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## Differential Calculus

### Course: Differential Calculus > Unit 5

Lesson 10: Connecting f, f', and f''- Calculus-based justification for function increasing
- Justification using first derivative
- Justification using first derivative
- Justification using first derivative
- Inflection points from graphs of function & derivatives
- Justification using second derivative: inflection point
- Justification using second derivative: maximum point
- Justification using second derivative
- Justification using second derivative
- Connecting f, f', and f'' graphically
- Connecting f, f', and f'' graphically (another example)
- Connecting f, f', and f'' graphically

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# Connecting f, f', and f'' graphically

Analyzing three graphs to see which describes the derivative of which other graph.

## Want to join the conversation?

- How can you graph acceleration with only the velocity graph?(4 votes)
- Hi Katelin,

Since acceleration is the derivative of velocity, you can plot the slopes of the velocity graph to find the acceleration graph.(14 votes)

- i keep getting confused with positive/ negative and increasing/ decreasing. i dont get when to use which method?(5 votes)
- A derivative is positive when the original function is increasing, and negative when the original function is decreasing. So you look at where the original function increases and decreases to tell you when the derivative is positive or negative.(3 votes)

- When looking at these graphs, how can you tell when the slope is either positive or negative. I'm having difficulty understanding how to draw the derivative of each line.(4 votes)
- The slope of the function is positive when the line is going up and vise versa. One example at3:20the line is going down so Sal starts his line in the negative side of the graph.(1 vote)

- how was the derivative found in this video?(2 votes)
- Although we were unable to find numerical values of the derivatives of any of these graphs, we can use extreme points of the original function graph (mins and maxes) to roughly plot out where the derivative equals zero (aka crosses the x-axis)(1 vote)

- At3:07, how did you tell that 3rd function is the derivative of the first function. Would I still be wrong if I say that the 1st function is the derivative of the 3rd function?(1 vote)
- f' is the derivative of f, and f'' is the second derivative of f, which is the first derivative of f'. Every order of derivative after is just the derivative of the function before that.(3 votes)

- What does the F stand for in graphing? Does it not stand for anything? I have been looking everywhere for an answer.(1 vote)
- f(x) just means "a function in terms of x" and it is the same as y, except f(x) is a function and must have only 1 y-value for each assigned x-values (in other words it must pass the "line test").(2 votes)

- Assuming that f is a polynomial, can I just pick the graphs of of f, f', and f'' visually by recalling that a derivative of a polynomial will produce an polynomial equation of a 1 lesser degree. That means the roots of f will have the greatest number of roots, then f', and lastly f''.

I this right?(1 vote)- It's true that f will have more roots than f' or f'', but they may have any number of
*real*roots, which are the only roots that will be visible in a graph. So the degree alone is not enough information.(2 votes)

- how do you find the derivative of a function?(2 votes)
- There are no points, so we can't find the numerical derivative function.(0 votes)

- It's also easy to rule out the graph on the left as f as the other graphs all have multiple roots. If the tangent slope of the first graph only hits 0 at one spot, so the graph of the derivative should only have 1 root crossing the x-axis.(0 votes)
- If the graph of f is a line, what is f'(x) and f"(x).(0 votes)
- Remember that the value of f'(x) anywhere is just the slope of the tangent line to f(x).

On the graph of a line, the slope is a constant. The tangent line is just the line itself. So f' would just be a horizontal line. For instance, if f(x) = 5x + 1, then the slope is just 5 everywhere, so f'(x) = 5. Then f''(x) is the slope of a horizontal line--which is 0. So f''(x) = 0.

See if you can guess what the third derivative is, or the fourth!(4 votes)

## Video transcript

- [Instructor] We have the
graphs of three functions here, and what we know is that one
of them is the function f, another is the first derivative of f, and then the third is the
second derivative of f. And our goal is to figure
out which function is which. Which one is f, which
is the first derivative, and which is the second? Like always, pause this video and see if you can work through it on your own before we do it together. All right, now let's do this together. The way I'm going to tackle
it is I'm gonna try to sketch what we can about the derivatives
of each of these graphs, or the functions
represented by these graphs. So in this first graph
here in this orange color, we can see that the slope
is quite positive here, but then it becomes less
and less and less positive, up until this point where the
slope is going to be zero, and then it becomes more and more and more and more negative. So the derivative of this
curve right over here, or the function represented by this curve is gonna start off reasonably
positive right over there. At this point, the derivative
is gonna cross zero, 'cause our derivative is zero there, slope of the tangent line. And then it's gonna get
more and more negative, at least over the interval that we see. So it might look, I don't
know, something like this. I don't know if it's a line or not. It might be some type of a curve. It would definitely have a
trend something like that. Now, we could immediately tell that this blue graph is not the derivative of this orange graph. Its trend is opposite. Over that interval, it's going from being
negative to positive, as opposed to going from
positive to negative. So we can rule out the blue graph as being the derivative
of the orange graph. But what about this magenta graph? It does look like it has the right trend. In fact, it intersects the
x-axis at the right place right over there. And at least over this interval, it seems it's positive from here to here. So it's positive. This graph is positive when the slope of the tangent
line here is positive. And this graph is negative when the slope of the tangent
line here is negative. Now, one thing that might
be causing some unease to immediately say that this last graph is the derivative of the first one is we're not used to situations where the derivative
has more extreme points, more minima and maxima
than the original function. But in this case, it could just be 'cause we don't see the
entire original function. So, for example, if this last
graph is indeed the derivative of this first graph, then what we see is our
derivative is negative right over here, but then right around here it
starts becoming less negative. So if that point corresponds
to roughly right over there, then over here our slope
will become less and less and less negative, and then at this point our
slope would become zero, which would be right around there. So for example, our graph
might look something like this, we just didn't see it. It fell off of the part of the graph that we actually showed. So I would actually say that this is a good candidate for being, the third function is a good candidate for being the derivative
of the first function. So maybe we could say that this is f and that this is f prime. Now, let's look at the second graph. What would its derivative look like? So over here our slope is quite negative, and it becomes less and
less and less negative until we go right over here
where our slope is zero. So our derivative would
intersect the x-axis right over there. It would start out negative, and it would become less
and less and less negative. And at this point it crosses the x-axis and it becomes more and more positive. So we see here our derivative becomes more and more positive. But then right around here it seems like it's getting less positive again. So it might look something like this where over here it's
becoming less positive again, less positive, less
positive, less positive. Right over here, our
derivative would be zero. Our derivative would
intersect the x-axis there. And then it just looks like it is, the slope is getting more
and more and more negative, so our derivative is gonna get more and more and more negative. Well, what I very
roughly just sketched out looks an awful lot like the
brown graph right over here. So this brown graph does indeed look like the derivative of this blue graph. So what I would say is
that this is actually f, and then this would be f prime. And then if this is f prime, the derivative of that is
going to be f prime prime. So that looks good. I would actually go with this. And if you wanted, just for safe measure, you could try to sketch out what the derivative of
this graph would be. Actually, let's just do that. So over here, the derivative of this, so right now we have a positive slope of our tangent line is getting
less and less positive. It hits zero right over there. So the derivative might
look something like this over the interval. Now, the slope of the
tangent line is getting more and more and more and more negative, right until about that point. So it's getting more and
more and more negative until about that point. And now it looks like
it's getting less and less and less and less negative, all the way until the derivative
goes back to being zero. And then it looks like
it's getting more and more and more and more positive. So the derivative of
this magenta curve looks like an upward opening U. And we don't see that over here, so we could feel good that its derivative
actually isn't depicted. So I feel calling the middle graph f, calling the left graph f prime, and calling the right graph
the second derivative.