Calculus help!!

[snoops123]

I have been struggling with this problem that my professor gave to me for hours now. Please help!

1. Sketch the graph of f(x)= x/(x^2+1) by using f' and f". Determine the domain, symmetry, intercepts, intervals of increase and decrease, local max. and min., concavity, and inflection points.

 

[HallsofIvy]

I have been struggling with this problem that my professor gave to me for hours now. Please help!

1. Sketch the graph of f(x)= x/(x^2+1) by using f' and f". Determine the domain, symmetry, intercepts, intervals of increase and decrease, local max. and min., concavity, and inflection points.

Have you determined what f' and f'' are yet?

 

[snoops123]

Have you determined what f' and f'' are yet?

Yes I have gotten f' and f".
f'(x)= (-x^2)+1/ (x^2+1)^2

f"(x)= 2x^5-4x^3-6x/ (x^2+1)^4

 

[HallsofIvy]

Good! Now, do you know what any of those words you are using mean? What is the "domain" of a function? What about "symmetry", "intercepts", etc.?

 

[mmm4444bot]

If you don't have a textbook, you can see two examples of how derivatives give information about function behavior HERE.

There are many examples on-line about how to use derivatives for finding extrema and concavity, too.

 

[lookagain]

Yes I have gotten f' and f".

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

f"(x)= 2x^5-4x^3-6x/ (x^2+1)^4

Not quite.

snoops, you are missing grouping symbols around your numerators, and the first pair of
parentheses as typed in the first fraction are not required.

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

f"(x) = (2x^5 - 4x^3 - 6x)/(x^2 + 1)^4

The amended second derivative is not simplified (read: reduced).

- - - - - - -- --- - - - - - - --- - - - - -- - - - - - - -- - - - - - - - --- -


\(\displaystyle f"(x) \ = \ \dfrac{(x^2 + 1)^2(-2x) \ - \ (-x^2 + 1)(2)(x^2 + 1)(2x)}{(x^2 + 1)^4} \)


\(\displaystyle f"(x) \ = \ \dfrac{-2x(x^2 + 1)^2 \ - \ 4x(-x^2 + 1)(x^2 + 1)}{(x^2 + 1)^4} \)


\(\displaystyle f"(x) \ = \ \dfrac{-2x(x^2 + 1)[x^2 + 1 \ + \ 2(-x^2 + 1)]}{(x^2 + 1)^4} \)


\(\displaystyle f"(x) \ = \ \dfrac{-2x(x^2 + 1)(x^2 + 1 - 2x^2 + 2)}{(x^2 + 1)^4} \)


\(\displaystyle f"(x) \ = \ \dfrac{-2x(x^2 + 1)(-x^2 + 3)}{(x^2 + 1)^4} \)


\(\displaystyle f"(x) \ = \ \dfrac{-2x(-x^2 + 3)}{(x^2 + 1)^3} \)