Question

f(x)=x^2 -1 / x^3
Determine the concavity of the function's curve, and find the inflection points.

Answer 1

did you find the second derivative?

Answer 2

no
just first derivative

Answer 3

if it was me, i would start with
\[f(x)=\frac{1}{x}-\frac{1}{x^3}\] and find the derivative twice

Answer 4

the first derivative was

Answer 5

what is second ?

Answer 6

like i said, since you have to find the second derivative, i would start with
\[f(x)=\frac{1}{x}-\frac{1}{x^3}\] so
\[f'(x)=-\frac{1}{x^2}+\frac{3}{x^4}\] making the second derivative
\[f''(x)=\frac{2}{x^3}-\frac{12}{x^5}\] snappier that way

Answer 7

course now you still have to subtract, giving
\[f''(x)=\frac{2x^2-12}{x^5}=\frac{2(x^2-6)}{x^5}\] and now the zeros are easy to find

Answer 8

ok when I find second derivative what next ?

Answer 9

find the critical points, where it is equal to zero or undefined.
it is undefined at \(x=0\) but you can ignore that because the function is also undefined there, so it would be a miracle if the derivative existed at 0
the zeros of \(x^2-6\) you find in your head

Answer 10

can you draw the solution ?

Answer 11

not sure what you mean
the zeros i.e. the inflection points are at \(x=-\sqrt6\) and \(x=\sqrt6\)

Answer 12

aha ok than ?

Answer 13

what else is there to do? you were asked for the inflection points right?

Answer 14

yes

Answer 15

ok thanks a lot :):)

Answer 16

yw
oh if you also need concavity, check for the signs of the second derivative, where is it positive or negative
that is all

Answer 17

ok :) thanks agine