# VCE Maths questions help (1 Viewer)

#### boredsatan

##### Member
Is the square root of 4 just 2 or both 2 and -2?

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#### HeroWise

##### Active Member
Square rooting is usually principal root, so omly +ve case

#### boredsatan

##### Member
Is this always the case that you only take the +ve?

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#### boredsatan

##### Member
2sin(2x-pi/6) = 1 for a domain of -pi<x<pi
we multiply the domain values of 2, so it becomes -2pi<2x<2pi
but then do we subtract the domain values by -pi/6
or since 2sin(2x-pi/6) = 2sin(2(x-pi/12)), do we subtract the domain values by -pi/12?
Thanks

#### boredsatan

##### Member
2sin(2x-pi/6) = 1 for a domain of -pi<x<pi
we multiply the domain values of 2, so it becomes -2pi<2x<2pi
but then do we subtract the domain values by -pi/6
or since 2sin(2x-pi/6) = 2sin(2(x-pi/12)), do we subtract the domain values by -pi/12?
Thanks
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#### fan96

##### 617 pages
Is this always the case that you only take the +ve?
It depends on the notation.

is a function that denotes taking the principal square root, which is always the positive one, for real .

However, for always has two solutions.

#### boredsatan

##### Member
if we had to find the derivative of sqrt(x^2+3), and evaluate it when x = 1, we get 1/(sqrt(4)), which so in this case, would it be 1/2 and 1/-2?
or just 1/2?

#### InteGrand

##### Well-Known Member
if we had to find the derivative of sqrt(x^2+3), and evaluate it when x = 1, we get 1/(sqrt(4)), which so in this case, would it be 1/2 and 1/-2?
or just 1/2?
Just 1/2.

#### boredsatan

##### Member
Just 1/2.
So in what cass is the square root both the +ve and -ve and in what cases is it only the +ve?

2sin(2x-pi/6) = 1 for a domain of -pi<x<pi
we multiply the domain values of 2, so it becomes -2pi<2x<2pi
but then do we subtract the domain values by -pi/6
or since 2sin(2x-pi/6) = 2sin(2(x-pi/12)), do we subtract the domain values by -pi/12?
Thanks

f(x) = sqrt(x+1), and has a restricted domain of [0, infinity)
g(x) = x^2+4x+3, and has a restricted domain of (-infinity, -3]
How would i find the range of f(g(x)) without a calculator?
f(g(x)) = sqrt(x^2+4x+4), so would I use the domain of g(x) as the domain of f(g(x))?

#### InteGrand

##### Well-Known Member
So in what cass is the square root both the +ve and -ve and in what cases is it only the +ve?

2sin(2x-pi/6) = 1 for a domain of -pi<x<pi
we multiply the domain values of 2, so it becomes -2pi<2x<2pi
but then do we subtract the domain values by -pi/6
or since 2sin(2x-pi/6) = 2sin(2(x-pi/12)), do we subtract the domain values by -pi/12?
Thanks

f(x) = sqrt(x+1), and has a restricted domain of [0, infinity)
g(x) = x^2+4x+3, and has a restricted domain of (-infinity, -3]
How would i find the range of f(g(x)) without a calculator?
f(g(x)) = sqrt(x^2+4x+4), so would I use the domain of g(x) as the domain of f(g(x))?

#### InteGrand

##### Well-Known Member
2sin(2x-pi/6) = 1 for a domain of -pi<x<pi
we multiply the domain values of 2, so it becomes -2pi<2x<2pi
but then do we subtract the domain values by -pi/6
or since 2sin(2x-pi/6) = 2sin(2(x-pi/12)), do we subtract the domain values by -pi/12?
Thanks
You don't need to do anything, because the domain is already given to us.

#### InteGrand

##### Well-Known Member
So in what cass is the square root both the +ve and -ve and in what cases is it only the +ve?

2sin(2x-pi/6) = 1 for a domain of -pi<x<pi
we multiply the domain values of 2, so it becomes -2pi<2x<2pi
but then do we subtract the domain values by -pi/6
or since 2sin(2x-pi/6) = 2sin(2(x-pi/12)), do we subtract the domain values by -pi/12?
Thanks

f(x) = sqrt(x+1), and has a restricted domain of [0, infinity)
g(x) = x^2+4x+3, and has a restricted domain of (-infinity, -3]
How would i find the range of f(g(x)) without a calculator?
f(g(x)) = sqrt(x^2+4x+4), so would I use the domain of g(x) as the domain of f(g(x))?

#### boredsatan

##### Member
What do you mean? Would the domain of f(g(x)) just be the domain of g(x), which in this case is restricted to (-infinity,-3]?