Finding decreasing interval given the function | AP Calculus AB | Khan Academy

So we have the function ( f(x) = x^6 - 3x^5 ) and we want to know over what intervals is ( f ) decreasing. We're going to do it without even having to graph ( y = f(x) ). The way we do that is we look at the derivative of ( f ) with respect to ( x ) and think about when that is less than zero. If the rate of change of ( f ) with respect to ( x ) is less than zero, well, over those intervals it will be decreasing.

So let's first take the derivative. So ( f'(x) ) is going to be equal to, just using the power rule here, it's going to be ( 6x^5 - 15x^4 ). Now, let's think about when this is going to be less than zero over what intervals ( 6x^5 - 15x^4 < 0 ).

So, we could factor out a ( 3x^4 ). So, ( 3x^4(2x - 5) < 0 ). Did I do that right? Let's see. If I were to distribute it, ( 32 = 6 ), ( x^4x = x^5 ), and ( 3*5 = 15 ), ( x^4 ) yep, that's right.

So if I'm taking the product of two things and I want it to be less than zero, well, there’s only one way for that to happen: either the first thing is positive and the second is negative, or the first is negative and the second is positive. So let's analyze that.

So, either ( 3x^4 < 0 ) and ( 2x - 5 > 0 ), or let me just put the or in a separate color here, or ( 3x^4 > 0 ) and ( 2x - 5 < 0 ).

So let's see. For ( 3x^4 < 0 ), well, if we divide both sides by three, this is just going to be ( x^4 < 0 ). Is there any way for something to the fourth power to be less than zero? Well, we're assuming we're dealing with real numbers here, and any real number to the fourth power is going to be greater than or equal to zero. So it's actually impossible for something to the fourth power to be less than zero. We can rule out this first case.

So we can rule out that first case right over there.

Now, we're only going to worry about the second case. So, ( 3x^4 > 0 ) will happen as long as ( x \neq 0 ). This is because for any other ( x ), this will be true. ( x ) could be negative; you take it to the fourth power, multiply it by three, it will be greater than zero. So this is really just the condition that ( x ) cannot be equal to zero.

Now, let's see the second one: ( 2x - 5 < 0 ). That means ( 2x < 5 ), and then ( x < \frac{5}{2} ). So as long as ( x < \frac{5}{2} ) and ( x \neq 0 ), this function will be decreasing.

If we wanted to write it in terms of intervals, we could say ( x < 0 ) or ( 0 < x < \frac{5}{2} ).

So ( x < 0 ) is all the negative values, and then we’re essentially just excluding zero and going all the way to ( \frac{5}{2} ).

Remember, all I did is I said, well, when is our first derivative negative? Because if the first derivative is negative, then the rate of change of ( f ) with respect to ( x ) is negative or ( f ) is decreasing as ( x ) is increasing.