- Slope of a line secant to a curve
- Secant line with arbitrary difference
- Secant line with arbitrary point
- Secant lines & average rate of change with arbitrary points
- Secant line with arbitrary difference (with simplification)
- Secant line with arbitrary point (with simplification)
- Secant lines & average rate of change with arbitrary points (with simplification)
- Secant lines: challenging problem 1
- Secant lines: challenging problem 2
Sal interprets an expression as the slope of a secant line between a specific point on a graph and any other point on that graph. Created by Sal Khan.
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- The second question: When is (f(x) - 6) / x at its greatest value?
It looks very much like we're supposed to replace x and f(x) with the coordinates from ONE of the points plotted, yet Sal's answer is about the slope between two points.
Why is that?(24 votes)
- We could answer the question by plugging in the values of x and f(x) for each of the six labeled points and see which one produces the greatest value. However, Sal is making use of an insight to answer the question more easily. The formula we're evaluating happens to be exactly the formula we would need to calculate the slope from point A to any of the other points. The numerator is the difference in y-values, and the denominator is equivalent to the difference in x-values (which would be x-0, which is of course equal to x). Because the formula has exactly the same form as the formula for the slope from point A, we can answer the question by looking for the point where the slope from point A is greatest, because the greatest slope will produce the greatest value for the slope formula. This saves us the trouble of calculating the formula for each of the points, because we can easily see which point will produce the greatest slope.(53 votes)
- Why couldn't A be the answer to the second part of the question, as A is 0?(17 votes)
- At point A, the numerator becomes 6-6=0, but the denominator also becomes 0. So instead of having the greatest value at A. the value becomes undefined. You can't have 0/0 and we must look for a different point.(37 votes)
- Do we not get to do some of this stuff ourselves? Why do we have practice sessions for the last two videos but not this one, which is harder than the other two combined.(9 votes)
- Hello! I have small question. :)
At1:53, Sal says that the tangent line to point B is horizontal. How can we know this? The line might have a non-zero slope so small as to be invisible at this scale.
This is actually something that jumps out at me every time (unlabeled) graphs and (discrete) tables of values are used to "prove" anything, especially with limits, and it always really confuses me.
Thanks in advance!
- You are shown f(x), and given 6 points to choose from and asked where does f(x)·f'(x)=0.
The question is telling you that one of the 6 labeled points will make this true.
By process of elimination, it must be B. Also, when you look at B, it appears to be at the vertex of a curve, where the derivative is 0, which confirms the process of elimination.
Now if the question asked "What is the derivative of f at B" then you could say "it appears as if f' could be 0 at B but unless I am given the formula of f, I can't say for certain". This would be fine - but that is not the case here.
The whole point here is to get an intuitive feel for visually assessing graphs of functions and their derivatives.(6 votes)
- Just to clarify, f(x) gives you the actual value at any point and f'(x) gives you the value of the slope of the tangent line for that point(4 votes)
- In the second question, couldn't you eliminate all the points further back than d because there are already higher x coord points and y coord points?(3 votes)
- Is there some point between D and E where the slope of the tangent line of that point is 0? It seems to be that would be the case for all functions that have a "wave" in them.(3 votes)
- Yes, as the slope changes from being positive to being negative, the slope reaches 0 for one point. The same is true if the slope of a function is moving from negative to positive and the interval in question is continuous. An interval in which a transition in the signs of the slope is made without the slope reaching zero would be discontinuous (for instance, a graph with a sharp peak or dip), because the slope would be approaching infinity instead of 0. Later, you will learn about "critical points", which are points where the derivative = 0. The critical points represent minimum or maximum points in the graph.(1 vote)
- At point E, when (x,f(x)) = (8,5), f(x)-6/x=(2,~2.8). The secant line here has a positive slope. Why is this (point E) not the answer?(2 votes)
- The secant line from A to E has a negative slope, not positive. in fact, all of the points have a negative slope from A to it, meaning that we're looking for the least negative slope. Since from point A to point D, the line is least slanted, it is the least negative, and thus, the greatest. The slope of the line from point A to point E is actually the second greatest because the line is second least slanted. The slope of the line from A to E is (f(x)-6)/x, or (5-6)/8, or -1/8.
I hope this helps!
P.S. I'm curious as to how you got the point (2, ~2.8), though. It doesn't seem relevant at all. How did you get it?(4 votes)
- plz tell would the slope B will be greatest in case, all secants were positive?................ sry stupid confirmation(0 votes)
- Did you mean the slope of tangent line at B, when the slopes of secant lines at A / C / D / E / F were positive?(5 votes)
We're asked, at which points on the graph is f of x times f prime of x equal to 0? So if I have the product of two things and it's equal to 0, that tells us that at least one of these two things need to be equal to 0. So first of all, let's see are there any points when f of x is equal to 0? So we're plotting f of x on the vertical axis. We could call this graph right over here, we could say this is y is equal to f of x. So at any point, does the y value of this curve equal 0? So it's positive, positive, positive, positive, positive, positive. But it is decreasing right over here. Well, it's decreasing here. Then it's increasing. Then it's decreasing. And it does get to 0 right over here, but that's not one of the labeled points. And they want us to pick one of the labeled points or maybe even more than one of these labeled points. So we're going to focus on where f prime of x is equal to 0. And we just have to remind ourselves what f prime of x even represents. f prime of x represents the slope of the tangent line at that value of x. So for example, f prime of 0-- which is the x value for this point right over here-- is going to be some negative value. It's the slope of the tangent line. Similarly, f prime of x, when x is equal to 4-- that's what's going on right over here-- that's going to be the slope of the tangent line. That's going to be a positive value. So if you look at all of these, where is the slope of the tangent line 0? And what does a 0 slope look like? Well, it looks like a horizontal line. So where is the slope of the tangent line here horizontal? Well, the only one that jumps out at me is point B right over here. It looks like the slope of the tangent line would indeed be horizontal right over here. Or another way you could think of it is the instantaneous rate of change of the function, right at x equals 2, looks like it's pretty close to-- if this is x equals 2-- looks like it's pretty close to 0. So out of all of the choices here, I would say only B looks like the derivative at x equals 2. Or the slope of the tangent line at B, it looks like it's 0. So I'll say B right over here. And then they had this kind of crazy, wacky expression here. f of x minus 6 over x. What is that greatest in value? And we have to interpret this. We have to think about what does f of x minus 6 over x actually mean? Whenever I see expressions like this, especially if I'm taking a differential calculus class, I would say well, this looks kind of like finding the slope of a secant line. In fact, all of what we know about derivatives is finding the limiting value of the slope of a secant line. And this looks kind of like that, especially if at some point, my y value is a 6 here. And this could be the change in y value. And if the corresponding x value is 0, then this would be f of x minus 6 over x minus 0. So do I have 0, 6 on this curve here? Well, sure. When x is equal to 0, we see that f of x is equal to 6. So what this is right over here-- let me rewrite this. This we could rewrite as f of x minus 6 over x minus 0. So what is this? What does this represent? Well, this is equal to the slope-- let me do some of that color-- this is equal to the slope of the secant line between the points, x, f of x, x, and whatever the corresponding f of x is. And we could write it as 0, f of 0 because we see f of 0 is equal to 6. This right over here is f of 0. In fact, let me just write that as 6. And the point, 0, 6. So let's go through each of these points and think about what the slope of the secant line between those points are and point A. This is essentially the slope of the secant line between some point x, f of x, and essentially point A. So let's draw this out. So between A and B you have a fairly negative slope. Remember we want to find the largest slope. So here it's fairly negative. Between A and C, it's less negative. Between A and D, it's even less negative. It's still negative, but it's less negative. And then between A and E, it becomes more negative now. And then between A and F, it becomes even more negative. So when is the slope of the secant line between one of these points and A the greatest? Or I guess we could say the least negative? Because it seems like they're always negative. It would be between point D and A. So when is this greatest in value? Well, when we're looking at point D. At point D, x is equal to 6 and it looks like f of x is like 5 and 1/2 or something. So this will turn into f of 6, which is 5 and 1/2 or maybe it's even less than that-- 5 and 1/3 or something, minus 6 over 6 minus 0. That's how we'll maximize this value. This is the least negative slope of the secant line.