Question

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

Answer 1

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]}

Answer 2

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.