Sketch the graph and show all local extrema and inflection points.

Question

f(x)= 1/x^2-2x-8

I believe the ans is "Relative max: [1,-1/9]" and No inflection points".

Answer 1

f(x) = x^-2 - 2x - 8
f ' (x) = -2x^-3 - 2
f '' (x) = 6x^-4

extrema:
-2/x^3 = 2
x^3 = -1
x = -1, f(-1) = 1 - 2(-1) - 8 = -5

at x = -1, f ''(-1) is > 0, so
(-1,-5) is a minimum

my answer confirmed by

http://www.wolframalpha.com/input/?i=plot++y+%3D+1%2Fx%5E2-2x-8


pts of inflection:
6/x^4 = 0
so solution, thus no points of inflection.

Answer 2

Good solution, but wrong problem.

f = 1/(x^2-2x-8) = 1/((x-4)(x+2))
f' = 2(x-1)/(x^2-2x-8)^2
f" = 6(x^2-2x+4)/(x^2-2x-6)^3

f'=0 at x=1
f" is never zero, so no inflection points.

We see there are vertical asymptotes at x=4 and x=-2
f'=0 at x=1
f"(1) < 0, so f(1) is a local max

take it to wolframalpha to see the graph.

Answer 3

Oh yeah. Forgot to mention that y=0 is the horizontal asymptote.