How do you sketch the graph #y=1/(1+x^4)# using the first and second derivatives?

1 Answer
Jan 10, 2017

#y(x)# is positive and even, infinitesimal for #x->+-oo#, with a maximum in #x=0# and two inflection points for #x=+-root(4)(3/5)#

Explanation:

To analyze the behavior of the function we start from considering the domain of definition and see that #y(x)# is defined and continuous in all of #RR# as the denominator never vanishes.

We can also see that #y(x)# is positive in its domain and it it even as

#y(-x) = y(x)#

so that its graph is symmetrical with respect to the #y#-axis.

At the limits of the domain #y(x)# is infinitesimal as:

#lim_(x->+-oo) y(x) = 0#

so the function is going to have #y=0# as asymptote on both sides.

Now we calculate the first and second derivatives:

#y'(x) = - (4x^3)/(1+x^4)^2#

#y''(x) = frac ( -12x^2(1+x^4)^2 +32x^6(1+x^4)) ((1+x^4)^4) = frac ( -12x^2(1+x^4) +32x^6) ((1+x^4)^3) = frac (4x^2( -3x^4 +8x^4-3) )((1+x^2)^3) = frac (4x^2(5x^4-3)) ((1+x^2)^3)#

We can see that the only critical point where #y'(x) = 0# is for #x=0# and that:

# x<0 => y'(x) >0#
# x>0 => y'(x) <0#

so that the function is monotone increasing for #x in (-oo,0)#, reaches a maximum for #x=0# and then is monotone decreasing for #x in (0,+oo)#.

The second derivative has three zeros in #x=0# and #x=+-root(4)(3/5)#. We know the first is a maximum, so #y(x)# has two inflection points in #x=+-root(4)(3/5)#. and:

#abs(x) > root(4)(3/5) => y''(x) > 0# and y(x) is concave up.
#abs(x) < root(4)(3/5) => y''(x) < 0# and y(x) is concave down.

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