VCE Methods Question Thread!

[yawho]

12a and b can be done in much shorter ways.

[TrueTears]

brightsky applied the algebraic definition, but i guess geometrically is fine too

[yawho]

interesting, how do you do it geometrically?

[TrueTears]

apply the geometric representation, odd and even functions

[yawho]

care to elaborate, it is not trivial

[TrueTears]

never said it was trivial but sure, divide through the rational function to arrive at a hyperbola, one for f(x) and one for f(-x), hyperbolas aren't even by nature, the only way to restrict a hyperbola is to look at its asymptotes, there are 2 ways to arrive at the conclusion that f(x) = p

one is to note that the equality f(x)=f(-x) is analogous to an intersection of the 2 curves for x E R\{-r,r} and the only intersection geometrically is when f(x) = p (horizontal asymptote)

another way is to notice that since a hyperbola is not even in nature, we have to find some part of it that is, and a horizontal line is clearly even, hence f(x) = p (again the horizontal asymptote)

[Deceitful Wings]

could you show me the working out for this question?

Solve |x²-4x| + 1 < 5

[pi]

Hint: Try solving solving |x^2 -4x| + 1 = 5 and then using a quick-sketch to find out what "domain for x" you are after. Remember that

[Deceitful Wings]

this is what i am getting
|x²-4x| <5-1
x²-4x <4 or -4
x²-4x-4< 0 or x²-4x+4<0
(x-2)² -8< 0 or (x-2)² < 0
x-2 < +- sqrt 8 or x <2
x < 2+- sqrt 2 or x < 2
x< 2+sqrt 2 or 2-sqrt 2 or x < 2

but i am pretty sure it's wrong because i am supposed to get a different domain :\
i know the answer if i draw it up, but i want to work it out algebraically, is it possible?

[kamil9876]

x2-4x <4 or -4

there's your mistake, it should be which is the same as saying AND .

So you can solve each seperate problem and then the final answer is the intersection.

[Deceitful Wings]

thanks so much!!! that totally makes more sense now!

[Deceitful Wings]

x2-4x <4 or -4

there's your mistake, it should be which is the same as saying AND .

So you can solve each seperate problem and then the final answer is the intersection.

wait... i am still stuck on this question.
i got x>2, x<2+2sqrt2 and x<2-2sqrt2
this can't be right?
i thought it would be
f(x)={ x²-4x, (x<2-2sqrt2,0) union (4,x<2+2sqrt2)
        { -x²-4x, [0,4]

how do i get from your answer to this answer?

P.S. sorry for my really noob computer skills, i would type the proper symbols (e.g. sqrt) if i knew how

[yawho]

never said it was trivial but sure, divide through the rational function to arrive at a hyperbola, one for f(x) and one for f(-x), hyperbolas aren't even by nature, the only way to restrict a hyperbola is to look at its asymptotes, there are 2 ways to arrive at the conclusion that f(x) = p

one is to note that the equality f(x)=f(-x) is analogous to an intersection of the 2 curves for x E R\{-r,r} and the only intersection geometrically is when f(x) = p (horizontal asymptote)

another way is to notice that since a hyperbola is not even in nature, we have to find some part of it that is, and a horizontal line is clearly even, hence f(x) = p (again the horizontal asymptote)

The problem with the explanation is that you assumed f(x)=p and verified that it satisfies f(x)=f(-x). Are there others that also satisfy f(x)=f(-x)?

How do you approach part b geometrically?

[TrueTears]

nope i didn't assume f(x) = p, where did i state or imply that? the second way to arrive at the conclusion that f(x) = p is perfectly valid, it is just simple geometric reasoning, what is your query regarding it?

part b) is even easier geometrically, again you have f(x) as p+(q-pr)/(x+r), since f(-x) = -f(x) means that geometrically the graph has rotational symmetry with respect to the origin, then this only occurs if the hyperbola has x = 0 and y = 0 as its asymptote, hence r = 0 and p = 0 -> f(x) = q/x

anything more you want me to elaborate on?

[yawho]

Part b was explained well.
Regarding part a: first explanation, how did you conclude the intersection of the two curves is f(x)=p?
Second explanation, what did you mean by some part of a hyperbola is even when it is an odd function?