Graphing helpppp...

Printable View

November 11th 2006, 11:37 PM
^_^Engineer_Adam^_^

Graphing helpppp...

Sketch the graph of the function and determine its domain and range:
1.) $F(x) = \frac{x^3 - 2x^2}{x-2}$

the answer from the back of our book is
domain: ${x|x is not equal to 2};$ - i dont know why

range: $[0,+ /infty )$

2.) $F(x) = |3x + 2|$

thank you very much
November 12th 2006, 12:47 AM
CaptainBlack

Quote:

Originally Posted by ^_^Engineer_Adam^_^

Sketch the graph of the function and determine its domain and range:
1.) $F(x) = \frac{x^3 - 2x^2}{x-2}$

the answer from the back of our book is
domain: ${x|x \mbox{ is not equal to }2};$ - i dont know why

range: $[0,+ \infty )$

The domain is the set on which the function is defined, as division by
zero is not allowed the function is not defined when $x=2$
as $x-2$ is zero.

When $x \ne 2$ we may write:

$F(x) = \frac{x^3 - 2x^2}{x-2}=\frac{x^2(x-1)}{x-2}=x^2$

which may take any value from $0$ and upwards, so the range is: $[0, \infty)$

So the graph of the function is identical to they of $y=x^2$ except that there
is a hole at $x=2$

RonL
November 12th 2006, 01:00 AM
CaptainBlack

Quote:

Originally Posted by ^_^Engineer_Adam^_^

2.) $F(x) = |3x + 2|$

First the domain of this function is all of $\mathbb{R}$ as it is defined for all real numbers.

As $F(x) \ge 0$ the smallest value that $F$ can take is $0$ and there is clearly no largest value that it can take. So its range is $[0,\infty)$ .

To sketch this we can rewrite the function as:

$<br /> F(x) = \left\{ {3x + 2,\ x \ge -2/3 \atop<br /> -3x-2,\ x < -2/3} \right.<br />$

So the graph is these two line segments with a join at $x=-2/3$

RonL
November 12th 2006, 01:25 AM
OReilly

1 Attachment(s)

Here is graph for F(x)=|3x+2|.
November 12th 2006, 06:38 AM
Soroban

Hello, ^_^Engineer_Adam^_^!

Quote:

Sketch the graph of the function and determine its domain and range:
$1)\;f(x) \:= \:\frac{x^3 - 2x^2}{x-2}$

To repeat what Captain Black said . . .

We see that $x \neq 2$ .
. . $\boxed{\text{Domain: all } x \neq 2}$ .or: $\boxed{(-\infty,\,2) \cup (2,\,\infty)}$

Since $x \neq 2$ , we have: . $f(x) \:=\:\frac{x^2(x - 2)}{x-2} \:=\:x^2$

We have a parabola, $y = x^2$ , with a "hole" at (2,4).

Code:

| * | * | * | * * | o(2,4) * | * ---------***-------- |

Hence: . $\boxed{\text{Range: all }y \geq 0}$ .or: $[0,\,\infty)$

Quote:

$2)\;f(x) \:= \:|3x + 2|$

We know that the graph of $y \:=\:|x|$ is a $V$ with its vertex at the origin.

The graph of $y \:=\:|3x|$ is the same $V$ but it "rises faster".

The graph of $y \:=\:|3x + 2| \:=\:\left|3\left(x + \frac{2}{3}\right)\right|$
. .is the same graph moved $\frac{2}{3}$ units to the left.

The graph looks like this:

Code:

\ | / \ |/ \ / \ /| \ / | \ / | --------*---+------ -2/3 |

Therefore: . $\begin{array}{cc}\text{Domain} & (-\infty,\,\infty) \\ \text{Range} & [0,\,\infty) \end{array}$