Problems from Concavity and convexity, inflection points of a function

Development:

We will solve the exercise by doing all the steps: to derive two times the function, equal the result to zero, then solve the equation and the resulting points will be the inflection points.

1. $f^{\prime }(x)=2x\Rightarrow f^{″}(x)=2$

   $f^{″}(x)=0\Rightarrow 2=0\Rightarrow$ There are no inflection points.

2. $f^{\prime }(x)=0\Rightarrow f^{″}(x)=0$

   $f^{″}(x)=0\Rightarrow 0=0$

   This is a particular case. We come to a situation that is not false but we do not find any concrete $x$. This means that all $x$ are inflection points.

3. $f^{\prime }(x)={\displaystyle \frac{2x}{x^2+1}}\Rightarrow f^{″}(x)={\displaystyle \frac{-2x^2+2}{(x^2+1{)}^2}}$

   $f^{″}(x)=0\Rightarrow {\displaystyle \frac{-2x^2+2}{(x^2+1{)}^2}}=0\Rightarrow -2x^2+2=0\Rightarrow x^2=1\Rightarrow x=1,x=-1$

   We have inflection points in $x=1$ and $x=-1$.

4. $f^{\prime }(x)=3x^2-4x\Rightarrow f^{″}(x)=6x-4$

   $f^{″}(x)=0\Rightarrow 6x-4=0\Rightarrow x={\displaystyle \frac{4}{6}}$

   We have inflection points in $x={\displaystyle \frac{4}{6}}$.

Solution:

1. There are no inflection points.
2. All the points are inflection points.
3. We have inflection points in $x=1$ and $x=-1$.
4. We have inflection points in $x={\displaystyle \frac{4}{6}}$.

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