Author Topic: Absolute - confused!! (Read 843 times) Tweet Share

Ematuro · « **on:** March 10, 2011, 09:10:30 pm »

f (x) = | x²− 3x |+ 2

What I don't understand.. is how you find the domains of it in the hybrid function. I get everything else.. and I know how to sketch and all but yeah finding the domain confuses me can anyone help?

xZero · « **Reply #1 on:** March 10, 2011, 09:15:53 pm »

hybrid...function? well the domain of f(x) is R

Darren · « **Reply #2 on:** March 11, 2011, 12:03:06 am »

It's not a hybrid, it's a modulus function. The graph would be flipped ( just add a negative sign ) for the values in between the x intercepts of the equation.

/0 · « **Reply #3 on:** March 11, 2011, 12:46:25 am »

Modulus functions are hybrid functions:

$f(x)=\Bigg\lbrace\begin{array}{rcl} (x^2-3x)+2 & \mbox{and} & x^2-3x \geq 0 \\\\ -(x^2-3x)+2 &\mbox{and}& x^2-3x<0\end{array}$

So you need to find where $x^2 -3x \geq 0$ and $x^2-3x<0$ . If you sketch a graph of $y=x^2-3x$ you can find the domains visually.

nacho · « **Reply #4 on:** March 11, 2011, 03:53:29 pm »

~~The only thing the modulus function changes is the range, the domain remains the same~~

edit: I retract this statement, sorry for any confusion.

Ematuro · « **Reply #5 on:** March 11, 2011, 04:27:41 pm »

Quote from: /0 on March 11, 2011, 12:46:25 am

Modulus functions are hybrid functions:

$f(x)=\Bigg\lbrace\begin{array}{rcl} (x^2-3x)+2 & \mbox{and} & x^2-3x \geq 0 \\\\ -(x^2-3x)+2 &\mbox{and}& x^2-3x<0\end{array}$

So you need to find where $x^2 -3x \geq 0$ and $x^2-3x<0$ . If you sketch a graph of $y=x^2-3x$ you can find the domains visually.

But is it possible to find the domains without sketching? and yeah what I don't get is how to solve those domains

/0 · « **Reply #6 on:** March 11, 2011, 06:17:01 pm »

You can find the domains without sketching, if you're prepared to do more work.

$\mathbf{Algebraically}$

To find $\lbrace x:x^2-3x \geq 0\rbrace$ algebraically, factor:

$x(x-3)\geq 0$

Note that if you multiply two numbers to give a non-negative number, they must both be non-positive or non-negative.

Case 1: Both are non-negative:
$x\geq 0$ and $(x-3)\geq 0\Rightarrow x \geq 3$

This gives the solution $x \geq 3$ .

Case 2: Both are non-positive:
$x\leq 0$ and $(x-3) \leq 0\Rightarrow x \leq 3$

This gives the solution $x\leq 0$ .

So the full solution is $x \geq 3$ or $x\leq 0$

Now for $\lbrace x:x^2-3x<0\rbrace$ simply take the complement, which is ( $x<3$ or $x> 0$ ) $\Rightarrow 0< x <3$ .

$\mathbf{Graphically}$

$y=x^2-3x$ has roots at $x=0,3$ . Since the coefficient of $x$ is positive, we know the parabola will be 'nose-down'.

So $\lbrace x:x^2-3x \geq 0\rbrace=\lbrace x:x \leq 0\mbox{ and } x\geq 3\rbrace$ and $\lbrace x^2-3x <0 \rbrace = \lbrace x:0<x<3\rbrace$

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Author Topic: Absolute - confused!! (Read 843 times) Tweet Share

Ematuro

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