# Intervals of increasing/decreasing Using Inequalities

• Cosmophile
In summary: Also, please remember to use the template provided when starting a new thread, and make sure to show your own attempted solution so we can see where you might be going wrong. Thanks!
Cosmophile

## 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

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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!

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

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.

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.

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

Cosmophile said:

## 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##?

When you go from $x^2 < 4$ to $x < \pm 2$ you introduce a problem ...
Cosmophile said:
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 \ ##​

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.

Cosmophile said:
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.

Cosmophile said:
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.

Last edited:
Cosmophile
Cosmophile said:
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.

## Q1: What is an interval of increasing/decreasing using inequalities?

An interval of increasing/decreasing using inequalities is a range of values on a number line where the function is either continuously increasing or decreasing. This means that for every value within the interval, the output of the function is either increasing or decreasing, respectively.

## Q2: How do you graph an interval of increasing/decreasing using inequalities?

To graph an interval of increasing/decreasing using inequalities, you first need to identify the direction of the inequality (greater than or less than). Then, plot the critical points, which are the points where the function changes from increasing to decreasing or vice versa. Finally, shade the appropriate region between the critical points to represent the interval.

## Q3: What is the difference between an open and closed interval of increasing/decreasing using inequalities?

An open interval does not include the endpoints, while a closed interval includes the endpoints. For example, the open interval (0, 5) means all values between 0 and 5, but not including 0 and 5. On the other hand, the closed interval [0, 5] includes the values 0 and 5.

## Q4: How do you solve an inequality to find the interval of increasing/decreasing?

To solve an inequality and find the interval of increasing/decreasing, you need to first isolate the variable on one side of the inequality. Then, determine the direction of the inequality based on the variable's coefficient. Finally, write the solution as an interval notation, using brackets for a closed interval and parentheses for an open interval.

## Q5: Can an interval of increasing/decreasing using inequalities have more than two critical points?

Yes, an interval of increasing/decreasing using inequalities can have more than two critical points. This means that the function can change from increasing to decreasing or vice versa multiple times within the interval. To graph this type of interval, you would need to plot all the critical points and shade the appropriate regions between them.

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