yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Finding decreasing interval given the function | Calculus | Khan Academy


3m read
·Nov 10, 2024

Let's say we have the function ( f(x) = x^6 - 3x^5 ).

My question to you is, using only what we know about derivatives, try to figure out over what interval or intervals this function is decreasing. Pause the video and try to figure that out.

All right, now let's do this together. We know that a function is decreasing when its derivative is negative.

Or another way to say it, it's going to be decreasing when ( f'(x) < 0 ).

So what is ( f'(x) )? Well, we could use the derivative rules and derivative properties. We know we apply the power rule to ( x^6 ), we bring the 6 out front, or multiply the one coefficient here times 6 to get ( 6x^5 ).

We decrement that exponent minus bring down the 5 times the -3 to get -15 ( x^{4} ). Now, we need to figure out over what intervals this is going to be less than 0.

Let's see how we can simplify this a little bit. Both of these terms are divisible by ( x^4 ) and they're both divisible by 3.

So let's factor out ( 3x^4 ). If you factor out ( 3x^4 ) here, you're left with ( 2x - 5 ).

So we have:

[
3x^4(2x - 5) < 0.
]

Any interval where this is true, we are going to be decreasing.

Now, how do we get this to be less than 0? Well, if I take the product of two things and it's less than 0, that means that they have to have different signs.

Either one's positive and the other is negative, or one's negative and the other is positive. So we have two situations: we could say either:

  1. ( 3x^4 > 0 ) and ( 2x - 5 < 0 ) (this is one situation).

Or,

  1. ( 3x^4 < 0 ) and ( 2x - 5 > 0 ) (and I'll do this one in a different color).

Actually, let me stay on the second case first. Are there any situations where ( 3x^4 ) can be less than 0?

You take any number, you take it to the 4th power, and even if it's negative, it's going to become positive. So you can't get a negative expression right over here.

So, actually, the second condition is impossible to obtain. You can't get any situation for any ( x ) where ( 3x^4 < 0 ).

So we can rule this one out. This is our best hope.

So under what conditions is ( 3x^4 > 0 )?

Well, if you divide both sides by 3, you get ( x^4 > 0 ). If you think about it, this is going to be true for any ( x ) value that is not equal to 0.

Even if you have a negative value there, if you have -1, you take it to the fourth power and it becomes a positive 1.

Only 0 will be equal to 0 when you take it to the fourth power. So one way you could say this is going to be true for any non-zero ( x ), or we could just say ( x \neq 0 ).

This is a little more straightforward. We add 5 to both sides, we get ( 2x < 5 ).

Dividing both sides by 2, you get ( x < \frac{5}{2} ).

So it might be tempting to say, all right, the intervals that matter are all the ( x )'s less than ( \frac{5}{2} ), but ( x ) cannot be equal to 0.

Now, is that the entire interval where our function is decreasing?

Let's think about what happens at 0 itself. We are decreasing over the interval from negative infinity all the way up to 0.

We're also decreasing from 0 to ( \frac{5}{2} ). So if we're decreasing right to the left of 0 and we're decreasing right to the right of 0, we're actually going to be decreasing at 0 as well.

So there's something interesting here. Even though the derivative at ( x = 0 ) is going to be equal to 0, we are still decreasing there.

The interval that we care about, the interval over which we're decreasing, is just ( x < \frac{5}{2} ).

We can see that by graphing the function. I graphed it on Desmos, and you can see here that the function is decreasing from negative infinity.

It's decreasing at a slower and slower rate. We get to 0, still decreasing to the left of 0, and then it continues to decrease to the right of 0.

So any value, any ( x ) value to the left of 0, the value of the function is going to be larger than ( f(0) ).

And ( x ) to the right of 0, the value of the function is going to be less than the function at 0.

It's actually decreasing through 0, even though the slope of the tangent line at 0 is 0.

Even though it's not negative, and then we keep decreasing. So we're decreasing for all values of ( x < \frac{5}{2} ), which you can see visually here.

More Articles

View All
Camp Khan Parent Webinar
Hi everyone, good afternoon or good evening, depending on where you’re joining us um in the country. My name is Roy, and I’m here to give you a quick overview of Camp Con, our new summer camp. Quick agenda here: we’re going to do intros real quickly, talk…
Will OpenAI Kill All Startups?
This is Michael Seibel with Dalton Caldwell, and today we’re going to talk about how OpenAI is going to kill all startups. This is our last video; might as well pack it in, we’re done. OpenAI is going to do this. They’re going to make the videos—next vide…
Evidence of evolution: anatomy | Evolution | Middle school biology | Khan Academy
[Instructor] About 3.5 billion years ago, single-celled organisms were the only life forms that existed on Earth. These organisms passed on their genetic material with slight changes to their descendants. And over long periods of time, these genetic cha…
Polar curve area with calculator
What we’re going to try to do is use our powers of calculus to find this blue area right over here. What this blue area is, is the area in between successive loops of the graph. The polar graph ( r(\theta) = 3\theta \sin(\theta) ) I’m graphing it in polar…
The Mission | Official Trailer | National Geographic Documentary Films
My friend John paid some pirates to go to an island to talk to people about Jesus, when he knew that he had no business doing that. John’s parents brought him up to be Christian. He was just, like, full of light. I had a little bit of a crush on him. You …
Slow Motion of an AK-47 Underwater (Part 1) - Smarter Every Day 95
Hey it’s me Destin. This week on Smarter Every Day, I’m gonna trick you into learning science using a gun and a high speed camera. You remember the old pistols underwater video? Well this week I’m gonna do it with a better high speed camera, and a bigger …