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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 5

Lesson 3: Determining intervals on which a function is increasing or decreasing# Finding decreasing interval given the function

Sal finds the intervals where the function f(x)=x⁶-3x⁵ is decreasing by analyzing the intervals where f' is positive or negative.

## Want to join the conversation?

- Why are the intervals open, not closed?(13 votes)
- I think the interval should include 0. Because the graph is continuous. Can you find an x<0 makes f(x)<=f(0) or an 2.5>x>0 makes f(x)>=f(0)? Of course there are no such points.

Since none, the graph is decreasing around point (0,0) regardless of f'(0)=0,which tells us the graph is "horizontal"

If you must say 0 shouldn't appear in the decreasing interval, you could say that the 'strictly decreasing interval' doesn't include 0.(3 votes)

- Couldn't you use sign analysis?(10 votes)
- Yes you can! The next videos include them!(14 votes)

- Why was x<0 included in the intervals for x that lead to a negative differential at3:48? Unless I missed something in the video, I didn't see any explanation for this.(8 votes)
- we are looking for intervals which f is decreasing.

it means we find intervals for f'(x) < 0

since our f'(x) = x^4*(6x-15)

for x<0 our f'(x) will always show negative value.

ex) for x = -1, f'(-1) = 1*(-6-15) = -21(2 votes)

- isn't the instantaneous speed at x = 0 is 0 so it isn't decreasing (at the moment).(8 votes)
- Is the interval gonna be written as ] 5/2, 0 [ U ] 0, -infinity[ ?(5 votes)
- I would write this as:
`(-∞, 0) ∪ (0, 5/2)`

. Note the use of square brackets is for closed intervals, i.e. the interval includes the numbers at each end, which is incorrect for this question.

A video addressing this is:

https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/introduction-to-interval-notation(6 votes)

- When we are solving 6x^5-15x^4>0, why can't we divide both sides by 3x^4 since x^4>0 (in this case x cannot be 0)? So we can solve it more easily.(2 votes)
- When we are solving equations for variables (which is essentially what we are doing here), if we divide by 3x^4, we remove possible answers. If we divided, we may forget that x can't = 0, and just be left with x <(5/2). As a general rule of thumb, never divide by a variable.(5 votes)

- I understand the function is decreasing for x between 0 and 2.5. I don't understand why x is decreasing for x<0. X^4 will always be positive. Also, if I graph the equation, the function is actually increasing for all values of x<0. Can anyone explain this?(3 votes)
- If you start at 0 and go towards negative infinity, then yes, all the values are increasing. However, we are talking about increasing in terms of slope, so we move from left to right. If you started at negative infinity and moved towards 0, then all the values would be decreasing and there slope of the tangent line will be negative. I hope this helps!(2 votes)

- At3:00, why couldn't you take the 4th root of both sides of x^4 < 0 and get x < 0?(2 votes)
- If you did this, you would have to use two inequalities for the +/- symbol, x<+0 and x>-0. This is the same thing as saying x<0 and x>0, or x=/=0.(4 votes)

- Hey I've tried graphing this and there's an interval, 0 and -1 and a little more than that, where f is decreasing? How would you explain this??

https://www.desmos.com/calculator

The slope goes to y=110 and then approaches infinity but just before that it has a negative slope, or decreasing.(3 votes) - My question are,

1)At2:50,why the first case is rule out ?

2)At3:21, why the second case for 3x^4 <0 , turns into x not equal to 0 ?

3)What does it mean at4:25?(1 vote)- Question 1)

See: https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-negative-numbers-multiply-and-divide/cc-7th-mult-div-negatives/v/why-a-negative-times-a-negative-is-a-positive

2)

given 3x^4 > 0, if x=0 then (3)0^4 is not greater than zero, it is equal to zero, so the inequality 3x^4 > 0 does not hold when x=0

see https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-greater-than-less-than/v/greater-than-and-less-than-symbols

3) The derivative is the slope of the curve at a point x=c.

If the point is either less than zero, or between zero and 5/2, the derivative evaluates to a negative number, which means the slope of the function evaluated at those points is negative, so the slope is negative, hence the function is decreasing in those intervals, which is what we were asked to find.

Keep Studying!(5 votes)

## Video transcript

- [Instructor] Let's
say we have the function F of X is equal to X to the sixth, minus three X to the fifth. And my question to you is using only what we
know about derivatives. Try to figure out over
what interval or intervals is this function decreasing? Pause the video and
try to figure that out. All right, now let's do this together. So 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 prime of X is less than zero. So what is F prime of X? Well, we could use the derivative rules and drew the properties we know. We apply the power rule
to X, to the sixth, We bring the six out front or multiply the one coefficient here, times six, to get six, X to the fifth detriment that exponent, minus bring the five times the three minus 15, X to the, we detriment the five, so X to the fourth. And we need to figure out over what intervals is this
going to be less than zero. And now let's see how we can
simplify this a little bit. Both of these terms are
divisible by X to the fourth, and they're both divisible by three. So let's factor out a
three X to the fourth times, you factor out of
three X to the fourth year, you're left with a two X, and then over here you have minus five has to be less than zero. Any interval where this is true, we are going to be decreasing. Now, how do we get this
to be less than zero? Well, if I take the product of two things and it's less than zero, that means that they have
to have different signs, either one's positive
and the other's negative or one's negative and
the other's positive. So we have two situations. So we could say either, either, three X to the fourth is greater than zero and, and two X minus five is less than zero. So that's one situation. I'll do some in a different color, or I'll do this one in a different color, three X to the fourth is less than zero and two X minus five is greater than zero. Actually let me stay on
the second case first. Are there any situations where three X to the fourth
can be less than zero? You take any number, you take it to the fourth power even if it's a negative, it's going to become a 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 three X to the
fourth is less than zero. So we can rule this one out. And so this is our best hope. So under what conditions
is three X to the fourth, greater than zero. Well, if you divide both sides by three, you get X to the fourth
is greater than zero. And if you think about it, this is gonna be true for any X value that is not equal to zero. Even if you have a negative value there, if you have a negative one, you take it to the fourth
power becomes a positive one. Only zero will be equal to zero when you take it to the fourth power. So one way we could say this is going to be true for any non zero X, or we could just say,
X does not equal zero, and, this is a little
bit more straightforward, we add five to both sides we get two X is less than five, divide both sides by two, you get X is less than five halves. So it might be tempting to say, all right, the intervals that matter are all the exes less than five halves, but X cannot be equal to zero. Now, is that the entire interval where our function is decreasing? Well, let's think about
what happens at zero itself. We're decreasing over the
interval from negative infinity all the way up to zero. And we're also decreasing
from zero to five halves. And so for decreasing
right to the left of zero, we're decreasing right
to the right of zero, we're actually going to
be decreasing at zero at we're also gonna be
decreasing at zero as well. So there's something interesting here, even though the derivative
at X equals zero, is going to be equal to zero, we are still decreasing. And so the interval that we care about the interval over which we're decreasing is just X is less than five halves. And 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 zero still
decreasing to the left of zero, and then it continues to
decrease to the right of zero. So any, any value, any X
value to the left of zero the value of the function is going to be larger than F of zero, and X to the right of zero. The value of the function is going to be less than the function at zero. So it's actually decreasing through zero, even though the slope of the
tangent line at zero is zero, even though it's non negative and then we keep decreasing so we're decreasing for all values of X, less than five halves, which you can see visually here.