AP®︎ Calculus AB (2017 edition)
Sal analyzes various graphs to find intervals where the derivative of the graphed function is positive or negative. Created by Sal Khan.
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- wait, what does the " f'(x) " exactly mean? >,<"(8 votes)
- F'(x) is the symbol used for the derivative of a function f(x). Derivative of a function is essentially the slope of the function at a particular x value. For example, a function may be defined as f(x)=2x^2+1. F'(x)=4x. If we wanted to find the slope at x=1, then f'(1)=4(1)=4(13 votes)
- What is the difference between f(x) and f'(x)?(5 votes)
- alzuubi1 is correct, however I would like to clarify that the derivative (f'(x)) is also a function (in most cases).(6 votes)
- Can you choose a neutral area?(5 votes)
- Nope because the slope at a neutral area is 0....It says f'(x) > 0 .... HAS to be positive....not f'(x) >= 0....(6 votes)
- So the function is positive if it's above zero?(3 votes)
- If f(x) > 0, the function is positive. However, make sure to make the distinction between a function and the derivative of a function, otherwise you might get confused when a function has a positive value, and its derivative a negative value, or vice-versa.(1 vote)
- Did anyone notice the notion of function increasing and decreasing depends probably in our (western) type of reading which is from left to right ?(2 votes)
- We can say that a function f is increasing on an interval when, for every x, y in the interval, x>y implies f(x)>f(y). We can do similar for decreasing functions. This is independent of our left-to-right convention.(3 votes)
- what are the rules for simplifying a deriviative(2 votes)
- As with any problem, there is no "correct" way to simplify a derivative. For example, who is to say that (x+1)(x-1) is more simplified than x-1^2? Really what you are looking for is to make the work easiest on yourself. If you need to find a second derivative, and you have the function d/dx = (x-1^1)/(x+1), it will be much less work to find the second derivative of the equivalent expression d/dx = x-1 because you avoid the quotient rule. However, if you don't need to do anything after finding the derivative there isn't really any need to simplify it.(2 votes)
- How come 'x should be in sitting in first or second quadrant"1:15when in second quadrant x is negative?(1 vote)
- What he said was 'f(x) should be sitting in the first or second quadrant'. f(x) is the y-value, which is positive in those quadrants.(3 votes)
- The part of the graph Sal is indicating at2:40looks to me like it's below 0 on the y-axis....so why is he saying the function is positive?(1 vote)
- so we are always assuming y to be the parameter by which x is changing. And also we assume the graph to be moving from right to left even though no arrows are present.?(1 vote)
- Usually, x is the independent variable, and y is the dependent variable. This means that you can independently choose an x-value, and the y-value (output) depends on which x you chose.(1 vote)
- I'm confused can someone please explain this concept(1 vote)
- The key to understanding this concept is to understand what exactly a slope is. If you have some function, the derivative of that function is the slope of the original function. So when the video is asking for an interval where the derivative is greater than 0, you must look for a slope that is increasing or getting more and more steep in a sense.
Another interesting note here is that if you have a function graphed, you can graph the derivative of that function by analyzing the slope of the original function at every few intervals.(1 vote)
A function f of x is plotted below. Highlight an interval where f prime of x with the first derivative of f with respect to x is greater than 0. So if our derivative, f prime of x, is greater than 0, that means that the slope of the tangent line is positive. That means that the function must be increasing whenever this is true. So where is the function increasing? So over here we see that as x increases, our function is going to smaller and smaller values. So our function is decreasing this entire place. Then it's decreasing at slower and slower rates all the way to the point right over here where the slope of the tangent line is flat. And then the function starts increasing. And it starts increasing at faster and faster and faster rates. So we could put this blue area anywhere from here to anywhere in this region here where the function itself is increasing. Notice, the function could even be negative, but the function is increasing. So even this would be a valid place. That would be a valid place. Anywhere where the function's rate of-- when x increases, the function is actually getting larger. So let's check our answer. And let's do a couple more of these. A function f of x is plotted below. Highlight an interval where f of x is greater than 0. So f of x should be sitting in the first or second quadrant right over here. It needs to be greater than 0, and f prime of x is less than 0. So f prime of x less than 0 means that the function is decreasing. The slope of the tangent line is negative. So let's think about it. The two areas where f of x is greater than 0 are this interval right over here and this interval right over here. But we also care about the function decreasing. This won't be valid, because the function is increasing here. It's increasing at slower and slower rates, but it is increasing. The function is going up as x goes up. If we go over here, the function is decreasing as x increases. So this seems to meet our constraints. This area right over here won't because the function is positive, but it's also increasing. So here the derivative is positive. Here the derivative is positive. Here the derivative is negative. So the function itself is positive, but the slope of the tangent line is negative. It's decreasing as x increases. Let's do one more. The function f is plotted below. Highlight an interval where f of x is greater than 0 and f prime of x is less than 0. So the same exact idea-- where are we positive but our function is actually decreasing? So it's positive here, but we can't pick that. It's positive here, but we can't pick that. The function is decreasing here, but the function isn't positive there. The derivative is negative, but the function isn't positive. So that wouldn't make it. So this is the only region that we can throw it in. And we got it right again.