Course: Differential Calculus > Unit 5Lesson 3: Intervals on which a function is increasing or decreasing
Finding increasing interval given the derivative
Sal is given that the derivative of function g, is g'(x)=x²/(x-2)³. He uses that to find the intervals where g is increasing, by looking for the intervals where g' is positive.
Want to join the conversation?
- In the previous video in the playlist 'CALCULUS DERIVATIVE APPLICATIONS' Sal put the derivative to be less than zero then just solved the inequality to find what x was in that case. Can't we just use that way for all problems of this kind? Which way is better?(16 votes)
- Yes, you can solve it the way he did in the past video, he only wanted to show you that that wasn't the only way. In the example shown in this video, you would set both the numerator and the denominator greater to zero so your whole fraction can be greater than zero, by doing that you get that the x in the numerator should be greater than zero and the x in the denominator should be greater than 2. Since you're working with both at the same time, your x value should be greater than 2 so the whole fraction ends up greater than zero. Hope I helped!(15 votes)
- g'(x) is the slope of the function g,right? so when we get g'(x)=0,it would mean that the slope is parallel to x axis.but if the function has to go on after that point where sloe is 0,in this casewhen x=0,there should be a turning of the graph.
for eg,if the fn was decreasing frm (-infinity) till x=0,and at x=0,the slope is 0,so from x=0,shouldnt the graph increase??
here at4:57the graph at both intervals are decreasing.how is that possible?(8 votes)
- When g'(x)=0, x is a critical number (c.n.) A c.n. is where the graph could have min/max (turning point where the graph change from decrease to increase or vice versa) but it does not guarantee that it will have a min/max. That's why we have to do what we call the first derivative test like Sal does in the video.
An example of this would be f(x)=x³ then f'(x)=x²
f'(x) = 0 at x = 0, but f(x)=x³ is increasing for all x because at x=0 the slope is 0 but it's neither a min or a max.(10 votes)
- at3:10, is there a difference between (negative infinity, 0) and [-negative infinity, 0]
is there a close interval contains negative infinity?(4 votes)
- Parentheses ( indicate non inclusive. Brackets [ indicate inclusive. So (-inf,0) means it is neither negative infinity or zero, but everything in between. [-inf,0] means that it could be zero or negative infinity along with everything in between. Since something can never equal infinity, we use parentheses by them. therefore, [-inf,0] is technically impossible. (-inf,0) also does not equal (-inf,0] because it could be zero in the second one but not the first one.(8 votes)
- At the top of the screen, it says that g(x) is defined for all real numbers, but integrating g'(x) and plugging in x=2, we get that g(2) is undefined. In the example of the video, g'(2) is also undefined, but a critical value requires that it be defined by g(x). So x=2 isn't a critical value, and we can't define and interval using 2 as an endpoint. I understand that the results work out nicely with g(x) defined in the problem and the actual g(x) (which is undefined at x=2), but in general, without being given that some function f is continuous over all real numbers, wouldn't it be invalid to assume that f(x) has critical values where f'(x)=0 or undefined without knowing f(x) or integration? And furthermore, doesn't that mean it's invalid to define an interval using these values, and therefore you can't actually know where f(x) is increasing or decreasing based on f'(x) without knowing f(x) or integration?(4 votes)
- What is the antiderivative of g'(x), i mean how do we integrate g'(x) = x^2/(x-2)^3 ?(2 votes)
- It is not necessary to use partial fractions to integrate x^2/(x-2)^3.
We can use the u-substitution u=x-2. Then du = 1dx = dx, and x=u+2. Therefore, we have
integral of x^2/(x-2)^3 dx = integral of (u+2)^2 / u^3 du
= integral of (u^2+4u+4) / u^3 du
= integral of (1/u + 4/u^2 + 4/u^3) du
= ln|u| - 4/u - 2/(u^2) + C
= ln|x-2| - 4/(x-2) - 2/(x-2)^2 + C.
We can check this answer by differentiating it, and verifying that the result is the original function.
(d/dx) of (ln|x-2| - 4/(x-2) - 2/(x-2)^2 + C)
= 1/(x-2) + 4/(x-2)^2 + 4/(x-2)^3
= [(x-2)^2 + 4(x-2) + 4]/(x-2)^3
= (x^2 - 4x + 4 + 4x - 8 + 4)/(x-2)^3
The answer ln|x-2| - 4/(x-2) - 2/(x-2)^2 + C checks!
Have a blessed, wonderful day!(4 votes)
- In the graph of f(x)= polynomial having f'(x) = undefined could indicate that there's a sharp point in the graph of f(x) right where the slope would simultaneously be decreasing and increasing i.e. undefined? But what if you have f'(x) undefined in a function where f(x) = rational function? What would this undefined x in f'(x) indicate about the behavior as the graph approaches this x and what would that look like?(2 votes)
- I don't think you can ever get a "sharp point" in a purely§ polynomial function (although a piecewise polynomial could). In fact since the derivative of a polynomial will always be another polynomial, I don't think they can have an undefined derivative anywhere ...
For a rational function, you do have situations where the derivative might be undefined — points where the original function is undefined i.e. has zero in the denominator.
f(x) = x³/(x-5)at
x=5— asymptotic discontinuity in the function
g(x) = x(x+2)(x-3)/(x+2)at
x=-2— point discontinuity in the function
§Definition of polynomial:
- The point x=0 is also the inflection point, correct?(2 votes)
- Why did we have to find the critical points–why didn't he just solve it like he did in the previous video? And what does this graph look like because I'm having trouble visualising it.(2 votes)
- Well, this is "finding increasing interval[s] given the derivative", right? What exactly are critical points? Critical points are x-values where the derivative of a graph is nonexistent or is equal to zero (Calculus: Graphical, Numerical, Algebraic (5th Edition) by Finney et. al.). So what is to the right and left of those? Well, since points to the right and left of those critical points do not fulfill the value needed for a critical point (0 or DNE), it is either increasing or decreasing, and your critical points help set the bounds for that.
You don't know what the graph looks like because you aren't given it. You do, however, know what the derivative graph looks like, and looking at that alone lets you know where the graph is increasing or decreasing, but it wants you to find the answer algebraically.(1 vote)
- Since the derivative is negative for values both less than zero and between 0 and 2, it follows that the function is decreasing over both those intervals. So, how is 0 a critical point?(2 votes)
- A critical point is when the derivative equals 0. And while it is always negative where you indicated, the derivative itself is increasing at one point.
A much easier example to see this is -x^2. if this were the derivative of something, this also has a critical point at (0,0). Do you understand why that point is a critical point?(1 vote)
- So for the interval(s) where g is decreasing, will it be from negative infinity to 2? Is 0 also included in the interval for which the function is decreasing, or not? I'm a little confused if we include 0 or not, since it is a critical point where the derivative is 0.(2 votes)
- 0 would not be included, because at 0 the derivative is 0 which means it is not increasing r decreasing. so there are two intervals where it is decreasing.
It might help to think of a more simple example. Look at x^3. It looks like it is always increasing, but exactly at 0 the slope is 0 (slope, derivative, whatever you want to call it.) that means there is no change it is completely horizontal. But it is only at that exact point.
It's similar for where the equation doesn't exist. if it doesn't exist then it can't be increasing or decreasing. there is nothing there.
Let me know if that doesn't make sense and I can try and explain differently.(1 vote)
- [Voiceover] Let g be a function defined for all real numbers. Also let g prime, the derivative of g, be defined as g prime of x is equal to x squared over x minus two to the third power. On which intervals is g increasing? Well, at first you might say, they don't even give us g. How do we figure out when g is increasing? Well, the answer is all we need is g prime which they do give us. And saying on which intervals is g increasing, that's equivalent to saying, on which intervals is the first derivative with respect to x on which intervals is that going to be greater than zero? If your rate of change with respect to x is greater than zero, if it's positive, then your function itself is going to be increasing. And so there's a couple of ways that we could approach this. You might just want to inspect kind of the structure of this expression and think about, well, when is that going to be greater than zero or we could do it a little bit more methodically. We could say, well, let's look at the critical points or the critical values for g. So critical, critical points for g and just to remind ourselves what critical points are, that is when g prime of x is equal to zero or g prime of x is undefined, is undefined, and we have videos on critical points or critical values and why those are relevant is those are the places, those are possible places where the sign could change, the sign of g prime could change. So when is g prime of x equal to zero? Well, the way to get g prime of x equal to zero is getting the numerator equal to zero and that's only going to happen if x squared is equal to zero or if x is equal to zero. So that's the only place where g prime of x is equal to zero and where is g prime of x undefined? Well, it's going to be undefined if the denominator becomes undefined. The denominator becomes undefined if the denominator is zero and so that's going to happen if x minus two is equal to zero, x minus two is equal to zero or x is equal to two. So we have two critical points or critical values here and what I'm going to do is I want to graph them. Let's put them on a number line and let's just think about what g prime is doing in the intervals between the critical points. So let's start at zero, one, two, three and then let's go to negative one and we have a critical point at, let me do that in magenta, we have a critical point at x equals zero right over there and we have a critical point at x equals, at x equals two right over there. And so let's think about what g prime is doing in the intervals between the critical values or on either side of the critical values. So let's think about, let's first think about this interval. Let me do it in this purple color. Let's think about the interval between, between negative infinity and zero. So if we think about this interval, so negative infinity and zero, that open interval, well, if we look at g prime, the numerator is still going to be positive. If you take any negative value squared, you're going to get a positive value so this is going to be positive. Now, what about the denominator? You take a negative number, you subtract two from it, you're still going to get a negative number and then you take it to the third power. Well, a negative number to the third power is going to be a negative number so that right over there is going to be negative. So you're going to have a positive divided by a negative so g prime is going to be negative so let me write that down. So on this interval, on this interval, I'll write it like this. g prime of x is less than zero or if we cared or if we want to know when it's decreasing, we would know it's definitely decreasing over that interval. Now, let's take the interval between zero and two right over here. So this is the interval from zero to two, the open interval. So what's going to go on with g prime of x here? Well, once again, x squared, anything greater than a zero and it says we're not including zero in this interval. Well, this is for sure going to be positive and so let's see, if we have x minus two where x is greater than zero but less than two. So if x, we could just say for example, if x was one, one minus two is negative one. We're still going to get negative values in this denominator right over here. So since we're still going to get negative values in this denominator, the denominator is still going to be, you take a negative value to the third power, well, you're going to still get a negative value so this is going to be negative. So you're still going to have g prime as less than zero so let me write that down. So you still have g prime of x is less than zero. And then let's take the interval above. Let's take the interval from two to infinity. Two to infinity. Well, the numerator is positive. It's always going to be positive for any x not being equal to zero and this denominator, you're taking values greater than two, subtracting two from it which is still going to give you a positive value. You take the third power, it's all going to be positive. It is all going to be positive. So this is the interval where g prime of x is greater than zero. So on which intervals is g increasing? Well, that's where g prime of x is greater than zero so it's going to be from two, from two to infinity or we could just write it like this. We could write x is greater than two. Either way, for either of these, g prime of x is greater than zero and your function g is going to be increasing.