# Math Help - Absolute max values..

1. ## Absolute max values..

It says find the absolute max values of

|12x - x^3| +1 x e (-1, 3]

I know how to find the absolute max values of normal functions but its the | | that is confusing me, every time I try my normal method, my answers are wrong,. What do the lines mean and how do I approach them?

2. Originally Posted by Dave J
It says find the absolute max values of

|12x - x^3| +1 x e (-1, 3]

I know how to find the absolute max values of normal functions but its the | | that is confusing me, every time I try my normal method, my answers are wrong,. What do the lines mean and how do I approach them?
I can understand the $|12x - x^3|$, what's messing me up is the $+1 x e (-1, 3]$. What is this?

I'll ignore the last half of this and just focus on the absolute value part.

We need to find out where $12x - x^3 = 0$.
$x^3 - 12x = 0$

$x(x^2 - 12) = 0$

So x = 0 or $x^2 - 12 = 0$

Thus the solutions are $x = 0, -\sqrt{12}, \sqrt{12}$.

Now we ask, where is $12x - x^3 = 0$ negative? We have 4 test intervals:
$(-\infty, -\sqrt{12})$: $12x - x^3 < 0$
$(-\sqrt{12}, 0)$: $12x - x^3 > 0$
$(0, \sqrt{12})$: $12x - x^3 < 0$
$(\sqrt{12}, \infty )$: $12x - x^3 > 0$

All of this leads to the following:
$|12x - x^3| = \left \{ \begin{array}{cc} -(12x - x^3); & x \leq -\sqrt{12} \\ 12x - x^3; & -\sqrt{12} < x \leq 0 \\ -(12x - x^3); & 0 < x \leq \sqrt{12} \\ 12x - x^3; & \sqrt{12} < x \end{array} \right.$

Now you can take your derivative on the "piece-wise defined" function, like you would any other function.

-Dan

3. Ahh the question is

|12x - x^3| +1

and
x is all real numbers from (-1, 3]

See I know about the derivative and such, but the +1 and the two | | throw me off. Im not sure how they interact with each other.

4. Originally Posted by Dave J
Ahh the question is

|12x - x^3| +1

and
x is all real numbers from (-1, 3]

See I know about the derivative and such, but the +1 and the two | | throw me off. Im not sure how they interact with each other.
Ah! The 20 W lightbulb in my head just turned on!

You do it in the same way as I showed below. You need to find out where the argument of the absolute value function is negative. Then just take the negative of that. Simpler example:
$f(x) = |x| + 1$

$f(x) = \left \{ \begin{array}{cc} -x + 1 & x \leq 0 \\ x + 1 & 0 < x \end{array} \right.$

-Dan

5. Oh I see how it works, thanks