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### Course: Calculus 1 > 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 (another example)

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

## Want to join the conversation?

- Sal's logic at4:00seems to be flawed, but that is not critical. What do I mean? The leftmost and the rightmost graphs look like derivatives of each other, and just two of them are not enough for figuring out which of them is f, f' or f'', but it's clear that the only possible derivative of the function in the middle graph is the function on the leftmost graph (and the middle one itself can't be a derivative of the other two), which necessarily makes the middle one f and the left one f'. Something like that, as it is often said.(9 votes)
- Sal's logic is not flawed. In fact, you are repeating what he has just stated. For the context of the question, of course the left graph is f' since it is the only candidate for the middle graph, thus making the third graph the most likely candidate for the second derivative or the derivative of the derivative (The derivative of the left most graph). Obviously, without the middle graph, the first and right most graphs can be arbitrarily picked to be the derivatives of each other.(4 votes)

- If the derivative of an exponential function, is also an exponential function (like the derivative of e^x equals e^x), how are you supposed to figure out which is the original function and which is the derivative just by looking at its graph?(6 votes)
- You can't. Unless the original function was something like 𝑒ˣ + 𝑐 for some nonzero constant 𝑐. In this case, the derivative is 𝑒ˣ and the original function is a vertical shift of its derivative.(9 votes)

- do we have a video that talks about going from f'-graph to a f-graph??(4 votes)
- I'm not sure if a video exists, but going from a f'-graph to an f-graph should be straight forward. Remember that the first derivative test determines local extrema. So on the f' graph the critical points are where f'(x) = 0 (note: the pertinent critical points are where f' crosses the x-axis). Additionally the first derivative test requires a sign change before and after f' crosses the x-axis for x values close to f'(x) = 0. A sign change of positive to negative on the f' graph around f'(x)=0 specifies a relative maximum at f(x). A relative minimum at f(x) is a negative to positive sign change on the f' graph around f'(x)=0. This information should be enough to identify the f-graph.

One thing to watch out for is holes in the f'-graph. This could indicate a point where f'(x) is undefined. This could appear on the f-graph as a sharp curve or a point discontinuity. f(x) may also be undefined. So could appear on the f-graph as a vertical asymptote at x. Typically these "identify the graph" problems specify that f is a differentiable function, so these "edge" cases are not an issue.(1 vote)

- for the derivative of f, how did you know it starts from the top in quadrant one? like why not Q4, how do u differentiate f to f'!(1 vote)
- The entirety of the previous unit, and the one before, are about going from f to f' (and f'')(1 vote)

- The right and left functions are fascinating. They are the derivate of each other. That means if you keep applying the derivate operator on one of them you will get the other function then the function itself again then the other one, and so on.(2 votes)
- That's not necessarily true. For example, if one function is f(x) = sin(x), its derivative would be cos(x). However, the derivative of cos(x) is -sin(x), not sin(x).(0 votes)

## Video transcript

- [Narrator] We have the
graph of three functions here. And we're told that one
of them is the function F, one is its' first derivative,
and then one of them is the second derivative. We just don't know which one is which. And so, like always, pause
this video and see if you can figure it out. Alright, now the way I'm
going to tackle it is I'm gonna look at each of these graphs and try to think, what would
their derivatives look like. So for this first one,
we can see our derivative right over here, our
slope of our tangent line. It would be a little bit
negative and then it gets more and more and more negative, and as we approach this vertical
asymptote, right over here, it looks like it's
approaching negative infinity. So, the derivative would
actually, over here, it would be a little bit less than zero, but then it would get more
and more and more negative and then it would approach
negative infinity. So, it would have a similar
shape, general shape, to the graph itself, at
least to the left of this vertical asymptote. Now, what about to the right
of the vertical asymptote? Right to the right of the
vertical asymptote it looks like the slope of the tangent line
is very negative, but then it becomes less and
less and less negative, and it looks like it is
approaching, it is approaching zero. So, on this side the derivative
starts out super negative and then it looks like it is
the derivative is going to asymptote towards zero,
something like that. So, based on what I just sketched, it looks like this right
graph is a good candidate for the derivative of this left graph. You might say, what's
wrong with this blue graph? Well this blue graph out here,
notice that it's positive. So if this were the
derivative of the left graph, then that means the left graph
would need a positive slope out here, but it doesn't
have a positive slope. It's a slightly negative
slope becoming super negative. And so, right here we're slightly negative and then we become very negative. And so maybe this is, let's call this F, and maybe this is F prime. This is F prime right over here. And now, let's look at this middle graph. What would its derivative do? So over here, our slope
is slightly negative, and then it becomes more
and more and more and more and more negative. And so the derivative
of this might look like, it has to be slightly negative, but then it gets more and more
and more and more negative as we approach that vertical asymptote. And then on the right side
of the vertical asymptote, our derivative is very positive here, and then it gets less and
less and less and less and less and less positive. And so, we start our derivative
would be very positive and then it would get
less and less and less and less positive. It looks like it might,
the slope might asymptoting towards zero, so our graph
might look something like that. Well the left graph right here looks a lot like what I just sketched out as a candidate derivative
for this blue graph, for this middle graph. And so I would say that this is F, then this is the derivative of that, which would make it F prime
and then we already established that this right graph is the
derivative of the left one, so if it's the derivative of F prime, it's not F prime itself,
it's the second derivative. So I feel pretty good about that. And just for good measure,
we could think about what the derivative of
this graph would look like. Here the slope is slightly negative, but then it gets more and more
and more and more and more negative, so the derivative
would have a similar shape here. And then here, our derivative
would be very positive, and it gets less and
less and less and less and less positive. And so, we start very
positive and then it gets less and less and less and
less and less positive. And so as a general shape, it
actually does look a lot like this first graph. But the reason why I'm not
gonna say that this first graph is the derivative of the right hand graph, is because this right hand graph
was the only good candidate that we had for the derivative
of the left hand graph. And so I feel pretty good
with what we selected, that this middle one is F, the left one is the first derivative, and the right one is the second derivative.