NEED HELP! With Defining Absolute Functions!

[brandoncasilla]

So the absolute value equation is 2|x+6|+12=4 how do you define it so its like y={x+2 if x>2 or -(x+2) if x<2???

 

[jpage447]

You must have typed/written the equation incorrectly. The original can't be 2|x+6|+12=4. This is why...

if we subtract 12 from both sides we get
2|x+6|= -8
since absolute value is always positive there is obviously something wrong with this equation because an absolute value CANNOT equal a negative number.

 

[brandoncasilla]

Ok nvm so how about this?

How do you define quadratic absolute equations like |x^2-9|=x^2-9

 

[daon2]

How do you define quadratic absolute equations like |x^2-9|=x^2-9

Are you trying to solve this equation? If so, do you know the solution set of the equation \(\displaystyle |y| = y\)?

 

[pka]

So the absolute value equation is 2|x+6|+12=4

Are you trying to solve \(\displaystyle 2|x+6|+12=4~?\)

If so, look before you leap.
Ask yourself, "what non-negative number can I add to 12 and get 4?"

The answer is not any. So there is no solution.

There is no solution to \(\displaystyle |x-4|+4.5=3\).

 

[HallsofIvy]

How do you define quadratic absolute equations like |x^2-9|=x^2-9

\(\displaystyle x^2- 9= (x- 3)(x+ 3)\). That is a quadratic function opening upward and is 0 at x= -3 and 3. So it is positive for x< -3, negative for -3< x< 3, and positive for x> 3.

For x< -3, then, [itex]|x^2- 9|= x^2- 9= x^2- 9[/tex] and that last equation is true for all x.

For -3< x< -3, \(\displaystyle |x^2- 9|= -(x^2- 9)= x^2- 9\) so that \(\displaystyle 2(x^2- 9)= 2(x- 3)(x+ 3)= 0\). That is true only for x= -3 and x= 3.

For x> 3, \(\displaystyle |x^2- 9|= x^2- 9= x^2- 9\) and, again, that is true for all x.

Putting those together, \(\displaystyle |x^2- 9|= x^2- 9\) is true for all x less than or equal to -3 and for all x greater than or equal to 3.

In general |f(x)|= f(x) if and only if \(\displaystyle f(x)\ge 0\).