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Find the following for function #f(x)=\ln(x^4+1)#...?

  • intervals of increase/decrease
  • local min/max values
  • intervals of concavity and inflection points

2 Answers
Nov 14, 2016

Answer:

For #x>0, y uarr and x<0, ydarr#.Local and global minimum f=0 at x=0. Concave/convex for |x|>/< #(3/4)^(1/4)#, and #x=+-(3/4)^(1/4)# are points of inflexion.

Explanation:

The graph below vividly answers all your questions. Yet, there is a

formal way of finding all.

#f>=ln 1>=0#

#f'=4x^3/(1+x^4) = 0# at x = 0.; f' > or < 0 according as x > or < 0.

So, for #x>0, y uarr and x<0, ydarr#

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

#=4x^2((3-4x^4))/(1+x^4)=0#, at #x = 0

and at #x = +-(3/4)^(1/4)=+-0.9306# nearly.

Also, by successive differentiation, it is seen that #f''' and f^(4) # are

also 0 at x = 0. But #f^(v)# is not 0. As f'' = f'''= 0 and the even order

derivative #f^(4) # is not 0, x=0 is not a point of inflexion.

At #x = +-(3/4)^(1/4), f''= 0# and f''' is not 0. So, these are points of

inflexion.

So, #f'' uarr# for #x in (-(3/4)^(1/4), (3/4)^(1/4))#, the graph is convex

and for# |x| > (3/4)^(1/4)#, it is concave.

Of course, at x = 0, it is flat. At #x=+-(3/4)^(1/4)#, it is neither concave

nor convex.

The tangent y = 0 does not cross the the curve here, at x = 0 but it

does at the points of inflexion # x = +-(3/4)^(1/4)#. The corrections in

this edition .are attributed to the nice observation in the comment by

Jim I hope that the answer is now alright, in all aspects.

graph{y-ln(1+x^4)=0 [-10, 10, -5, 5]}

Nov 14, 2016

Answer:

See below.

Explanation:

#f(x) = ln(x^4+1)# has domain #(-oo,oo)#

#f'(x) = (4x^3)/(x^4+1)#

The denominator is always positive, so the sign of #f'# agrees with the sign of #4x^3# which is negative on #(-oo,0)# and positive on #(0,oo)#.

So, #f# is decreasing on #(-oo,0)# and increasing on #(0,oo)#.

#f# has a local minimum of #f(0) = 0#. It has no other critical numbers, so no other local extreme values.

#f''(x) = ((12x^2)(x^4+1)-(4x^3)(4x^3))/(x^4+1)#

# = (12x^6+12x^2-16x^6)/(x^4+1)^2#

# = (-4x^6+12x^2)/(x^4+1)^2#

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

#f''(x)# is never undefined and is #0# at #0# and at #+-root(4)3#

On #(-oo,-root(4)3)#, #f''(x) < 0# so #f# is concave down. (A.k.a. "concave".)

On #(-root(4)3,0)#, #f''(x) > 0# so #f# is concave up. (A.k.a. "convex".)

On #(0, root(4)3)#, #f''(x) > 0# so #f# is concave up. (A.k.a. "convex".)

On #(root(4)3, oo)#, #f''(x) < 0# so #f# is concave down. (A.k.a. "concave".)

The inflection changes at #(-root(4)3, ln4)# and at #(root(4)3, ln4)# . Those are the only inflection points.