# 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

Nov 14, 2016

For $x > 0 , y \uparrow \mathmr{and} x < 0 , y \downarrow$.Local and global minimum f=0 at x=0. Concave/convex for |x|>/< ${\left(\frac{3}{4}\right)}^{\frac{1}{4}}$, and $x = \pm {\left(\frac{3}{4}\right)}^{\frac{1}{4}}$ are points of inflexion.

#### Explanation:

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

formal way of finding all.

$f \ge \ln 1 \ge 0$

$f ' = 4 {x}^{3} / \left(1 + {x}^{4}\right) = 0$ at x = 0.; f' > or < 0 according as x > or < 0.

So, for $x > 0 , y \uparrow \mathmr{and} x < 0 , y \downarrow$

$f ' ' = 4 {x}^{2} \left(\frac{3}{1 + {x}^{4}} - 4 {x}^{4} / {\left(1 + {x}^{4}\right)}^{2}\right)$

$= 4 {x}^{2} \frac{\left(3 - 4 {x}^{4}\right)}{1 + {x}^{4}} = 0$, at #x = 0

and at $x = \pm {\left(\frac{3}{4}\right)}^{\frac{1}{4}} = \pm 0.9306$ nearly.

Also, by successive differentiation, it is seen that $f ' ' ' \mathmr{and} {f}^{4}$ 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 $x = \pm {\left(\frac{3}{4}\right)}^{\frac{1}{4}} , f ' ' = 0$ and f''' is not 0. So, these are points of

inflexion.

So, $f ' ' \uparrow$ for $x \in \left(- {\left(\frac{3}{4}\right)}^{\frac{1}{4}} , {\left(\frac{3}{4}\right)}^{\frac{1}{4}}\right)$, the graph is convex

and for$| x | > {\left(\frac{3}{4}\right)}^{\frac{1}{4}}$, it is concave.

Of course, at x = 0, it is flat. At $x = \pm {\left(\frac{3}{4}\right)}^{\frac{1}{4}}$, 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 $x = \pm {\left(\frac{3}{4}\right)}^{\frac{1}{4}}$. 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

See below.

#### Explanation:

$f \left(x\right) = \ln \left({x}^{4} + 1\right)$ has domain $\left(- \infty , \infty\right)$

$f ' \left(x\right) = \frac{4 {x}^{3}}{{x}^{4} + 1}$

The denominator is always positive, so the sign of $f '$ agrees with the sign of $4 {x}^{3}$ which is negative on $\left(- \infty , 0\right)$ and positive on $\left(0 , \infty\right)$.

So, $f$ is decreasing on $\left(- \infty , 0\right)$ and increasing on $\left(0 , \infty\right)$.

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

$f ' ' \left(x\right) = \frac{\left(12 {x}^{2}\right) \left({x}^{4} + 1\right) - \left(4 {x}^{3}\right) \left(4 {x}^{3}\right)}{{x}^{4} + 1}$

$= \frac{12 {x}^{6} + 12 {x}^{2} - 16 {x}^{6}}{{x}^{4} + 1} ^ 2$

$= \frac{- 4 {x}^{6} + 12 {x}^{2}}{{x}^{4} + 1} ^ 2$

$= \frac{- 4 {x}^{2} \left({x}^{4} + 3\right)}{{x}^{4} + 1} ^ 2$

$f ' ' \left(x\right)$ is never undefined and is $0$ at $0$ and at $\pm \sqrt[4]{3}$

On $\left(- \infty , - \sqrt[4]{3}\right)$, $f ' ' \left(x\right) < 0$ so $f$ is concave down. (A.k.a. "concave".)

On $\left(- \sqrt[4]{3} , 0\right)$, $f ' ' \left(x\right) > 0$ so $f$ is concave up. (A.k.a. "convex".)

On $\left(0 , \sqrt[4]{3}\right)$, $f ' ' \left(x\right) > 0$ so $f$ is concave up. (A.k.a. "convex".)

On $\left(\sqrt[4]{3} , \infty\right)$, $f ' ' \left(x\right) < 0$ so $f$ is concave down. (A.k.a. "concave".)

The inflection changes at $\left(- \sqrt[4]{3} , \ln 4\right)$ and at $\left(\sqrt[4]{3} , \ln 4\right)$ . Those are the only inflection points.