# Thread: VCE Maths questions help

1. ## Re: VCE Maths questions help

Originally Posted by boredsatan
Bump!
Bump!

2. ## Re: VCE Maths questions help

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?

3. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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?
bump

4. ## Re: VCE Maths questions help

Originally Posted by boredsatan
bump
Have you found and simplified f(g(x))? Once you do this, you should be able to do the question.

5. ## Re: VCE Maths questions help

Originally Posted by InteGrand
Have you found and simplified f(g(x))? Once you do this, you should be able to do the question.
f(g(x)) = sqrt(x^2+4x+4), so would I use the domain of g(x) as the domain of f(g(x))?

6. ## Re: VCE Maths questions help

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

Bump

8. ## Re: VCE Maths questions help

Originally Posted by boredsatan
Bump
Bump

9. ## Re: VCE Maths questions help

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

10. ## Re: VCE Maths questions help

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

bump

12. ## Re: VCE Maths questions help

Originally Posted by boredsatan
bump

bump

14. ## Re: VCE Maths questions help

Originally Posted by boredsatan
bump

15. ## Re: VCE Maths questions help

Originally Posted by boredsatan
bump

16. ## Re: VCE Maths questions help

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

17. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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
bump

18. ## Re: VCE Maths questions help

Originally Posted by boredsatan
Is this always the case that you only take the +ve?
It depends on the notation.

$f(x) = \sqrt x$ is a function that denotes taking the principal square root, which is always the positive one, for real $x$.

However, $x^2 = k$ for $k > 0$ always has two solutions.

19. ## Re: VCE Maths questions help

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?

20. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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.

21. ## Re: VCE Maths questions help

Originally Posted by InteGrand
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))?

22. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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))?
$\noindent For any positive real number x, the symbol \sqrt{x} refers to the \emph{positive} square root of x.$

23. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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.

24. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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))?
$\noindent In general, the domain of f\circ g would be the set of all x values such that g(x) is in the domain of f. In this case, what is the domain of f and which values x in the domain of g make g(x) land in the domain of f?$

25. ## Re: VCE Maths questions help

Originally Posted by InteGrand
$\noindent In general, the domain of f\circ g would be the set of all x values such that g(x) is in the domain of f. In this case, what is the domain of f and which values x in the domain of g make g(x) land in the domain of f?$
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]?

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