anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Directrix
  • 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
  • phi
Do you see the equation editor in the bottom left? that would be good to use
anonymous
  • anonymous
\[\sqrt{x+3}/(x+8)(x-2)\]

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anonymous
  • anonymous
f(x)= ^
phi
  • 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
  • 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
  • anonymous
no not sure.
phi
  • 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
  • anonymous
if you add -3 to both sides then you get x>-3?
phi
  • 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
  • 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
  • anonymous
oh I see it, wow your good.
phi
  • 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
  • anonymous
(2+8)(2-2)
anonymous
  • anonymous
2-2=0
anonymous
  • anonymous
0 times anything = 0
phi
  • 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
  • anonymous
can we do another?
phi
  • phi
ok
anonymous
  • anonymous
Verify the identity. cos 4x + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x
phi
  • phi
It looks painful. Do you have the "half-angle" formula for cos 4x ?
anonymous
  • anonymous
isn't it =+-square root 1+ cos theta/ 2
phi
  • 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
  • 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
  • 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
  • phi
\[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = ? \]
anonymous
  • anonymous
2-2sin^2(a) ?
phi
  • phi
\[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = 1 - 2 \sin^2(2x) \]
anonymous
  • anonymous
so 1-2sin^2(2x) 1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x
phi
  • phi
yes, and if we add +2sin^2(2x) to both sides, we can simplify that a bit
anonymous
  • anonymous
1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x +2sin^2x +2sin^2x ---------------------------------------------- ?
phi
  • phi
yes
anonymous
  • 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
  • phi
oops, you should add 2 sin^2(2x) (not 2sin^2(x) )
phi
  • 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
  • phi
It looks like we should use the same rule as before , but now on cos(2x)
anonymous
  • anonymous
1-2sin^2(x)
phi
  • phi
yes, put that in for cos(2x)
anonymous
  • anonymous
1 + cos(2x) = 2 - 2 sin^2(x) ------------------------------------------------ 1-2sin^2(x)= -2sin^2(x)
phi
  • 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
  • phi
the left side is \[ 1 + 1 - 2\sin^2(2x) \]
anonymous
  • 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
  • phi
yes, both sides are obviously the same.
anonymous
  • anonymous
can we do another problem, not like this one though.
phi
  • phi
one more, but please make it a new post.
anonymous
  • anonymous
ok hold on.

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