ssNake's Methods 1/2 Q's.

[REBORN]

But can't you be like the period is 2pi and just sketch a reflected (in the x-axis) cos graph for the specified domain and you get 4 sols...

[b^3]

if you sketch it for the domain -3pi/2 to 2pi and find the intercepts with the axis then you get 5 solutions (zero counts) not four, think of it this way lets just say a is 1 and b is sqrt(2). then just solve the equation cos(x)=-1/srt(2) for -3pi/2 to 2pi. then you get which is four solutions. this would be equivalent to drawing y=cos(x) and y=-1/sqrt(2) and finding the intercepts.

[REBORN]

Argh I didn't count the zero -.-

How about this Q?

Q: The number of sols of the eq b=a sin(x) where domain is [-2pi,pi/2] with a,b real positive numbers and a>b

[xZero]

sin(x)=b/a, since a>b and they are both positive, b/a must be smaller than 1 but greater than 0 => 0<b/a<1. This implies that x must be in quadrant 1 and 2, since the domain is from -2pi to pi/2, we have quadrant 1,2 and 1 again. Thus there's 3 solutions.

Hope im right

[REBORN]

You're right.

Who said it has to be smaller than 1? :S

I don't get it still...

[xZero]

a is bigger than b, so lets say that a = b+1. b/a -> b/(b+1). which means that this fraction must be smaller than 1.

[b^3]

and if it wasn't smaller than one, there wouldn't be any solutions because the values of sin and cos cannot be greater than one since they are the x and y values in a unit circle of radius one. that was what I was trying to get at before.

[REBORN]

Let's take this example: 8^2/3

Is this equal to this: cuberoot of 8squared | (cuberoot eight) then sq it by 2.

Argh can't do latex....hope that made sense :S

[luken93]

Yeah, easiest way it to root by the denominator, then power by the numerator.

As another example, 64^(2/3)
= (64^(1/3))^2
= 4^2
= 16

After a while. you'll spot them almost immediately.

[RobM8]

Evaluate what ever is easiest inside of the brackets first.
In this case I would just cuberoot 8 to get 2 and then square it to get 4.

then sq it by 2.

I would be impressed if you manage to square it by anything other than 2.

[REBORN]

y=2^2x

Is that a dilation of factor 1/2 from the y-axis?

[REBORN]

How do you find the equation of the asymptote of an exponential?

eg: y = 2 x 2^x/3

[b^3]

if you think of it in transformations, there is no y shift so it still just the y=0 of the original exponential graph.

[taiga]

For asymptotes just think about portions of the equation that can not equal something no matter what real number you chuck up the top.

So for:
 2^x/3

Can you see how no matter what x is, there is no way in hell that you can make the overall portion 0, and there is also no number you can chuck onto there to allow that portion to be negative either.

[REBORN]

Solve for x given that:

-log3x-1(1/32)=5

*3x-1 is the base*

I don't know what to do due to the negative.

Normally I'd do (3x-1)^5 = 1/32 .......but the negative must change things? how?