Need help with Mathematics.

[Vladislav]

I'm doing my extra weekend school homework and I'm really struggling with some problems:


Given that f(x)=3^x+3^-x,

a) show that f(2x)=[(fx)]^2-2

b)making use of the resut in (a)
show that f(2x)[f(2x)+4]={[f(x)]^2}^2-4


I have heaps more and help would be greatly appreciated.

Also can anyone explain how to find the domain and range of a function?

[ComradeRed]

In a function f(x)=y, all numbers of x which form a set X={x} is the domain. The range is the set of all the y, that is Y={y}.



Given that f(x)=3^x+3^-x,

a) show that f(2x)=[(fx)]^2-2
Well, plug it in!

3^(2x) + 3^-{2x} = 3^{2x} + (1/3^{2x})

Then we just manipulate it into a more bastardized fraction:

((3^{2x})^2 + 1)/3^{2x} = 3^{2x} + (1/3^{2x}) = 9^x + 9^-x.

Alternatively we may abuse complex numbers such that:

3^{2x} + (1/3^{2x}) = (sqrt{3}^{x} + (i/sqrt{3^2x}))(sqrt{3}^{x} - (i/sqrt{3^2x}))

and we could pursue this path.

I'm pulling a blanck, being as tired as I am (;)), but I would think that tinkering would be needed to solve it ;) I'll play with it all day tomorrow.

[Tormented by Treachery]

Well Vlad, I know how to do it using calculus, finding D and R, but I don't know if you have used calc yet... If so, take the derivative, set it equal to zero, and then the two numbers you get will contain minimum(s) or maximum(s), which will tell you your Range. I think your D is all reals, but I'm probably wrong, since it is 2 am :)

[ComradeRed]

Given that f(x)=3^x+3^-x,

a) show that f(2x)=[(fx)]^2-2 Lemme take another stab at it.

I first of all define a constant J=3^x + i*3^-x. where i is the squareroot of -1.

We can redefine f(2x)=J`J for the notation "N`" denoting the complex conjugate of a complex variable N.

That is to say J`=3^x - i*3^-x; note that the complex part is now subtracted (that is what a complex conjugate does: it changes the sign of the complex part leaving everything else the same). If we carry it through we get:f(2x) = 9^2x + 9^-2x.
Again, I use LaTeX notation.

Now, we can abuse this to figure it out :)

Firstly, we need to discover what the hell f(x) is in this new notation. The problem is that J is complex :(

But suppose the answer is correct. That means

f(2x) = J`J = [(fx)]^2-2

which means J`J = [(fx)]^2-2. First of all, this is the correct notation? Because it's equal to 1 if it is (2-2=0, x^0=1 for all x).

I think what would be meant is [f(x)^2]^-2 ? If this is the case, we get something very interesting!

f(x)^2 = (3^x + 3^-x)^2 = 9^2x + 9^-2x + 6.

This in turn raised to the aforementioned power yields:

[f(x)^2]^-2 = sqrt(9^2x + 9^-2x + 6) = plus-or-minus(3^x + 3^-x).

With f(2x) = 9^2x + 9^-2x. If your teacher were lazy enough, I would think he was trying to say that:

f(x)^2 = (3^x + 3^-x)^2 = 9^2x + 9^-2x (commutative relationship)

and that equals what we want. Frankly I can't understand the [(fx)]^2-2. I'm not sure if it is written incorrectly or what... :huh:

[Vladislav]

Thanks for the help. I have figured it out.
Nah ,Tormented by Treachery I haven't done calculus yet.

Thank you once more.