Finding decreasing interval given the function | Calculus | Khan Academy

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,

2. ( 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.