Question

Help with these last two questions

Answer 1

Q19 is the same kind of problem as Q18
(don't allow a divide by 0, which happens for only one x value we don't allow)

Q20, we need the x+3 to be 0 or bigger
x+3 >= 0
to "solve" for x, add -3 to both sides

Answer 2

for Q20 it would be -3<=0

Answer 3

?

Answer 4

x+3 >= 0
add -3 to both sides:
x+3-3 >= 0 - 3

on the left side 3-3 is 0
on the right side 0 -3 is -3
x+0 >= -3
or
x>= -3

Answer 5

So that would be the answer to Q20

Answer 6

I just notice, if x is -3
what do we get for \(\sqrt{x+3}\)
?

Answer 7

well it would be 0?

Answer 8

But zero cant be a sq rt

Answer 9

right?

Answer 10

square root of zero is ok (we get 0)

but, in this problem it that causes a *divide by 0*

Answer 11

but, in this problem it causes a *divide by 0*

Answer 12

yes

Answer 13

so the answer is
x > -3

Answer 14

and then i was confused on 19

Answer 15

i know that we already did something like this but I'm confused

Answer 16

there is no square root in Q19
so all we have to worry about is divide by 0

are there any x's that cause the bottom x+6 to be 0 ?

Answer 17

yes i think

Answer 18

can you figure out what x makes
x+6= 0
?

Answer 19

you can use algebra to solve for x. add -6 to both sides

Answer 20

x=-6?

Answer 21

yes. and if we use x=-6 in
\[ \frac{1}{x+6} \]
what do we get ?

Answer 22

0

Answer 23

we get 1/0 (one divided by 0 )
that is "undefined" (people don't know what it means, so we don't allow it)

so the answer to Q19 is
all values of x are allowed *except* x=-6
we write
\[ (-\infty, +\infty) , x \ne -6\]

Answer 24

thanks