# Study of the maximum and minimum points of the function.

Does the function f (x) = x ^ 3 + x ^ 2 + x + 1 at x = 0 have some maximum or minimum point? In other exercises to make the 1st derivative, you can factor terms and give two values ​​ax but in this function I could not factor. Making the first derivative is left f'(x) = 3x ^ 2 + 2x + 1, but I can not factor. I hope they help me because it is very messy.

2voto

Anon User Points 0
Bluenile, the derivative that you calculated is ok. For quadratic polynomials, there is a very direct way of finding the roots and is using the following expression: 
\begin{align}&{\color{blue}a}x^2+{\color{red}b}x+{\color{green}c}=0\\&x_{1,2}=\frac{-{\color{red}b}\pm \sqrt{{\color{red}b}^2-4\cdot {\color{blue}} \cdot {\color{green}c}}}{2\cdot {\color{blue}a}}\\&in this case we have that:\\&a=3\\&b=2\\&c=1\\&\text{so the expression becomes:}\\&x_{1,2}=\frac{-{\color{red}2}\pm \sqrt{{\color{red}2}^2-4\cdot {\color{blue}3} \cdot {\color{green}1}}}{2\cdot {\color{blue}3}}=\\&\frac{-2 \pm \sqrt{-8}}{6}\end{align}As you can see from this expression became the root of a negative number which means that it does not have real roots.

Taking advantage of the use of the geogebra graph the function, and I think you will realize that you do not have ends (which if you have is a point of inflection at x=-1/3)

Salu2

2voto

Anon User Points 0
So... 
\begin{align}&f(x)=x^3+x^2+x+1\\&\\&f'(x)=3x^2+2x+1\\&\\&\text{The question is if x=0 is an extreme. If we calculate f'(0) we have that is equal to 1}\\&\text{either x=0 is not an endpoint. Then you ask if the function f has extremo}\\&\\&f'(x)=3\left(x^2+\frac{2}{3}x+\frac{1}{3}\right)\\&\\&f'(x)=3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}+\frac{1}{3}\right)\\&\\&f'(x)=3\left(x^2+\frac{2}{3}x+\frac{1}{9}+\frac{2}{9}\right)\\&\\&f'(x)=3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{2}{3}\\&\\&f'(x)=3\left(x+\frac{1}{3}\right)^2+\frac{2}{3}\\&\\&f'(x)\geq \frac{2}{3} , \forall x \in\mathbb{R}\\&\\&\text{let us Remember this, if x=x_0 is an endpoint to a function f:X\subset \mathbb{R}\to \mathbb{R}, then}\\&\text{f'(x_0)=0 so by comparison, our function does not have extreme }\forall x \in \mathbb{R}, \text{is more}\\&\text{the function is always increasing}\\&\\&\end{align}

### TheXpert.space

TheXpertSpace is an online community of people who give and receive advice.
You can check other people responses or create a new question if it's not already solved