Question

Help please

Answer 1

focus on the "composite function"
in Q17
\[ G(F(x)) = \sqrt{6-x} \]
the domain are the "allowed" x values
you start by assuming all x values are allowed (used)
then look for any x's that you should exclude

Answer 2

you exclude (don't allow) x values that cause "trouble"

what kind of trouble?
1) don't allow *divide by zero*
2) don't allow square root of a negative number

Answer 3

So take out the "x"

Answer 4

you look at
\[ \sqrt{6-x}\]
some x's cause trouble (you get square root of a negative number)
for example, if x was 10 you would get \(\sqrt{6-10} = \sqrt{-4} \)
and we don't allow that.

any idea what x's cause trouble ?

Answer 5

we can use algebra to figure out which x's are ok
we want
6-x >= 0 (that says 6-x is 0 or positive)
then add x to both sides
6-x+x >= 0+x
6 >= x
or
\( x\le 6\)
are all the good x's

Answer 6

so the answer for Q17 is
\[ (-\infty , 6] \]

Answer 7

Oh okay so this would be the answer for question 17

Answer 8

Oh okay thanks

Answer 9

So for 18 it would be the same process right?

Answer 10

for Q18

the function is \( \frac{1}{3x} \)

of the list of "troubles":
1) don't allow *divide by zero*
2) don't allow square root of a negative number

which one could we run into with this function ?

Answer 11

the first one?

Answer 12

yes, we don't have a square root
but we do divide by 3x
we *do not* want 3x to be 0 (because then we would have 1 divided by 0, not allowed)

the answer for Q18 is
all real numbers are allowed, except for when
3x= 0
what value of x is that ? any idea ?

Answer 13

0?

Answer 14

yes
using algebra
3x= 0
divide both sides by 3
\[ \frac{3}{3} x = \frac{0}{3} \\ x=0\]
so x=0 is the value we do not allow

the answer is
\[ (-\infty, +\infty), x \ne 0 \]

Answer 15

Thank you so much @phi you really do help me