# How would I graph 10x/X^2+1 in detail using derivatives?

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### 1 Answer

Given `f(x)=(10x)/(x^2+1)` :

`f'(x)=(10(x^2+1)-10x(2x))/((x^2+1)^2)=-(10(x^2-1))/((x^2+1)^2)`

`f''(x)=(20x(x^2-3))/((x^2+1)^3)`

(1) From f(x) we know the y-intercept is 0.

(2) The function is a rational function whose domain is `RR`

(3) Any local extrema occur at critical points. Since f'(x) is defined everywhere, we need only find where f'(x)=0.

A fraction is zero if the numerator is zero (and the denominator is nonzero) so:

`-10x^2+10=0=>x=+-1` . Using the first derivative test we find that x=1 is a local maximum, while x=-1 is a local minimum.So (1,5) is a maximum, (-1,-5) is a minimum.

(4) We can check for concavity and inflection points using the second derivative test.

`20x(x^2-3)=0=>x=0,+-sqrt(3)` . These are all inflection points, and the concavity is :

down on `(-oo,-sqrt(3))`

up on `(-sqrt(3),0)`

down on `(0,sqrt(3))`

up on `(sqrt(3),oo)`

(4) The only x-intercept is 0.

(5) The graph:

** The third derivative is rarely used for graphing purposes. In this case, it can tell the rate of change of the concavity.

The name of the third derivative in acceleration problems is the jerk, which gives the rate of change of the acceleration. I haven't seen any other applications of the third derivative **