Concavity and convexity, inflection points of a function

Find the inflection points of the following functions.

$f(x)=x^2+1$
$f(x)=1$
$f(x)=\ln(x^2+1)-x$
$f(x)=x^3-2x^2$

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.

$f'(x)=2x \Rightarrow f''(x)=2$

$f''(x)=0 \Rightarrow 2=0 \Rightarrow$ There are no inflection points.

$f'(x)=0 \Rightarrow f''(x)=0$

$f''(x)=0 \Rightarrow 0=0$

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.

$f'(x)=\dfrac{2x}{x^2+1} \Rightarrow f''(x)=\dfrac{-2x^2+2}{(x^2+1)^2}$

$f''(x)=0 \Rightarrow \dfrac{-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 $x=-1$.

$f'(x)=3x^2-4x \Rightarrow f''(x)=6x-4$

$f''(x)=0 \Rightarrow 6x-4=0 \Rightarrow x=\dfrac{4}{6}$

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

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

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