Determine the local maximum and local minimum points for the function $f(x) = 2x^3 + 3x^2 - 12x + 3$. Determine the local maximum and local minimum points for the function $f(x) = 2x^3 + 3x^2 - 12x + 3$. 
I know that the local maximum and local minimum points largest and smallest value of the function, but I do not understand how to find the points from the equation. Assistance would be greatly appreciated.
 A: To find the critical points of $f$, you should differentiate and set the derivative equal to 0.
$f(x) = 2x^3 + 3x^2 - 12x + 3 \implies f'(x) = 6x^2 + 6x - 12$
Then, $f'(x) = 0 \iff 6x^2 + 6x- 12 = 0 \iff x^2 +x -2$.
Using Bhaskara, you find the roots of this polynomial ($-2$ and $1$).
Now evaluate $f(-2)$ and $f(1)$ to decide which one is the maximum and which one is the minimum.
You should also note that
$\lim_{x \rightarrow -\infty} f(x) = -\infty$ and $\lim_{x \rightarrow +\infty} f(x) = +\infty$
A: 
I know that the local maximum and local minimum points largest and smallest value of the function ...

Actually, a local maximum or minimum occurs, respectively, at a point where the function reaches its maximum or minimum value in an open interval.  
Definition. A function $f$ is said to have a relative (or local) maximum at a point $c$ in an open interval $(a, b)$ if $f(x) \leq f(c)$ for each $x \in (a, b)$.  A function $f$ is said to have a relative (or local) minimum at a point $c$ in an open interval $(a, b)$ if $f(x) \geq f(c)$ for each $x \in (a, b)$. A relative maximum or minimum is called a relative extremum of the function.
The function does not necessarily reach its maximum or minimum value at a relative extremum.  
For example, the function 
$$g(x) = x + \frac{1}{x}$$
has a relative maximum at $x = -1$ since $g(x) \leq g(-1) = -2$ for $x \in (-\infty, 0)$ and a relative minimum at $x = 1$ since $g(x) \geq g(1) = 2$ for $x \in (0, \infty)$.  Since $g(-1) < g(1)$, the relative maximum $x = -1$ is not an absolute (global) maximum of the function $g$.  Similarly, since $g(1) > g(-1)$, the relative minimum $x = 1$ is not an absolute (global) minimum of the function $g$.


... I do not understand how to find the points from the equation.

The relative extrema of a function $f$ occur at points where $f'(x) = 0$ or $f'(x)$ does not exist.  Points at which $f'(x) = 0$ are called critical points.
Differentiating the function $f(x) = 2x^3 + 3x^2 - 12x + 3$ yields
\begin{align*}
f'(x) & = 6x^2 + 6x - 12\\
      & = 6(x^2 + x - 2)\\
      & = 6(x + 2)(x - 1)
\end{align*}
Since $f'$ is a polynomial function, the derivative is defined everywhere.  Setting the derivative equal to zero yields the critical points $x = -2$ or $x = 1$.  
When the derivative is positive in an interval, the function is increasing over that interval.  When the derivative is negative in an interval, the function is decreasing over that interval.
Since the derivative $f'(x) = 6(x + 2)(x - 1)$ is a quadratic polynomial, it is continuous.  Thus, it can only change sign at a point where it is equal to zero.  Thus, the derivative can only change sign at the critical points $x = -2$ and $x = 1$. 
In the interval $(-\infty, -2)$, $f'(x) = 6(x + 2)(x - 1) > 0$ as you can tell by substituting the test point $-3$ for $x$.  Hence, the function $f$ is increasing on the interval $(-\infty, -2]$.  
In the interval $(-2, 1)$, $f'(x) < 0$ as you can tell by substituting the test point $0$ for $x$.  Hence, the function $f$ is decreasing on the interval $[-2, 1]$.  
In the interval $(1, \infty)$, $f'(x) = 6(x + 2)(x - 1) > 0$ as you can tell by substituting the test point $2$ for $x$.  Thus, the function $f$ is increasing on the interval $[1, \infty)$.  
We showed above that $f$ is increasing on $(-\infty, -2]$ and decreasing on $[-2, 1]$.  Since the function increases to the left of $-2$, is defined at $x = -2$, and decreases to the right of $-2$, the function has a relative maximum at $x = -2$.
We showed above that $f$ is decreasing on the interval $[-2, 1]$ and increasing on $[1, \infty)$.  Since $f$ is decreasing to the left of $1$, defined at $x = 1$, and increasing to the right of $1$, $f$ has a relative minimum at $x = 1$.
We have applied the First Derivative Test.
First Derivative Test.  Let $f$ be a function that is continuous on $[a, b]$ and differentiable on $(a, b)$ except possibly at a point $c$.
(a)  If $f'(x) > 0$ for all $x < c$ and $f'(x) < 0$ for all $x > c$, then $f$ has a relative maximum at $c$.
(b) If $f'(x) < 0$ for all $x < c$ and $f'(x) > 0$ for all $x > c$, then $f$ has a relative minimum at $c$.
Since 
$$\lim_{x \to \infty} f(x) = \infty$$ 
and 
$$\lim_{x \to -\infty} f(x) = -\infty$$ 
the relative extrema are not absolute extrema.
A: If the function $f$ is differentiable, the extrema-- maxima, minima and saddle points-- occur when $f'(x) = 0$.
If at those points if $f''(x) > 0$, $(x,f(x))$ is a local minimum.  If $f''(x) < 0$, $(x,f(x))$ is a local maximum.  If $f''(x) = 0$, $(x, f(x))$ is a saddle point (which might, or might not, be a maximum of minimum or neither).
So 
1)find out what $f'(x)$ is.  (It will be a quadradic).
2)Solve for $f'(x) = 0$.  (There will be two solutions, $x_0$ and $x_2$.)
3) find out what $f''(x)$ is.  (It will be a linear equation.)
4) Plug in $f''(x_0)$ and $f''(x_1)$ and see whether they are positive, negative, or zero.
