Maths Methods 3/4 Help Thread 2011

[Water]

log e (4e^3x)

would equate to 3x + log e (4) ...

I"m pretty sure, x would have no restrictions then.


@matty, DAMNIT, I was too late ):

[pi]

I read it as being multiplied by x. Clearly this is not the case, and there is no restriction.

My bad, I failed at the first time

[panicatthelunchbar]

hi, can someone please help me with this?!

cos-squared x + sin x = 1,    [0,2pie]

sorry, dont know latex or whatever everyone uses :/

thanks

[brightsky]

cos^2(x) + sin^2(x) = 1
so cos^2(x) = 1 - sin^2(x)
so the equation becomes
1 - sin^2(x) + sin x = 1
let sin(x) = u
-u^2 + 1 + u = 1
-u^2 + u = 0
u(1 - u) = 0
u = 0 or 1 - u = 0
sin x = 0 or 1 - sin x = 0
from the first one, we get x = -2pi, -pi, 0, pi, 2pi in the given domain
from the second one, sinx = 1
so x = -3pi/2, pi/2 in the given domain

[luffy]

cos^2(x) + sin^2(x) = 1
so cos^2(x) = 1 - sin^2(x)
so the equation becomes
1 - sin^2(x) + sin x = 1
let sin(x) = u
-u^2 + 1 + u = 1
-u^2 + u = 0
u(1 - u) = 0
u = 0 or 1 - u = 0
sin x = 0 or 1 - sin x = 0
from the first one, we get x = -2pi, -pi, 0, pi, 2pi in the given domain
from the second one, sinx = 1
so x = -3pi/2, pi/2 in the given domain

- The domain is .... (just mentioning it xD)

[panicatthelunchbar]

so cos^2(x) = 1 - sin^2(x)
so the equation becomes
1 - sin^2(x) + sin x = 1

How did you do that?

[xZero]

sin^2(x) + cos^2(x) = 1, cos^2(x) = 1 - sin^2(x). Sub this into the original equation

[brightsky]

cos^2(x) + sin^2(x) = 1
so cos^2(x) = 1 - sin^2(x)
so the equation becomes
1 - sin^2(x) + sin x = 1
let sin(x) = u
-u^2 + 1 + u = 1
-u^2 + u = 0
u(1 - u) = 0
u = 0 or 1 - u = 0
sin x = 0 or 1 - sin x = 0
from the first one, we get x = -2pi, -pi, 0, pi, 2pi in the given domain
from the second one, sinx = 1
so x = -3pi/2, pi/2 in the given domain

- The domain is .... (just mentioning it xD)

oh woops, thought it was [-2pi, 2pi] for some strange reason. :p then yeah just take the solutions in [0, 2pi].

[kefoo]

1. The General solution to 2cos(3x) + 1 = 0 is:
 - first i rearranged it to cos 3x = -1/2
 - then went 3x = 2kpi - pi/3 and got stuck, prob got it wrong here.
2. It is known that the graph of the function with domain R and rule h(x) = g(x) - f(x) where g(x) = ax and f(x) = 2^x has an x-axis intercept when x = 3. What is the value of a
3. log10(y) = log10(x) + 1

[brightsky]

1)
2cos(3x) + 1 = 0
cos(3x) = -1/2
since we know that cos(pi/3) = 1/2
then cos(pi - pi/3) = -1/2
cos(2pi/3) = -1/2
by symmetry, cos(pi + pi/3) = -1/2 as well, that is, cos(4pi/3) = -1/2
the rest of the solutions are found by adding 2pi*n, where n E Z (rotating it around the whole circle)
so 3x = 2pi/3 + 2pi*n or 4pi/3 + 2pi*n
3x = (2pi + 6pi*n)/3 or (4pi + 6pi*n)/3
x = (2pi + 6pi*n)/9 or (4pi + 6pi*n)/9

2)
h(x) = ax - 2^x
it has x-intercept at x=3, so h(3) = 3a - 2^3 = 0
3a - 8 = 0
a = 8/3

3)
i'm not sure what you want to do with that equation but:
log(y) = log(x) + 1
log(y) - log(x) = 1
log(y/x) = 1
y/x = 10

edit: arithmetic error

[luken93]

1) cos(3x) = -1/2
3x = arccos(-1/2)
3x = pi +- pi/3 (Think about symmetry)
x = pi/3 +- pi/9 + 2n/3pi
brightsky's method is probs easier

2) h(3) = 0
a(3) - 2^(3) = 0
3a - 8 = 0
a = 8/3

3) What are we solving for?
If y;
log10(y) = log10(x) + log10(10)
log10(y) = log10(10x)
y = 10x
If x, rearrange

[luken93]

Quick Q, I was under the assumption that to make f(g(x)) definable, you had to restrict the domain of g(x) to be a subset of the domain of f(x).

But so it turns out, for all the questions I've encountered, the domain of f(g(x)) = the domain of g(x) intersecting with the implied domain of f(g(x)).

So the question beckons, what's the point of restricting the domain of g(x) to make f(g(x)) definable when you can find the domain once you've "composed" them. Am I missing something fundamental?

[xZero]

to make f(g(x)) definable, the range of g(x) must be a subset of the domain of f(x), hence we must restrict the domain of g(x), which will become the domain of f(g(x))

[luken93]

Oh wait, I think I've answered my own question there..

[luken93]

to make f(g(x)) definable, the range of g(x) must be a subset of the domain of f(x), hence we must restrict the domain of g(x), which will become the domain of f(g(x))

Yeah I understand that part, but when you find the domain of f(g(x)) (without considering the restrictions on g(x)), it is the intersection of the implied domains of f(g(x)) and g(x)