Question involving higher derivatives

[physicsernaw]

Homework Statement

Which of the following satisfy (f^k)(x) = 0 for all k >= 6?

a) f(x) = 7x^4 + 4 + x^-1
b) f(x) = sqrt(x)
c) f(x) = x^(9/5)
d) f(x) = x^3 - 2
e) f(x) = 1 - x^6
f) f(x) = 2x^2 + 3x^5

Homework Equations

None, but given as a problem in a chapter where the topic is higher order derivatives.

The Attempt at a Solution

I think the answer is e) but it's true for all k not just k>=6 and I don't know how finding the answer relates to higher order derivatives or how I'd use higher order derivatives to find the solution

k >= 6, for x = 1 & -1

(1-1)^k = 0
0^k = 0, lol

Edit: Hmmm maybe I'm reading the problem wrong? Is it asking which function is always = 0 for all x, and all k >= 6?

 

[Dick]

Homework Statement

Which of the following satisfy (f^k)(x) = 0 for all k >= 6?

a) f(x) = 7x^4 + 4 + x^-1
b) f(x) = sqrt(x)
c) f(x) = x^(9/5)
d) f(x) = x^3 - 2
e) f(x) = 1 - x^6
f) f(x) = 2x^2 + 3x^5

Homework Equations

None, but given as a problem in a chapter where the topic is higher order derivatives.

The Attempt at a Solution

I think the answer is e) but it's true for all k not just k>=6 and I don't know how finding the answer relates to higher order derivatives or how I'd use higher order derivatives to find the solution

k >= 6, for x = 1 & -1

(1-1)^k = 0
0^k = 0, lol

You misunderstand the problem. The kth derivative is supposed to be identically zero (zero for ALL values of x, not just some). What's the 6th derivative of 1-x^6?

 

[eumyang]

Homework Statement

Which of the following satisfy (f^k)(x) = 0 for all k >= 6?

a) f(x) = 7x^4 + 4 + x^-1
b) f(x) = sqrt(x)
c) f(x) = x^(9/5)
d) f(x) = x^3 - 2
e) f(x) = 1 - x^6
f) f(x) = 2x^2 + 3x^5

Homework Equations

None, but given as a problem in a chapter where the topic is higher order derivatives.

The Attempt at a Solution

I think the answer is e) but it's true for all k not just k>=6 and I don't know how finding the answer relates to higher order derivatives or how I'd use higher order derivatives to find the solution

k >= 6, for x = 1 & -1

(1-1)^k = 0
0^k = 0, lol

That doesn't make sense. These are supposed to be higher-order derivatives, not functions raised to an exponent. For choice (e), the 1st derivative (k = 1) is
$f'(x) = -6x^5$,
which is clearly not zero.

The question is, for which function(s) will the sixth- and higher-order derivatives be zero?
When is
$f^{(6)}(x) = 0, f^{(7)}(x) = 0, f^{(8)}(x) = 0$
(and so on)?

 

[physicsernaw]

Ohhh. I thought the question was asking for f to the kth power, not the kth derivative of f

XD

Thanks.