Identifying f, f', and f'' based on graphs

Let ( f ) be a twice differentiable function. One of these graphs is the graph of ( f ), one is of ( f' ), and one is of the second derivative of ( f' ). Match each function with its appropriate graph.

So, I encourage you to pause the video and try to figure out which of these is ( f ), which of these is ( f' ), and which of these is ( f'' ).

All right, now let's work through this together. There's a couple of ways to think about it. One way to think about it is if these were graphs of polynomials, then ( f ) would be a higher degree polynomial.

So you have ( f ); if you were to take the derivative of that, you would get ( f' ), which is this derivative with respect to ( x ). If you take the derivative with respect to ( x ) of that, you get ( f'' ).

So, whatever degree polynomial this is, if it is a polynomial, this will be a lower degree, and then this would be an even lower degree polynomial.

So, let's see if we can make sense of these graphs using that lens. One way to think about it is higher degree polynomials are going to have more minima or more potential minimum or maximum points. It looks like ( B ) here has the most minima and maxima; ( A ) has the second most, and then ( C ) has the third most.

So, my first hypothesis is that what we have right over here ( B ) that this is the graph of ( f ), that ( A ) is the graph of ( f' ), and that ( C ) is the graph of the second derivative ( f'' ).

So, that's just my hypothesis. Now let's see if it actually makes sense. What I can do is I can look at points where I know that the derivative would be zero and then look at the derivative to confirm or what I think is the derivative to confirm that they are. I could also look at trends.

So, let's say if this is ( f ), once again that was just my initial guess that this is going to be the function ( f ).

So, let's see; I have a slope of ( 0 ), a slope of a tangent line of ( 0 ) right over there. So slope of ( 0 ) right over there; I have a slope of ( 0 ) right over there. I have a slope of ( 0 ) right over there; I have a slope of ( 0 ), a horizontal tangent line right over there.

So, let's see if this is the derivative. Then this function should be equal to zero at this ( x )-value, this ( x )-value, this ( x )-value, and this ( x )-value, because the slope of the tangent line at those ( x )-values looks like zero.

So, this function is equal to zero here, here, here, and here. Well, these seem to coincide—this coincides with this, this coincides with that, and that coincides with that, and then that coincides with that.

So, I'm already feeling pretty good about it. So now let's see if this ( f'' ), what I think is ( f'' ), is indeed the derivative of ( f' ).

So, it's got a minimum point right over here. It looks like the slope of the tangent line here is zero. Let me do this in a different color. So this looks like we have a slope of ( 0 ) there. It looks like we have a slope of ( 0 ) right over there, and it looks like we have a slope of ( 0 ) right over there.

So that point for that ( x )-value, that ( x )-value, and that ( x )-value, and we see that this function is indeed equal to zero right over here, right over here, and right over here at the exact same ( x )-values.

So, based on what I just said, I would feel pretty good if I was under time pressure. I already feel pretty good about how I match things up.

But you could also look at trends between points. So, for example, let me pick out a trend. Between this maximum point and this minimum point, at first, the slope is decreasing. So, the slope is getting more and more negative, and you see from here to here is indeed getting more and more negative.

Then, it starts getting more and more positive starting right about here. So, that trend seems to be consistent.

You could look for these trends as well, but I find looking at the minimum and maximum where you see horizontal tangent lines or where the slope of the tangent line is zero—that those are the easiest to test for your derivatives or your second derivatives.