The derivative (1 Viewer)

[Estel]

Suppose f(x)=ax^2+bx+c
f(y)=0
f'(y)=0
does that mean there is a double root at x=y?
and suppose that f"(y)=0
does that mean there is a triple root at x=y?

Ta.

 

[CM_Tutor]

Originally posted by Estel
Suppose f(x)=ax^2+bx+c
f(y)=0
f'(y)=0
does that mean there is a double root at x=y?

Yes

and suppose that f"(y)=0
does that mean there is a triple root at x=y?

Ta.

No, as this would mean a = 0, in which case the function is linear, and has at most one root.

 

[Estel]

Thanks

but if you had something like ax^3 + bx^2 etc...
f, f', f"(y)=0 would that mean x=y is 3 roots?

And would have to go through all that if you had got to the part x^2=0 or x^3=0?

 

[CM_Tutor]

The result you are looking for is a 4u polynomial result, usually referred to as the multiple root theorem. It states that if x = alpha is a root of multiplicity k of the polynomial P(x) = 0, then x = alpha is a root of multiplicity (k - 1) of the polynomial P'(x) = 0.

Thus, if P(a) = P'(a) = 0 and P''(a) <> 0, then x = a is a double root of P(x) = 0, and (x - a)² is a factor of P(x) = 0.

Alxo, if P(a) = P'(a) = P''(a) = 0 and P'''(a) <> 0, then x = a is a triple root of P(x) = 0, and (x - a)³ is a factor of
P(x) = 0.

 

[velox]

sorry, what is f" ?

 

[:: ck ::]

second derivative...

 

[Estel]

CM_Tutor, do you have a proof on hand for that multiple root theorem? I don't have a 4U textbook...

 

[CM_Tutor]

Let P(x) = (x - a)^kQ(x) where Q(x) is a polynomial in x, a is a constant and k is a positive integer.

Further, x = a is a root of P(x) = 0 of multiplicity k.

We seek to prove that a is a root of multiplicity k - 1 of P'(x) = 0. This means it is necessary to prove that
(x - a)^k-1 is a factor of P'(x).

P'(x) = d/dx P(x) = Q(x) * k(x - a)^k-1 * 1 + (x - a)^k * Q'(x), using the Product Rule.
= (x - a)^k-1[Q(x) * k * 1 + (x - a)Q'(x)]
= (x - a)^k-1[kQ(x) + (x - a)Q'(x)]

Clearly, x = a is a root of P'(x) = 0 of multiplicity k - 1, as required.

Estel: This is a standard 4u question - All 4u students should be able to write out this proof in an exam.

 

[Estel]

You are just brilliant.
The proof is understandable and short enough to make it worthwhile using...