1. ## #Times f(x)=0

f(x) = x4-4x3-8x2+4

How many times does f(x) = 0? (excuse the bad formulation)

Hint: Gather as much information from the derivative of f(x) as possible. We are only interested in the number of times f(x) equals zero.

Okay so differentiating the function is simple: f ' (x) = 4x3-12x2-16x, but what kind of information do I need to extract in order to determine how many times f(x)=0?

2. ## Re: #Times f(x)=0

Hi Cinnaman,
a few tipps:
f'(x)=0 => x1=-1, x2= 0; x3=4 the event. max- and min points of f(x)
If f''(x1) < 0 => x1 is max. point =>MAX1( x1|f(x1))
If f''(x1) > 0 => x1 is min. point =>MIN1( x1|f(x1))
analog with x2 und x3.
EXTx= Extrem x-point;
xp<xn;
If sign of f(EXTxp) is not = f(EXTxn) => there is a x, beetwen xp und xn with f(x)=0

3. ## Re: #Times f(x)=0

Ah, that makes sense, thank you.

I figured out that there is no x where f(x)=0 between x=-1 and x=0, but there is one between x=0 and x=4.

Is the last step now to take the limit as x approaches inf / -inf and use them as extreme points aswell?

4. ## Re: #Times f(x)=0

Hello, Cinnaman!

$f(x) \:= \:x^4 - 4x^3 - 8x^2 + 4$

How many times does $f(x) = 0$ ?

Hint: Gather as much information from the derivative of $f(x)$ as possible.
We are only interested in the number of times f(x) equals zero.

Set $f'(x) = 0$ and solve.

. . $4x^3 - 12x^2 - 16x \:=\:0 \quad\Rightarrow\quad 4x(x+1)(x-4) \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}1,0,4$

There are horizontal tangents at: $(\text{-}1,1),\:(0,4),\:(4,\text{-}124)$

The Second Derivative Test give us: . $\begin{Bmatrix}(\text{-}1,1) & \text{min.} \\ (0,4) & \text{max.} \\ (4,\text{-}124) & \text{min.} \end{Bmatrix}$

The graph looks like this.
Code:
               |
*        |
-*-
*     * | *          *
*   *  |  *
-*-   |
-------------+---♥-------♥---
|
|    *     *
|     *   *
|      -*-
|
Therefore, the graph crosses the x-axis twice.