# Intervals of increasing/decreasing Using Inequalities

1. May 15, 2015

### Cosmophile

1. The problem statement, all variables and given/known data

Identify the open intervals on which the function $f(x) = 12x-x^3$ is increasing or decreasing

2. Relevant equations

$f(x)=12x-x^3$

$\frac {df}{dx} = 12-3x^2 = -3(x^2 - 4)$

3. The attempt at a solution

I'm reading out of two textbooks. One is a Larson/Edwards text, which says to use test values upon intervals and whatnot. However, my other textbook (Serge Lang), as well as the MIT 18.01 lecture videos I am watching, don't use this method and instead utilize inequalities. I prefer this method, but am having a difficult time exercising it, as my high school did very, very little with inequalities, so my experience is minimal. Here is what I've got:

$f$ is increasing when $f'(x) > 0$

$-3(x^2 - 4) > 0$

$(x^2 - 4) < 0$

$x^2 < 4$

$x < \pm 2$
$x < 2$, $x< -2$ = $-x > 2$

Here is where I get stuck. I'm not sure how to make sense of this answer. I know that $f$ is increasing on the interval $(-2, 2)$ and decreasing on $(-\infty, -2)(2, \infty)$, but don't know how to translate my answer to say this. I'm sure I'm either making a simple mistake or am just missing a bit of know-how regarding these types of problems. Any help is greatly appreciated; thanks in advance!

2. May 15, 2015

### geoffrey159

You've made a mistake, think again about what you wrote after x^2 < 4

3. May 15, 2015

When you go from $x^2 < 4$ to $x < \pm 2$ you introduce a problem: notice that $-5 < -2$ is true but $(-5)^2 < 4$ is not.

$$x^2 - 4 < 0 \Rightarrow (x-2)(x+2) < 0$$

and look at the different cases to solve the inequality.

4. May 15, 2015

### Cosmophile

I had noticed this, and immediately thought about absolute values, which is when I realized I was truly lost. I appreciate the insight!!

5. May 15, 2015

### Ray Vickson

Think about the graph $y = x^2$. For what x-region do you have $y < 4$?

6. May 15, 2015

### SammyS

Staff Emeritus
However, you can work with $\displaystyle\ x^2 < 4 \$ more directly.

Take the square root of both sides. (Yes, that preserves the inequality.)

$\displaystyle\ \sqrt{x^2\ } < \sqrt{4\ } \$​

Strictly speaking, $\displaystyle\ \sqrt{x^2\ } = \left|x\right| \$, so you have the following inequality.

$\displaystyle\ \left|x\right|<2 \$​

7. May 15, 2015

### Cosmophile

I ended up doing this:

$x^2 < 4$

$|x| < 2$, giving me $-2 < x < 2$

I'm currently working on:​

$f(x) = \cos\frac {x}{2}, 0 < x < 2\pi$​
So far, I have:

$\frac {df}{dx} = -\frac{1}{2} \sin\frac {x}{2}$

$-\frac{1}{2} \sin\frac {x}{2} > 0$

$\sin\frac {x}{2} > 0$​

$\arcsin\frac {x}{2} > \arcsin 0$​

$\frac {x}{2} > 0$

$x > 0$

Again, I'm confused. I know that, on the interval, f(x) is never increasing.

8. May 15, 2015

### SammyS

Staff Emeritus
Start a new thread for a new problem, unless what you're adding is just a very closely related extension of the original problem.

In this case, best to start a new thread , and state the problem completely in the body of the opening post - no matter how descriptive the title of the thread.

9. May 15, 2015

### Ray Vickson

Last edited: May 15, 2015
10. May 15, 2015

### Staff: Mentor

As SammyS said, please start a new thread for new problems. It keeps thread discussions from getting mixed up between different problems.