Intervals of increasing/decreasing Using Inequalities

[Cosmophile]

Homework Statement

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

Homework Equations

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

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

The Attempt at a Solution

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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!

 

[geoffrey159]

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

 

[statdad]

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

Start with
<br /> x^2 - 4 &lt; 0 \Rightarrow (x-2)(x+2) &lt; 0 <br />

and look at the different cases to solve the inequality.

 

[Cosmophile]

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

Start with
<br /> x^2 - 4 &lt; 0 \Rightarrow (x-2)(x+2) &lt; 0<br />

and look at the different cases to solve the inequality.

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

[/B]
$f(x)=12x-x^3$

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

The Attempt at a Solution

[/B]
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!

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

 

[SammyS]

When you go from x^2 &lt; 4 to x &lt; \pm 2 you introduce a problem ...

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

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 \$​

 

[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.

 

[SammyS]

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.

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.

 

[Ray Vickson]

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:

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

********************
Stop! The inequality $-\sin(u) > 0$ is the same as $+\sin(u) < 0$. Changing the sign on both sides of an inequality reverses the direction of the inequality. This is not hard---just draw a picture.

********************

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

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

$\arcsin\frac {x}{2} > \arcsin 0$​

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

$x > 0$

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

 

[Drakkith]

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.

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