# Problems from Concavity and convexity, inflection points of a function

Find the inflection points of the following functions.

1. $$f(x)=x^2+1$$
2. $$f(x)=1$$
3. $$f(x)=\ln(x^2+1)-x$$
4. $$f(x)=x^3-2x^2$$
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### 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. $$f'(x)=2x \Rightarrow f''(x)=2$$

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

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

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

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.

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

$$f''(x)=0 \Rightarrow \dfrac{-2x^2+2}{(x^2+1)^2}=0 \Rightarrow -2x^2+2=0 \Rightarrow x^2=1 \Rightarrow x=1, \ x=-1$$

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

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

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

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

### 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 $$x=\dfrac{4}{6}$$.
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