Mathematics Extension 1 (1 Viewer)

[markclyon]

<o>>I have trouble understanding the following question. If anyone can help I would appreciate it.</o>>

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<o>>A function f is called odd if f (-x)=-f(x) for all x.</o>>
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<o>>a) Prove that every odd function is zero at x=0.</o>>
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<o>>b) Prove that every odd polynominal P(x) is divisible by x.</o>>
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<o>>c) The polynominal P(x) is know to be monic, to be an odd function, and to have a root at x=-5. Show that P(x) has degree no less than 3.</o>>
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<o>>d) Find a polynominal Q(x) of degree 3 with the properties given in (c). Are there any other p</o><o>olynominals of degree 3 with these properties?</o>>

 

[cutemouse]

(a) From the property of an odd function, odd functions have point symmetry about the origin, ie. Odd functions pass through (0, 0), thus at x=0 f(0)=0.

(b) Using the factor theorem, if P(x) is divisible by x, then P(0)=0, which is true using (a) if P(x) is odd.

 

[The Nomad]

Prove every odd function is zero at x = 0:

f(-x) = -f(x) for x = 0, so f(0) = -f(0).
2f(0) = 0, so f(0) = 0, proven as required.

Someone else can do the rest for now.

 

[cutemouse]

Prove every odd function is zero at x = 0:

f(-x) = -f(x) for x = 0, so f(0) = -f(0).
2f(0) = 0, so f(0) = 0, proven as required.

I don't think that is a substainial proof.

You need to explain why f(0)=0

 

[sexisash]

a) An odd function is defined by f(x) = -f(-x).
At x = 0, f(0) = -f(-0)
f(0) + f(0) = 0 (since f(0) = f(-0))
2f(0) = 0
f(0) = 0, as required.

b) Take any odd polynomial P(x).
By the remainder theorem, P(x) is divisible by (x + 0) if P(0) = 0.
From (a), we know that for all odd functions, P(0) = 0.
Hence any odd polynomial P(x) is divisible by x.

c) If P(x) is a monic, the coefficient of the highest power is 1.

If P(x) is odd, all the powers of x must be odd.

If P(x) has at least one real root, the lowest power of x must be 1.
So

is possible.
But P(x) != x, since P(-5) = 0.
Hence the lowest degree of P(x) must be 3.

d) Let the polynomial be given by:

Now from (a), Q(0) = 0:

And since a root exists at x=-5, Q(-5) = 0:

From (a), Q(0) = 0:

Hence Q(x) given by:

There is one other polynomial of degree 3 with these properties, and it is given by the same polynomial with a leading coefficient of -1:

This would produce the same graph, but with a different y-intercept and opposite concavity.

 

[addikaye03]

I don't think that is a substainial proof.

You need to explain why f(0)=0

Hmm that's the proof i would have used. That's just an algebriac proof. Yours makes an assumption of the proof, it's not really that correct, i can't really explain, but it's not a complete derivation.

 

[annabackwards]

Hmm that's the proof i would have used.

Same here. If 2f(0) = 0 then by dividing by 2 on both sides, f(0) = 0

 

[shaon0]

I don't think that is a substainial proof.

You need to explain why f(0)=0

Dude you're so fail...Your proofs based on an observation rather than a logical process.

 

[cutemouse]

Hmm that's the proof i would have used. That's just an algebriac proof. Yours makes an assumption of the proof, it's not really that correct, i can't really explain, but it's not a complete derivation.

Okay...

The property of an odd function is that it has point symmetry about the origin. I guess you could almost treat this as a 'definition'... Thus it must pass through (0, 0)...

BTW, I see how it works now. I didn't see the algebraic approach before. Both approaches are good I guess, but I think mine's better

 

[shaon0]

Okay...

The property of an odd function is that it has point symmetry about the origin. I guess you could almost treat this as a 'definition'... Thus it must pass through (0, 0)...

BTW, I see how it works now. I didn't see the algebraic approach before. Both approaches are good I guess, but I think mine's better

geometrically it has symmetry of rotation around the origin, that would be better def but algebraic's more safer

 

[cutemouse]

geometrically it has symmetry of rotation

Yeah point symmetry, same thing.

Usually the geometric approach is easier anyway for most questions (like in Conics).