A Thread For Questions

[TrueTears]

Ohhh, it's not related with

Remember for the function, the period is where n is the coefficient of what's in front of the x.

So here the the coefficient is 2, thus period is

[Gloamglozer]

Ah right ok.  That makes so much more sense now. 

*face palm*  I think I think people reading this for the first time will probably think that I'm not a genuine Methods student lol.

And just one more question, with general solutions, all you do is:

1.  For sin and cos, you add to the period.
2.  For tan, you add to the period.

Right?

[TrueTears]

Ah right ok.  That makes so much more sense now.  

*face palm*  I think I think people reading this for the first time will probably think that I'm not a genuine Methods student lol.

And just one more question, with general solutions, all you do is:

1.  For sin and cos, you add to the period.
2.  For tan, you add to the period.

Right?

Nearly correct, you don't add it to the period.

For cos and sin functions the way I do the general solutions is:

Let one of the solutions be and other other solution be

Thus the general solution would be where or where

This applies to both sin and cos functions.

For tan function you only need to find one solution, let this solution be c.

Thus general solution would be where

[QuantumJG]

Few questions:
and for x e [0,infinite), then is defined over what domain?

I thought since they are both defined over [0,3], that would be the new domain but it is [0,9]
So I graphed the new function and it is from (-infinite,9] but why is that limited at 0? edit: is 0 the lowest because
How do I work out these questions?

2. and for x e R.
a) show g(f(x)) exists - done
b) write down the rule for g(f(x)) and state domain - done BUT - I got the domain by saying they were both defined over [0,pi] ? It worked for this Q but not for others.
c) Find the range - How do I get this? I tried to do what I did with the domain (-infinite,1] for one and [-3,3] for the other, giving me [-3,1]. range is [-8,1]
And say if you have the domain and sub it into the equation to find the y-value, is there any way to see whether this is the lowest/highest y-value?

3. Find all values of k such that equation has exactly one solution.

I tried discriminant and ended up with K = -0.25 but then also got k>0?
Basically, I let u = 2^x so I had

used discriminant to get 1+4k = 0, k = -1/4
and if I use the quadratic formula, I get
so doesn't   have to be greater than 0? If I solve it, then I get k > 0

Q1)

It would probably be best if you left .

This states that x > 0. As for why x can now extend up to 9 that does seem a bit weird, but, I think because for f(x)'s domain being there because they are the only values that keeps it in the real field. Now x can equal 9 and the function will still exist.

So with this x is an element of [0,9]

Q2)

b) so g o f = 1 - 9sin^2(2x)

if x in f(x) is restricted to 0 < x < pi, then g o f is restricted to 0 < x < pi, since, these are the only values that will map out a value.

What do you mean by "others"?

c) The range is simply the value restriction for g o f. With your graphics calculator graph 'g o f' in 0 < x < pi and find out the maximum and minimum value g o f will give and this is [-8,1].

3)

here you simply take the discriminant and state:

1 + 4k = 0 => k = -0.25

what you continued on doing was pretty weird.

basically with your quadratic formula you have:

discriminant < 0, Complex solution zone

discriminant = 0

discriminant > 0

so for x to be defined the discriminant must be equal to 0 or greater than 0, but, we are told that the discriminant must equal zero.

If we didn't have any restrictions we could say that there will be atleast one solution to the given equation if K is ≥ -0.25.

with exponentials x ϵ ℝ.

[TrueTears]

QuantumJG can you help me with a question?

For what values of k, where k is a real constant does the equation have two distinct solutions?

[No one else answer this Q, it's for QuantumJG]

[QuantumJG]

QuantumJG can you help me with a question?

For what values of k, where k is a real constant does the equation have two distinct solutions?

[No one else answer this Q, it's for QuantumJG]

4^x = (2^x)^x

=> (2^x)^2 - 5(2^x) - k =0

let u = 2^x

=> u^2 - 5u - k = 0

therefore for 2 solutions => discriminant > 0

=> 25 + 4k > 0

therefore k > -25/4

but lim(x -> -infinity)(4^x - 5(2^x)) = 0

=> 0 > k > -25/4

[TonyHem]

Few questions:
and for x e [0,infinite), then is defined over what domain?

I thought since they are both defined over [0,3], that would be the new domain but it is [0,9]
So I graphed the new function and it is from (-infinite,9] but why is that limited at 0? edit: is 0 the lowest because
How do I work out these questions?

2. and for x e R.
a) show g(f(x)) exists - done
b) write down the rule for g(f(x)) and state domain - done BUT - I got the domain by saying they were both defined over [0,pi] ? It worked for this Q but not for others.
c) Find the range - How do I get this? I tried to do what I did with the domain (-infinite,1] for one and [-3,3] for the other, giving me [-3,1]. range is [-8,1]
And say if you have the domain and sub it into the equation to find the y-value, is there any way to see whether this is the lowest/highest y-value?

3. Find all values of k such that equation has exactly one solution.

I tried discriminant and ended up with K = -0.25 but then also got k>0?
Basically, I let u = 2^x so I had

used discriminant to get 1+4k = 0, k = -1/4
and if I use the quadratic formula, I get
so doesn't  have to be greater than 0? If I solve it, then I get k > 0

Q1)

It would probably be best if you left .

This states that x > 0. As for why x can now extend up to 9 that does seem a bit weird, but, I think because for f(x)'s domain being there because they are the only values that keeps it in the real field. Now x can equal 9 and the function will still exist.

So with this x is an element of [0,9]

Q2)

b) so g o f = 1 - 9sin^2(2x)

if x in f(x) is restricted to 0 < x < pi, then g o f is restricted to 0 < x < pi, since, these are the only values that will map out a value.

What do you mean by "others"?

c) The range is simply the value restriction for g o f. With your graphics calculator graph 'g o f' in 0 < x < pi and find out the maximum and minimum value g o f will give and this is [-8,1].

3)

here you simply take the discriminant and state:

1 + 4k = 0 => k = -0.25

what you continued on doing was pretty weird.

basically with your quadratic formula you have:

discriminant < 0, Complex solution zone

discriminant = 0

discriminant > 0

so for x to be defined the discriminant must be equal to 0 or greater than 0, but, we are told that the discriminant must equal zero.

If we didn't have any restrictions we could say that there will be atleast one solution to the given equation if K is ≥ -0.25.

with exponentials x ϵ ℝ.

Q1) I'm not sure on what you wrote :/
I've read in the textbook that for f(g(x)), the domain of g(x)  ( after it has been restricted) is the domain of f(g(x)) but I can't 'see' the reason for it. If I restrict the range of g(x) to [0,3], then the domain is [0,9] so would that mean I need to restrict the domain of the inner function to be within the subset of the outer function and then find the domain of that restricted g(x)?

2) For others, I mean like question 1. With Q2, I just looked for the largest domain that they were both defined and then took that as the new domain.

2B) Exam 1 but yeah I got an explanation for this one already so all good.

3) Still not exactly sure here.
Like with TT's question below, I found K > -25/4 the 1st time and figured that would be it. But theres the second restriction with it being > 0 and the way I understood it was

u = -b+minussqrt{b^2-4ac}/2a
and then I had 2 = loge( whatever), and the stuff within the log had to be greater than 0 and the discriminant had to be > 0.

So for my one above, I got K = -1/4 but I figured I'd also have to find that "second" restriction/solution so if I use the quadratic, am I meant to leave it as U or get x on it's own? Is the 1+sqrt{1+4k}/2 > 0 not relevent to this question because I just need to find when it equals 0? Is k = -1/4 the only solution? thanks!

[TrueTears]

QuantumJG can you help me with a question?

For what values of k, where k is a real constant does the equation have two distinct solutions?

[No one else answer this Q, it's for QuantumJG]

4^x = (2^x)^x

=> (2^x)^2 - 5(2^x) - k =0

let u = 2^x

=> u^2 - 5u - k = 0

therefore for 2 solutions => discriminant > 0

=> 25 + 4k > 0

therefore k > -25/4

but lim(x -> -infinity)(4^x - 5(2^x)) = 0

=> 0 > k > -25/4

Cool, so for TonyHem's, why didn't you also consider the quadratic formula?

There's 2 conditions to satisfy. Not just the discriminant.

[khalil]

You know how
What are the retrictions to this rule. I know if = it would not work

[TrueTears]

You know how
What are the retrictions to this rule. I know if = it would not work

What you mean when m = x - 1?

[khalil]

eg. In(x-1)^2....does this equal to 2In(x-1)?

[TrueTears]

[khalil]

I graphed the botttom two, they're not the same

[kamil9876]

I think he is worried about domain. For example:

[TrueTears]

I graphed the botttom two, they're not the same

Yeah they are.