With function f(x)=[(x^2 -1)/x^3] find f'(x),f''(x),domain, asymptotes and such..?

Question

With function f(x)=[(x^2 -1)/x^3].
(note: the function is x^2 subtract 1 over x^3)

Find the following (with steps if possible):
1) f'(x),
2) f''(x),
3) domain,
4) intercepts,
5) symmetry,
6) vertical asymptotes
7) f(x)'s behavior as 'x' gets large
8) intervals of increase and decrease (Max & Mins)
9) Concavity
10) summary of Analysis (for graphing)
11)Fully labeled sketch, or fully descripti

Answer 1

f(x) = (x^2 - 1) / x^3 = x^(-1) - x^(-3).
1) f'(x) = (-1)x^(-2) - (-3)x^(-4) = -x^(-2) + 3x^(-4).

2) f"(x) = -(-2)x^(-3) + 3(-4)x^(-5) = 2x^(-3) - 12x^(-5).

3) The domain is R\{0}.

4) x-intercepts when x^2 - 1 = 0 <=> x = ±1.
No y-intercept since 0 is not in the domain.

5) f(-x) = ((-x)^2 - 1) / ((-x)^3) = (x^2 - 1) / (-x^3) = -f(x), so f is antisymmetrical.

6) Vertical asymptotes at x = 0. As x -> 0, x^2 - 1 -> -1. So as x->0+ (i.e. from above), y->∞, and as x->0- (i.e. from below), y ->-∞.

7) lim x->∞ f(x) = lim x->∞ (1/x - 1/x^3) = 0, similarly for x->-∞.

8) f'(x) = -x^(-2) + 3x^(-4) = (-x^2 + 3) / x^4
Critical points are when x = ±√3 or x = 0.
x^4 is always positive (for x ≠ 0) and -x^2 + 3 is positive <=> x ∈ (-√3, √3).
So the intervals are as follows:
Decreasing on (-∞, -√3]
(minimum at -√3)
Increasing on [-√3, 0)
(undefined, vertical asymptotes at x = 0 as above)
Increasing on (0, √3]
(maximum at √3)
Decreasing on [√3, ∞).

9) f"(x) = 2x^(-3) - 12x^(-5) = 2(x^2 - 6) / x^5.
So the inflection points are at ±√6.
Terminology varies a bit here, so review this in light of your teacher's definitions:
On (-∞, -√6] the function is concave (or concave downwards).
On [-√6, 0) the function is convex (or concave upwards).
On (0, √6] the function is concave (or concave downwards).
On [√6, ∞) the function is convex (or concave upwards).

I'll let you summarise this and produce a sketch yourself.