A Thread For Questions

[ngRISING]

what should all the circle and 3d shapre forumla's should we know?

[dekoyl]

What 3D shapes are there in methods? I don't think there are any 3D Shapes.

[m@tty]

Generally for related rate problems. eg A cone is being filled with water at a rate of ......

[ngRISING]

What 3D shapes are there in methods? I don't think there are any 3D Shapes.

oh is it? hmmph,
what shape formula's do i use when i try to find the rate of something leaking out of a sphere?
sorry is a sphere a 3D shape? if so i think its in methods. is it not? -.-  

[TrueTears]

"Shape formula?" I haven't heard of such a term before =S

[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.

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!

After careful analysis I found a flaw in my answer! With this graph it can either have that parabolic kind of shape or that cubic kind of shape.

I ask you to graph (just to understand this graph can follow 2 trends):

• 2^x + 2^-x
• 2^x - 2^-x

so as a matter of fact k>-0.25 will allow that particular function to pass through y = 1 once.

The solution is k > -0.25

[kamil9876]

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!

It is relevant and good to check. The thing that you have confused is the difference between AND/OR. In order for there to be a unique solution, the solution given by the quadratic formula for must be positive AND discriminant be zero. If one of these conditions fails for some value of k, there is no unique solution for that k. k=-0.25 is the only value that satisfies BOTH conditions. The other set of values only satisfies one condition but fails for discriminant=0.

[TrueTears]

Generally for related rate problems. eg A cone is being filled with water at a rate of ......

The OP asked for what are some formulas yet you give a case where you can use a 3D shape. nice...

[dekoyl]

oh is it? hmmph,
what shape formula's do i use when i try to find the rate of something leaking out of a sphere?
sorry is a sphere a 3D shape? if so i think its in methods. is it not? -.-  

Oh.  I get you. For stuff like that, you use the volume of a sphere, V = and then use your related rates knowledge/chain rules. It's usually quite basic and you can find it on the formula sheet.

[NE2000]

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

Yeah they are.

No they're not.

The reason is that in x can take on any value < 1 (but not 1) as the square will turn that positive and hence you can evaluate the log. In you can't. The solution is what kamil did and use absolute values so that you don't just halve the number of solutions.

[TrueTears]

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

Yeah they are.

No they're not.

The reason is that in x can take on any value < 1 (but not 1) as the square will turn that positive and hence you can evaluate the log. In you can't. The solution is what kamil did and use absolute values so that you don't just halve the number of solutions.

No the implied domain is stated within itself.

It's like saying do I really have to mean everytime I write the equation?

[NE2000]

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

Yeah they are.

No they're not.

The reason is that in x can take on any value < 1 (but not 1) as the square will turn that positive and hence you can evaluate the log. In you can't. The solution is what kamil did and use absolute values so that you don't just halve the number of solutions.

No the implied domain is stated within itself.

It's like saying do I really have to mean everytime I write the equation?

khalil graphed the bottom two, presumably on a graphics. When you do that, the graphs are different. And I gave the reason why. The point was to answer khalil's question. When you are graphing you need to distinguish between that and . Otherwise there goes a mark because if you are asked to graph the former many people would just graph the latter.

[TrueTears]

Yes that's right, but when I answered, I answered in terms of mathematical reasoning, if a domain is implied you don't need to write out the domain, unless there is a restriction placed on the graph which there isn't. So when you graph it you ignore the other parts of the graph. Thus those 2 are exactly the same.

[TonyHem]

If

If I work out the inverse of f first, , and then change x for 2a, then I get the right answer but if I work out f(2a) then inverse, it doesn't work out. I end up with then end up with

What did I do wrong? :/ Thanks
ans:

ps: Whats the code for cube-root in tex?

[TrueTears]

The question is in fact combining 2 question into one. Consider it as the following:

a) Find

b) Find

What would you do in this case? You'd go a) then b). So you should always work out the rule first before subbing in.

Subbing in 2a at first should also work, What I would do is say let X = 2a and then swap X and y. Then at the end sub 2a back in for X. Although I much prefer the method above.

LaTeX code: ^3\sqrt{x}