boredsatan
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Is the square root of 4 just 2 or both 2 and -2?
BumpBump
can someone please answer my questions?bump
bump pleasebump
bumpbump please
bump2sin(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
It depends on the notation.Is this always the case that you only take the +ve?
Just 1/2.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?
So in what cass is the square root both the +ve and -ve and in what cases is it only the +ve?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?
Also can someone please help me with the below questions?
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))?
You don't need to do anything, because the domain is already given to us.Also can someone please help me with the below questions?
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
So in what cass is the square root both the +ve and -ve and in what cases is it only the +ve?
Also can someone please help me with the below questions?
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))?
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]?