Odd & Even Functions (1 Viewer)

[kpq_sniper017]

This is a question I got in my first Prelim. 3U paper last year (but I just came across it again).

Given f(x) = 1-2x+x², find functions g(x) and h(x) such that f(x) = g(x)+h(x), where g(x) is an even function and h(x) is an odd function.

Just by looking at it, you can see g(x) is x²+1 and h(x) is -2x. But is there any algebraic way of solving this question? I don't know that just plucking the two functions out of the air is worth 2 marks.

 

[Xayma]

Well look at f(-x), it becomes 1+2x+x² x²+1 remains constant and 2x is timesed by -1. Therefore g(x)=x²+1 and h(x)=-2x

 

[Estel]

g(x) is even, and therefore forms all even power components of f(x), and h(x) is odd, and therefore forms all odd compondents of f(x)? Or am I just waaaay off the mark(s)? [pardon the pun]

 

[Xayma]

No Estel you would be right as long as you remember all constants are part of an even function.

 

[Estel]

No I meant I was unsure if it was sufficient as justification for a bald answer.

I hate questions like that... reminds me of science.

 

[Affinity]

One size fits all
g(x) = (1/2)[ f(x) + f(-x) ]
h(x) = (1/2)[ f(x) - f(-x) ]

works for any f(x)

 

[kpq_sniper017]

where'd you learn that affinity??
is it an ext 1 "application" - coz i haven't learnt that "one size fits all rule".

 

[CM_Tutor]

It's really easy to prove:

f(x) = g(x) + h(x) _____ (1), where g(x) is even and h(x) is odd.
So, f(-x) = g(-x) + h(-x) = g(x) - h(x) _____ (2)

(1) + (2): f(x) + f(-x) = g(x) + h(x) + g(x) - h(x) = 2 * g(x)
So, g(x) = (1 / 2) * [f(x) + f(-x)]

(1) - (2): f(x) - f(-x) = g(x) + h(x) - [g(x) - h(x)] = g(x) + h(x) - g(x) + h(x) = 2 * h(x)
So, h(x) = (1 / 2) * [f(x) - f(-x)]

 

[kpq_sniper017]

is it a rule that i should have knwon off-hand, because it's not in any textbooks that i can find, and my teacher never told the class.

btw, with that question, to get full marks, would u have to go through CM_Tutor's proof, or could you just state the rule and go from there?

 

[Estel]

Another question I have that goes with pcx's is are you allowed to use sums to products in your trig identity questions? And with the auxillary method in trig, do you need to equate coefficients or can you just shove your method in .. i.e. R- blah a- blah, done (I'm not so sure considering I see 4 marks allocated to a simple question in the HSC, are they expecting something?)

Confusing what you are and aren't allowed to assume... otherwise that sin3x/sinx etc question that keeps popping up could be done in 3 lines

 

[CM_Tutor]

Originally posted by pcx_demolition017
is it a rule that i should have knwon off-hand, because it's not in any textbooks that i can find, and my teacher never told the class.

It's useful only in limited circumstances in Extn 1, but it does have some useful Extn 2 applications, so no harm in knowing it, but no great harm in not knowing it.

btw, with that question, to get full marks, would u have to go through CM_Tutor's proof, or could you just state the rule and go from there?

No need for proof, as question said "find" functions. You could, in a case like this question, simply write:

g(x) = 1 + x² is an even function, as g(-x) = 1 + (-x)² = 1+ x² = g(x).
h(x) = -2x is an odd function, as h(-x) = -2(-x) = -(-2x) = -h(x).

Since it is clear that g(x) + h(x) = 1 + x² + -2x = 1 - 2x + x² = f(x), it follows that g(x) = 1 + x² and h(x) = -2x are the required functions.

Originally posted by Estel
Another question I have that goes with pcx's is are you allowed to use sums to products in your trig identity questions? And with the auxillary method in trig, do you need to equate coefficients or can you just shove your method in .. i.e. R- blah a- blah, done (I'm not so sure considering I see 4 marks allocated to a simple question in the HSC, are they expecting something?)

Confusing what you are and aren't allowed to assume... otherwise that sin3x/sinx etc question that keeps popping up could be done in 3 lines

I would not assume the answers to the auxilliary angle method. As for sums-to-products and products-to-sums, I never bothered to learn them, and simply derive them on the spot. (Sorry, that might not be much help... I think you could assume them if needed, esp. in Extn 2, which is where they usually appear.)

 

[Estel]

Can you help me cut out some of my working? I don't want to derive the whole auxillary method.... for 2 marks... ewwww

Write in form Rcos(x+A)
R is positive
3cosx+rt3sinx
= Rcos(x+A)
= RcosxcosA - RsinxsinA

Therefore, equating coefficients:
RcosA = 3
RsinA = -rt3
R^2(cos^2A+sin^2A)=12
R=rt12

cosA = 3/rt12
sinA = -rt3/rt12
So A lies in 4th quadrant.
Given condition: A is positive.
A = 330 degrees

Therefore
3cosx+rt3sinx = rt12cos(x+330)

ummm, what can I cut out
It's like turning 1+1=2 into a 3rd year university dissertation.

My old answer:
3cosx+rt3sinx
R = rt12
tanA=rt3/3
3cosx+rt3sinx = rt12cos(x-30)
rt12cos(x+330)

 

[Affinity]

omg.

3cos(x) + sqrt(3)sin(x)
= (2*sqrt(3))*[ (sqrt(3)/2)*cos(x) + (1/2)sin(x) ]
= 2sqrt(3)*[cos(pi/6)cos(x) + sin(pi/6)sin(x)]
= 2sqrt(3)*[cos(x - pi/6)]
= 2sqrt(3)*[cos(x + 11pi/6)]