the domain and range of each of the following functions:
a) y = 3x + 4
b) y = $3(x-4)^2 +5$
c) y = √ 2x-10
(the root is over all the numbers for c)
d) y = √ $-x^2 + 4$
(the square root sign is over all the numbers for d) as well
e) y = 2 |1− x |−17
f) $y = x/ (x^2-x-6)$

Can someone help me find these plus explain why for d) e) and f)
You can think of the domain as all the numbers that x is 'allowed' to be.

As in c)... You know that you can't take the square root of a negative number and get a real number. So, x can only be those numbers that make the radicand greater than or equal to zero. In other words, solve

$2x-10\geq0$

and you will have your domain.

3. (f) You cannot include in the domain where the function is undefined, therefore areas where the dominator of a rational function is equal to zero should not be included in the domain.

So the domain is (-∞, -2) U (-2, 3) U (3, ∞)

4. Thanks
I've figured out a) b) and c)

but i still cant figure out how to complete the following:

d) y = √-x^2 +4
e) y = 2 |1-x|-17
f) y= $x/ (x^2-x-6)$

pls help!

5. Originally Posted by foreverbrokenpromises
Thanks
I've figured out a) b) and c)

but i still cant figure out how to complete the following:

d) y = √-x^2 +4
e) y = 2 |1-x|-17
f) y= $x/ (x^2-x-6)$

pls help!
For d apply the principle VonNemo19 outlined in post 2.

d) $-x^2+4 \geq 0 \: \: \rightarrow \: \: 4-x^2 \geq 0$

$(2-x)(2+x) \geq 0$.

Spoiler:
$-2 \leq x \leq 2$

e) What values of x are not allowed in this linear equation?

Spoiler:
There are none so x is all the real numbers

f) $x^2-x-6=(x+2)(x-3)$. Remember that any value that makes the denominator equal to 0 is not allowed

Spoiler:
$x \in \mathbb{R} \: , \: x \neq -2,3$
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