Question

How do you sketch the graph by determining all relative max and min, inflection points, finding intervals of increasing, decreasing and any asymptotes given `f(x)=(4x)/(x^2+1)`?

Answer 1

`f(x)` starts from `y=0` and keeps decreasing until `x=-1` where it has a minimum for `y=-2`
`f(x)` increases for `x in (-1,1)` reaching a maximum for `x=1` at `y=2`
Finally, for `x>1` `f(x)` is strictly decreasing and approaches `y=0` from positive values.

Explanation:

We start observing that `f(x)` is defined and continuous for all `x in RR` and calculate the first and second derivative:

`f(x) = (4x)/(x^2+1)`

`f'(x) = frac (4(x^2+1)-8x^2) ((x^2+1)^2)=-4(x^2-1)/((x^2+1)^2)`

`f''(x) = -4 frac ( 2x(x^2+1)^2 - 4x(x^2-1)(x^2+1)) ((x^2+1)^4)= -4 frac ( 2x(x^2+1) - 4x(x^2-1)) ((x^2+1)^3)=-8 frac ( x^3+x - 2x^3+2x) ((x^2+1)^3)=(8x(x^2-3))/((x^2+1)^3)`

Now we can determine:

1) `lim_(x->+-oo) (4x)/(x^2+1) = 0`

So `f(x)` has `y=0` as horizontal asymptote on both sides.

2) `f'(x) = 0` for `x=+-1` so there are two critical points.
We can also see that the denominator of `f'(x)` is always positive, so the sign of the derivative os the same of the numerator `-4(x^2-1)` that is:

`f'(x) < 0` for `x in (-oo,-1)`
`f'(x) > 0` for `x in (-1,1)`
`f'(x) < 0` for `x in (1,+oo)`

3) `f''(x) = 0` for `x=0` and `x=+-sqrt(3)` so there are three inflection points, and:

`f''(x) < 0` for `x in (-oo,-sqrt(3))`
`f''(x) > 0` for `x in (-sqrt(3),0)`
`f''(x) < 0` for `x in (0,sqrt(3))`
`f''(x) > 0` for `x in (sqrt(3),+oo)`

We can conclude that:

`f(x)` starts from `y=0` and keeps decreasing until `x=-1` where it has a minimum for `y=-2`
`f(x)` increases for `x in (-1,1)` reaching a maximum for `x=1` at `y=2`
Finally, for `x>1` `f(x)` is strictly decreasing and approaches `y=0` from positive values.
`f(x)` is concave down for `x in (-oo,-sqrt(3))` and `x in (0,sqrt(3))`, concave up elsewhere and has three inflection points in
`x=0` and `x=+-sqrt(3)`

graph{(4x)/(x^2+1) [-10, 10, -5, 5]}