Will METAL!!
Find the domain of the given function.
f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two.
x > 0
All real numbers
x ≥ -3, x ≠ 2
x ≠ -8, x ≠ -3, x ≠ 2

- anonymous

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- Directrix

Would you write this out in symbols? I am not following what this equation is:
f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two.

- phi

Do you see the equation editor in the bottom left? that would be good to use

- anonymous

\[\sqrt{x+3}/(x+8)(x-2)\]

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## More answers

- anonymous

f(x)= ^

- phi

meanwhile, the domain are all "legal" x values. In other words, all x values that do not cause trouble.
What kinds of trouble?
we don't want square roots of negative numbers. if an x value causes that, we don't allow x to be that value
no divide by 0: if an x value causes us to divide by 0, we don't allow that.
Use those rules to decide which x values are not allowed, and that tells you which are allowed.

- phi

to answer, first look at \( \sqrt{x+3}\)
any idea what x values might cause us to have the square root of a negative number ?

- anonymous

no not sure.

- phi

negative means less than zero
if
x+3 < 0 then that means x+3 is negative. And we don't want that
we want
x+3 >=0 (that means x+3 is 0 or positive)
add -3 to both sides , what do you get ?

- anonymous

if you add -3 to both sides then you get x>-3?

- phi

you mean x>=-3
in other words, if we want x+3 to be zero or bigger we need
x+3 >= 0
and that means x >= -3
(otherwise we would have a square root of a negative number, and we don't want that)

- phi

so
\( x\ge -3 \)
is part of the answer.
next, look that the "bottom"
(x+8)(x-2)
what happens if x is -8 ?

- anonymous

oh I see it, wow your good.

- phi

the bottom becomes
(-8+8)(-8-2)
and -8+8 is 0
0 times anything is 0
in other words the bottom becomes 0, and we would be dividing the top by 0
we don't allow divide by zero.
so we can't let x be -8
luckily , x must be >= -3 (to keep the top legal), so we are ok (-8 is smaller than -3, so we won't allow it)
but we have
(x+8)(x-2)
what happens if x= 2 ?

- anonymous

(2+8)(2-2)

- anonymous

2-2=0

- anonymous

0 times anything = 0

- phi

yes, and we have zero. we don't want zero in the bottom
so we have to say x is not allowed to be 2
all together, the allowed x values are
\[ x\ge-3 , x\ne 2\]
which means x is -3 or bigger, but not 2

- anonymous

can we do another?

- phi

ok

- anonymous

Verify the identity.
cos 4x + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x

- phi

It looks painful. Do you have the "half-angle" formula for cos 4x ?

- anonymous

isn't it =+-square root 1+ cos theta/ 2

- phi

there are 3 different versions
\[ \cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a)-1 = 1 - 2 \sin^2(a) \]

- phi

because in your problem you only have sin on the right side, I would use the last version
\[ \cos(2a) = 1 - 2 \sin^2(a) \]
can you use that to replace cos(4x) in your problem ?

- phi

match up cos 4x with cos 2a
write cos 4x as cos (2* 2x)
and we match cos (2 * a)
a is 2x
in your problem

- phi

\[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = ? \]

- anonymous

2-2sin^2(a) ?

- phi

\[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = 1 - 2 \sin^2(2x) \]

- anonymous

so 1-2sin^2(2x)
1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x

- phi

yes, and if we add +2sin^2(2x) to both sides, we can simplify that a bit

- anonymous

1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x
+2sin^2x +2sin^2x
----------------------------------------------
?

- phi

yes

- anonymous

1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x
+2sin^2x +2sin^2x
----------------------------------------------
1(2x) + cos 2x = 2-2 sin^2 (2x)
I'm not sure

- phi

oops, you should add 2 sin^2(2x) (not 2sin^2(x) )

- phi

1-2sin^2(2x) + cos 2x = 2 - 2 sin^2(2x) - 2 sin^2 x
+2sin^2(2x) +2 sin^2(2x)
----------------------------------------------
1 + cos(2x) = 2 - 2 sin^2(x)

- phi

It looks like we should use the same rule as before , but now on cos(2x)

- anonymous

1-2sin^2(x)

- phi

yes, put that in for cos(2x)

- anonymous

1 + cos(2x) = 2 - 2 sin^2(x)
------------------------------------------------
1-2sin^2(x)= -2sin^2(x)

- phi

you replace cos(2x) with 1-2sin^2(x)
like this
1 + cos(2x) = 2 - 2 sin^2(x)
1 + (1-2sin^2(2x)) = 2 - 2 sin^2(2x)
now simplify the left side

- phi

the left side is
\[ 1 + 1 - 2\sin^2(2x) \]

- anonymous

1 + (1-2sin^2(2x)) = 2 - 2 sin^2(2x)
-------------------------------
1+1-2sin^2(2x) = 2-2sin^2(2x)
--------------------------------
2-2sin^2(2x) = 2-2sin^2(2x) yay!

- phi

yes, both sides are obviously the same.

- anonymous

can we do another problem, not like this one though.

- phi

one more, but please make it a new post.

- anonymous

ok hold on.

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