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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 decreasingIncreasing & decreasing intervals review
AP.CALC:
FUN‑4 (EU)
, FUN‑4.A (LO)
, FUN‑4.A.1 (EK)
Review how we use differential calculus to find the intervals where a function increases or decreases.
How do I find increasing & decreasing intervals with differential calculus?
The intervals where a function is increasing (or decreasing) correspond to the intervals where its derivative is positive (or negative).
So if we want to find the intervals where a function increases or decreases, we take its derivative an analyze it to find where it's positive or negative (which is easier to do!).
Want to learn more about increasing/decreasing intervals and differential calculus? Check out this video.
Example 1
Let's find the intervals where f, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 3, x, squared, minus, 9, x, plus, 7 is increasing or decreasing. First, we differentiate f:
f, prime, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 6, x, minus, 9
Now we want to find the intervals where f, prime is positive or negative.
f, prime intersects the x-axis when x, equals, minus, 3 and x, equals, 1, so its sign must be constant in each of the following intervals:
Let's evaluate f, prime at each interval to see if it's positive or negative on that interval.
| Interval | x-value | f, prime, left parenthesis, x, right parenthesis | Verdict |
|---|---|---|---|
| x, is less than, minus, 3 | x, equals, minus, 4 | f, prime, left parenthesis, minus, 4, right parenthesis, equals, 15, is greater than, 0 | f is increasing. \nearrow |
| minus, 3, is less than, x, is less than, 1 | x, equals, 0 | f, prime, left parenthesis, 0, right parenthesis, equals, minus, 9, is less than, 0 | f is decreasing. \searrow |
| x, is greater than, 1 | x, equals, 2 | f, prime, left parenthesis, 2, right parenthesis, equals, 15, is greater than, 0 | f is increasing. \nearrow |
So f is increasing when x, is less than, minus, 3 or when x, is greater than, 1 and decreasing when minus, 3, is less than, x, is less than, 1.
Example 2
Let's find the intervals where f, left parenthesis, x, right parenthesis, equals, x, start superscript, 6, end superscript, minus, 3, x, start superscript, 5, end superscript is increasing or decreasing. First, we differentiate f:
f, prime, left parenthesis, x, right parenthesis, equals, 6, x, start superscript, 5, end superscript, minus, 15, x, start superscript, 4, end superscript
Now we want to find the intervals where f, prime is positive or negative.
f, prime intersects the x-axis when x, equals, 0 and x, equals, start fraction, 5, divided by, 2, end fraction, so its sign must be constant in each of the following intervals:
Let's evaluate f, prime at each interval to see if it's positive or negative on that interval.
| Interval | x-value | f, prime, left parenthesis, x, right parenthesis | Verdict |
|---|---|---|---|
| x, is less than, 0 | x, equals, minus, 1 | f, prime, left parenthesis, minus, 1, right parenthesis, equals, minus, 21, is less than, 0 | f is decreasing. \searrow |
| 0, is less than, x, is less than, start fraction, 5, divided by, 2, end fraction | x, equals, 1 | f, prime, left parenthesis, 1, right parenthesis, equals, minus, 9, is less than, 0 | f is decreasing. \searrow |
| start fraction, 5, divided by, 2, end fraction, is less than, x | x, equals, 3 | f, prime, left parenthesis, 3, right parenthesis, equals, 243, is greater than, 0 | f is increasing. \nearrow |
Since f decreases before x, equals, 0 and after x, equals, 0, it also decreases at x, equals, 0.
Therefore, f is decreasing when x, is less than, start fraction, 5, divided by, 2, end fraction and increasing when x, is greater than, start fraction, 5, divided by, 2, end fraction.
Want to join the conversation?
Is this also called the 1st derivative test?(19 votes)
In summation, it's the 1st derivative test. Specifically, it's the 'Increasing/Decreasing test':
Increasing/Decreasing test:
If f'(x) > 0 on an interval, then f is increasing on that interval
If f'(x) < 0 on an interval, then f is decreasing on that interval
First derivative test:
If f' changes from (+) to (-) at a critical number, then f has a local max at that critical number
If f' changes from (-) to (+) at a critical number, then f has a local min at that critical number
If f' has no sign changes at that critical number, then f' has no local min nor max at the critical number.(67 votes)
I'm finding it confusing when a point is undefined in both the original function and the derivative. While not mentioned in the video on critical points, it's mentioned in the comments and practice problems that a point is not a critical point if it's undefined in both the derivative and in the original function.
On the other hand, in the practice problems, we're given something like:f'(x) = ((x-1)^2) / (x-4)
and asked to find the intervals over which the original function is increasing. The question states that the original function is undefined at x = 4. According to the definition, x = 4 should not be a critical point because it's undefined in both the derivative and the original function. However, it is a point of interest as f'(x) > 0 only when x > 4. If we don't consider x = 4 we won't find the right answer.
Is this an issue with the definition of critical points, the practice problem itself, or this method of finding increasing or decreasing intervals?
If it's the practice problem, I could imagine that maybe it's impossible for a function with that derivative to be undefined at 4 (though it seems unlikely.)
If it's this method, it seems like we need to consider points that aren't strictly critical points as per the definition.
I think a little more clarity around this particular case in this section and the one before would be helpful.(18 votes)- I found the answer to my question in the next section. Under "Finding relative extrema (first derivative test)" it says:
When we analyze increasing and decreasing intervals, we must look for all points where the derivative is equal to zero and all points where the function or its derivative are undefined. If you miss any of these points, you will probably end up with a wrong sign chart.
I'll leave my question here because I think it's confusing for this section to only discuss critical points and not to mention this.(23 votes)
- for the notation of finding the increasing/decreasing intervals of a function, can you use the notation Union (U) to express more than one interval?(4 votes)
Yes. For example, the function -x^3+3x^2+9 is decreasing for x<0 and x>2. Another way we can express this: domain = (-∞,0) U (2, +∞). This is known as interval notation.
(8 votes)
- Is x^3 increasing on (-∞,∞) or is it increasing on two open intervals and is increasing on (-∞,0)U(0,∞)?(4 votes)
f(x) = x³ is increasing on (-∞,∞).
A function f(x) increases on an interval I if f(b) ≥ f(a) for all b > a, where a,b in I.
If f(b) > f(a) for all b>a, the function is said to be strictly increasing.
x³ is not strictly increasing, but it does meet the criteria for an increasing function throughout it's domain = ℝ(4 votes)
What does it mean to say that the slope of a function is increasing or decreasing? Not when the function is increasing or decreasing, but the slope. Is it the same thing? I'm having some trouble with calculus homework that is treating it as if they aren't the same thing.(3 votes)- I think that if the problem is asking you specifically whether the slope of the tangent line to the function is increasing or decreasing, then it is asking whether the second derivative of the function is positive or negative.
When we want to know if the function is increasing or decreasing, we take the derivative of the function and check if the derivative (slope of the tangent) is positive or negative. But if we want to know whether that derivative is increasing or decreasing (whether the slope is increasing or decreasing), we'd take its derivative. The derivative of the "slope" would be the second derivative of the original function.
I'm betting we get to this a bit later when we start talking about using second derivatives to analyze functions.(4 votes)
- Using only the values given in the table for the function, f(x) = x3 – 3x – 2, what is the interval of x-values over which the function is decreasing?
(–4, 1)
(–4, –1)
(–1,1)
(–1, 2)(2 votes)
𝑓(−4) < 𝑓(−1), so 𝑓 can not be decreasing over (−4, −1) and thereby not over (−4, 1) either.
𝑓(−1) = 𝑓(2), so 𝑓 can not be decreasing over (−1, 2)
𝑓(−1) > 𝑓(1), so it is possible that 𝑓 is decreasing over (−1, 1)(2 votes)
- How do we decide if y=cos3x increasing or decreasing in the interval [0,3.14/2](2 votes)
We can tackle the trigonometric functions in the same way we do polynomials or rational functions! We take the derivative of y, giving us dy/dx = -3sin3x. Then, we find where this derivative is equal to zero or is undefined - this tells us all the possible x-values where the derivative might change from positive to negative, or negative to positive. Then we figure out where dy/dx is positive or negative. Since we know functions are increasing where their derivatives are positive, and decreasing where their derivatives are negative, we can then use this knowledge to figure out if the function is increasing or decreasing.(1 vote)
- for the number line we must do for all the x or the value of crtitical number that is in the domain?(2 votes)
- We only need to look at the critical values of x; that is, whether or not the function's derivative changes signs at those points, so that we can figure out if the derivative is positive or negative on its domain.(1 vote)
- Are there any factoring strategies that could help me solve this problem faster than just plug in and attempt? (3x^2 + 8x -5) The answer is (3x-5)(-x+1)(1 vote)
- This is yr9 math. The section you have posted is yr11/yr12.
Remember quadratic equations.
(x+a)(x+b) = x^2 +(a+b)x + ab. Use this to complete the factorisation.
Other methods includes completing the square and the quadratic formula.(2 votes)
If a graph has positive and negative slopes on an interval, but the y value at the end of the interval is higher than y value at the beginning, is it increasing on the interval?(1 vote)
Given that you said "has negative slope", no. It is increasing perhaps on part of the interval.(1 vote)
