# Test f for concavity?

## $f$ 2 times differentiable in $\mathbb{R}$ with: ${\left(f ' \left(x\right)\right)}^{3} + 3 f ' \left(x\right) = {e}^{x} + \cos x + {x}^{3} + 2 x + 7$ $\forall$$x$$\in$$\mathbb{R}$

Jan 16, 2018

$f$ is convex in $\mathbb{R}$

#### Explanation:

Solved it i think.

$f$ is 2 times differentiable in $\mathbb{R}$ so $f$ and $f '$ are continuous in $\mathbb{R}$

We have ${\left(f ' \left(x\right)\right)}^{3} + 3 f ' \left(x\right) = {e}^{x} + \cos x + {x}^{3} + 2 x + 7$

Differentiating both parts we get

$3 \cdot {\left(f ' \left(x\right)\right)}^{2} f ' ' \left(x\right) + 3 f ' ' \left(x\right) = {e}^{x} - \sin x + 3 {x}^{2} + 2$ $\iff$

$3 f ' ' \left(x\right) \left({\left(f ' \left(x\right)\right)}^{2} + 1\right) = {e}^{x} - \sin x + 3 {x}^{2} + 2$

• $f ' {\left(x\right)}^{2} \ge 0$ so $f ' {\left(x\right)}^{2} + 1 > 0$

$\iff$ $f ' ' \left(x\right) = \frac{{e}^{x} - \sin x + 3 {x}^{2} + 2}{3 \left({\left(f ' \left(x\right)\right)}^{2} + 1\right) > 0}$

We need the sign of the numerator so we consider a new function

$g \left(x\right) = {e}^{x} - \sin x + 3 {x}^{2} + 2$
, $x$$\in$$\mathbb{R}$

$g ' \left(x\right) = {e}^{x} - \cos x + 6 x$

We notice that $g ' \left(0\right) = {e}^{0} - \cos 0 + 6 \cdot 0 = 1 - 1 + 0 = 0$

For x=π $\implies$ g'(π)=e^π-cosπ+6π=e^π+1+6π>0

For x=-π g'(-π)=e^(-π)-cos(-π)-6π=1/e^π+cosπ-6π=1/e^π-1-6π<0

We finally get this table which shows the monotony of $g$ Supposed ${I}_{1} = \left(- \infty , 0\right]$ and ${I}_{2} = \left[0 , + \infty\right)$
$g \left({I}_{1}\right) = g \left(\left(- \infty , 0\right]\right) = \left[g \left(0\right) , {\lim}_{x \rightarrow - \infty} g \left(x\right)\right) = \left[3 , + \infty\right)$

$g \left({I}_{2}\right) = g \left(\left[0 , + \infty\right)\right) = \left[g \left(0\right) , {\lim}_{x \rightarrow + \infty} g \left(x\right)\right) = \left[3 , + \infty\right)$

because

• ${\lim}_{x \rightarrow - \infty} g \left(x\right) = {\lim}_{x \rightarrow - \infty} \left({e}^{x} - \sin x + 3 {x}^{2} + 2\right)$

$| \sin x | \le 1$ $\iff$ $- 1 \le - \sin x \le 1$ $\iff$

${e}^{x} + 3 {x}^{2} + 2 - 1 \le {e}^{x} + 3 {x}^{2} + 2 - \sin x \le {e}^{x} + 3 {x}^{2} + 2 + 1$ $\iff$

${e}^{x} + 3 {x}^{2} + 1 \le {e}^{x} - \sin x + 3 {x}^{2} + 2 \le {e}^{x} + 3 {x}^{2} + 3 \iff$

${e}^{x} + 3 {x}^{2} + 1 \le g \left(x\right) \le {e}^{x} + 3 {x}^{2} + 3$

• Using the squeeze/sandwich theorem we have

${\lim}_{x \rightarrow - \infty} \left({e}^{x} + 3 {x}^{2} + 1\right) = + \infty = {\lim}_{x \rightarrow - \infty} \left({e}^{x} + 3 {x}^{2} + 3 x\right)$ Therefore, ${\lim}_{x \rightarrow - \infty} g \left(x\right) = + \infty$

• ${\lim}_{x \rightarrow + \infty} g \left(x\right) = {\lim}_{x \rightarrow + \infty} \left({e}^{x} - \sin x + 3 {x}^{2} + 2\right)$

With the same process we end up to

${e}^{x} + 3 {x}^{2} + 1 \le g \left(x\right) \le {e}^{x} + 3 {x}^{2} + 3$

However, ${\lim}_{x \rightarrow + \infty} \left({e}^{x} + 3 {x}^{2} + 1\right) = + \infty = {e}^{x} + 3 {x}^{2} + 3$

Therefore, ${\lim}_{x \rightarrow + \infty} g \left(x\right) = + \infty$

The range of $g$ will be:

${R}_{g} = g \left({D}_{g}\right) = g \left({I}_{1}\right) \cup g \left({I}_{2}\right) = \left[3 , + \infty\right)$

• $0 \notin {R}_{g} = \left[3 , + \infty\right)$ so $g$ has no roots in $\mathbb{R}$
$g$ is continuous in $\mathbb{R}$ and has no solutions. Therefore, $g$ preserves sign in $\mathbb{R}$

That means

{(g(x)>0" , "xεRR),(g(x)<0" , "xεRR):}

Thus, g(π)=e^π-sinπ+3π^2+2=e^π+3π^2+2>0

As a result $g \left(x\right) > 0$, $x$$\in$$\mathbb{R}$

And $f ' ' \left(x\right) > 0$ , $x$$\in$$\mathbb{R}$

$\to$ $f$ is convex in $\mathbb{R}$

Jan 16, 2018

See below.

#### Explanation:

Given $y = f \left(x\right)$ the curve curvature radius is given by

$\rho = {\left(1 + {\left(f '\right)}^{2}\right)}^{\frac{3}{2}} / \left(f ' '\right)$ so given

${\left(f '\right)}^{3} + 3 f ' = {e}^{x} + \cos x + {x}^{3} + 2 x + 7$ we have

$3 {\left(f '\right)}^{2} f ' ' + 3 f ' ' = {e}^{x} + 3 {x}^{3} - \sin x + 2$ or

$f ' ' \left(1 + {\left(f '\right)}^{2}\right) = \frac{1}{3} \left({e}^{x} + 3 {x}^{3} - \sin x + 2\right)$ or

$\frac{1}{f ' ' \left(1 + {\left(f '\right)}^{2}\right)} = \frac{3}{{e}^{x} + 3 {x}^{3} - \sin x + 2}$ or

$\rho = {\left(1 + {\left(f '\right)}^{2}\right)}^{\frac{3}{2}} / \left(f ' '\right) = \frac{3 {\left(1 + {\left(f '\right)}^{2}\right)}^{\frac{5}{2}}}{{e}^{x} + 3 {x}^{3} - \sin x + 2}$

now analyzing $g \left(x\right) = {e}^{x} + 3 {x}^{3} - \sin x + 2$ we have

$\min g \left(x\right) = 0$ for $x \in \mathbb{R}$ so $g \left(x\right) \ge 0$ and then the curvature in

$\rho = \frac{3 {\left(1 + {\left(f '\right)}^{2}\right)}^{\frac{5}{2}}}{{e}^{x} + 3 {x}^{3} - \sin x + 2}$ doesn't changes sign so we conclude that $f \left(x\right)$ epigraph is convex in $\mathbb{R}$