onur369's Methods Question Thread :)

[onur369]

His talking about my question -.-

[brightsky]

its an absolute function, thats not right.

i believe just sketch the normal function f(x) = 2sin(x), find f|x| = 2sin|x| --> just wipe out everything to the left of the y-axis and reflect the stuff on the right, and then translate across to the right pi/2 units.

or more generally
2sin|x - pi/2|
= 2 sin(x - pi/2), x - pi/2 >= 0, x >= pi/2
= 2 sin(pi/2 - x), x - pi/2 < 0, x < pi/2

sketch from there

[luken93]

its an absolute function, thats not right.

The -------- seperates yours and onur's q?

[Respect]

His talking about my question -.-

he?

[onur369]

Hey guys, doing some prac exams:

I attached the initial question but its following questions are as followed:

a) Find the area, A, of rectangle XYZW in terms of a.

b) Find the maximum value of A and the value of a for which this occurs.

[xZero]

prac exam already? nice work!

a) To find the area, we need the length and the width. The length of the retangle is from the point X(-a,0) to W(a,0) so if we find the difference in their x-value, it will give us the length. Hence the length is a-(-a)=2a. The width is the difference in y-value from X(-a,0) to Y(-a,b) which is b. However we want to express the area in terms of a so we must find a relationship between a and b. Note the equation y=9-3x^2, where y=b and x=a. Thus b=9-3a^2. Area = LxW, Area=2a x (9-3a^2), Area=18a-6a^3


b)Now diff the area, find the maximum value for a

[onur369]

Find the implied domain of sqrt(5-sqrt(1-x))

[onur369]

answer says -24 lessthanequalto x  lessthenequalto 1

[gossamer]

You need to consider both the "inside" and "outside" square roots.

For there to be a real solution for the equation, sqrt(1-x) must be smaller or equal to 5 ( i.e. (-inf,5] ). But the sqrt can't be equal to a real negative number so it must be equal to [0,5]

You should be able to get the answer from there

[luken93]

My mistake



For it to work, let

Implied domain would then be

For it to exist, domain of u must be [-inf, 5]

Hence the max is 5







Hence, the domain is [-24, 1]

[onur369]

Thanks alot,

I have another question:

If g: [k,4] -> R, g(x) = -(x-2)^2 +3

a) Find the smallest value for k such that g^-1 exists
b) Find the rule that defines g^-1(x), giving the domain.

[brightsky]

sort of similar method by yeah:
1-x => 0, x <= 1
5-sqrt(1-x) >= 0, sqrt(1-x) =< 5, 1-x =< 25 (note this is justified because both sides of the inequation are positive), x => -24
Hence we have -24 =< x =< 1

notation: => is greater or equal to, etc.

[brightsky]

sketch the graph. you only want one side of the parabola. you should be able to see it from there.

(the answer you get should be 2.)

as for the rule, just swap the "y" and "x" around then make the domain [-1, 3] --> the range of the original function g(x)

[onur369]

Hi guys i need urgent help with this question:

Find exact solutions of equation:
sin(3x)-cos(3x)=0 for 0 less than equal to x less than equal to pi

[kamil9876]

Just let , hence . And so the equation is equivalent to:

, solve for y over the domain, then remember to divide by 3 to get the x values.