1. cubic- proof

How i can proof that third degree polynomial has a point of inflection.

Grettings. Ignis

2. Originally Posted by ignis
How i can proof that third degree polynomial has a point of inflection.

Grettings. Ignis
Show that the derivative has a turning point.

3. Hello, ignis!

How i can prove that third-degree polynomial has a point of inflection?

I assume you mean a third-degree polynomial function.

The general cubic function is: .$\displaystyle f(x) \:=\:ax^3 + bx^2 + cx +d,\;\;a \ne 0$

The first derivative is: .$\displaystyle f'(x) \:=\:3ax^2 + 2bx + c$

The second derivative is: .$\displaystyle f''(x) \:=\:6ax + 2b$

A point of inflection occurs where $\displaystyle f''(x) \,=\,0$

. . $\displaystyle 6ax + 2b \:=\:0 \quad\Rightarrow\quad x \:=\:-\dfrac{b}{3a}$

Therefore, there is an inflection point at: .$\displaystyle \left(-\dfrac{b}{3a},\:\dfrac{2b^3-9abc + 27a^2d}{27a^2}\right)$

4. Originally Posted by Soroban
Hello, ignis!

I assume you mean a third-degree polynomial function.

The general cubic function is: .$\displaystyle f(x) \:=\:ax^3 + bx^2 + cx +d,\;\;a \ne 0$

The first derivative is: .$\displaystyle f'(x) \:=\:3ax^2 + 2bx + c$

The second derivative is: .$\displaystyle f''(x) \:=\:6ax + 2b$

A point of inflection occurs where $\displaystyle f''(x) \,=\,0$

. . $\displaystyle 6ax + 2b \:=\:0 \quad\Rightarrow\quad x \:=\:-\dfrac{b}{3a}$

Therefore, there is an inflection point at: .$\displaystyle \left(-\dfrac{b}{3a},\:\dfrac{2b^3-9abc + 27a^2d}{27a^2}\right)$
It should be noted that the condition f''(x) = 0 is a necessary but not sufficient condition. f(x) will only have a point of inflection if f''(x) = 0 AND f'(x) has a turning point at the value of x given by the solution to f''(x) = 0.

Consideration of the graph of f(x) = x^4 illustrates why ....