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# Analyzing a function with its derivative

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.10 (EK)
,
FUN‑4.A.9 (EK)

## Video transcript

We've got the function f of x is equal to x to the third power minus 12x plus 2. And what I want to do in this video is think about at what points does my function f take on minimum or maximum values? And to figure that out I have to first figure out what are the critical points for my function f. And then which of those critical points do we achieve a minimum or maximum value? And to determine the critical points we have to find the derivative of our function because our critical points are just the point at which our derivative is either equal to 0 or undefined. So the derivative of this thing right over here, we're just going to use the power rule several times, and then I guess you can call it the constant rule. But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x is either undefined or 0. Well this thing is defined for all values of x. So the only places we're going to find critical points is when this thing is equal to 0. So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f of 2, or let me be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime negative 2 is also, same exact reason, is also equal to 0. So we can say-- and I'll switch colors here-- that f has critical points at x equals 2 and x equals negative 2. Well that's fair enough. But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12. So this is the point y is equal to negative 12. So this is, we're graphing y is equal to f prime of x. So it looks something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well over here our derivative is crossing from being positive, we have a positive derivative, to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative, that was our criteria for a critical point to be a maximum point. Over here we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function to have a minimum value at a critical point. So a minimum. And I just want to make sure we have the correct intuition. If our function, if some function is increasing going into some point, and at that point we actually have a derivative 0-- the derivative could also be undefined-- but we have a derivative of 0 and then the function begins decreasing, that's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation we're going into the point, the function has a negative slope, we see we have a negative slope here-- so the function might look something like this. And then right at this point the function is either undefined or has 0 slope. So in this case it has 0 slope. And then after that point, let me do it right under it. So going into it we have a negative slope. And then right over here we have a 0 slope. Which I could draw it even better than that. So if we were to imagine going into it we have a negative slope, right at that point we have a 0 slope, and then we have a positive slope. So the function begins increasing. That's why we say we have a minimum point right over there. So what I did it right over here is to try to conceptualize what the function itself could look like given the derivative, in this case switching from a positive derivative to a negative derivative, across that critical point, or going from a negative derivative to a positive derivative. That's why this is the criteria for a maximum point, this is the criteria for a minimum point. Well with that out of the way, can we use this intuition that we just talked about to at least try to sketch the graph of f of x? So let's try to do it. And it's just going to be a sketch, it's not going to be very exact. But at least it'll give us a sense of the shape of what f of x looks like. So my best attempt. So it might not be drawn completely to scale. So it's my x-axis, this is my y-axis. We know we have a critical point at x is equal to positive 2. And we have a critical point at x is equal to negative 2. We know just from inspection that the y-intercept right here, if the graph of y is equal to f of x, when x is 0 f of x is 2. So we're going to hit right over-- I don't want to draw this completely to the same scale as the x-axis. So let's say that this is 2 right over here. So this is where we're going to cross. This is going to be our y-intercept. And so we said already that we have a maximum point at x is equal to negative 2. So what is f of negative 2? f of negative 2 is equal to 8 or negative 8, let me be careful. It's negative 8. And then we're going to have 12 times negative 2, which is negative 24. But then we're going to add it. So we're subtracting negative 24. So this is plus 24. And then we finally have plus 2. So negative 8 plus 24 plus 2, so that's going to be negative, let's see, negative 8 plus 24 is 16, plus 2 is 18. So f of negative 2 is equal to 18. And I'm not drawing it completely to scale. But let's say that this is 18 right over here. So this is the function. This is the point negative 2 comma 18. And we know that it's a maximum point. The derivative going into that point is negative. The derivative going into that point is negative-- sorry, the derivative going into that point is positive. So we are increasing. The slope is positive. And then after we cross that point, the slope becomes negative. The derivative cross the x-axis, the slope becomes negative. I actually want to use that same color. It looks like this. And then, of course, the graph is going to cross the-- it's going to have a y-intercept, something like that. And then, as we approach 2, we are approaching another critical point. Now what is f of 2? f of 2 is going to be equal to positive 8 minus 24 plus 2. So this is 10 minus 24, which is equal to negative 14. So let's say this is the point negative 14 right over here. Actually I can draw it a little bit. Let's say this is negative 14. So this is f of 2 right over there. And we saw already that the slope is negative as we approach it. So our function is decreasing as we approach it. And then right there the slope is 0. We figured that out earlier, that's how we identified it being a critical point. And then the slope is increasing after that, the derivative is positive. The slope is increasing. So this is our sketch of what f of x could look like given that these are the critical points. And we were able to identify 2 as a minimum point. So this was a minimum value. The function takes on a minimum value when x is equal to 2. And the function took on a maximum value when x was equal to negative 2.