# 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)#?

##### 1 Answer

#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

Now we can determine:

1)

So

2)

We can also see that the denominator of

#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 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]}