Answer the Confusing Question: Calculate f'(x) at x=2

[Iclaudius]

Hello friends,

I have this question which stumps me because i have no idea what its actually asking for :S - it reads:

calculate the value of the first derivative of f(x) = x^(2) -3x - 1/x at x = 2.
In which interval does it lie?

(a) -x <= x < 0
(b) 0 <= x < 4
(c) x => 4
(d) x < -2

I figure interval is related to the domain - but if we evaluate f ' (2) we get 3/4. Where this value of f ' (x) will be constantly changing (and thus no interval would exist)? So I ask what the heck are they asking of me?

Thanks in advance

 

[HallsofIvy]

First, I do NOT get 3/4 for f'(2)
f'(x)= 2x- 3+ 1/x^2 so f'(2)= 4- 3+ 1/4= 1+ 1/4= 5/4, not 3/4. Did you for get to multiply by the "n" in $(x^n)'= nx^{n-1}$ which, here, is -1?

As for the question itself, it's not very well phrased but I think that the point is that the function and its derivative do not exist at x= 0 and so exist in the two intervals $(-\infty, 0)$ and $(0, \infty)$. x= 2 lies in the second of those intervals.

 

[Iclaudius]

Hiya Hallsofivy thanks for your reply - lazy mistake on my part