Justification using first derivative | AP Calculus AB | Khan Academy

The differentiable function f and its derivative f prime are graphed. So let's see. We see the graph of y is equal to f of x here in blue, and then f prime we see in this brownish orangish color right over here. What is an appropriate calculus based justification for the fact that f is decreasing when x is greater than three?

So we can see that that actually is indeed the case. When x is greater than 3, we see that our function is indeed decreasing as x increases; the y value, the value of our function, decreases. So, calculus based justification: without even looking at the choices—I could look at the derivative and we're going to be decreasing if the slope of the tangent line is negative, which means that the derivative is negative. We can see that for x greater than 3, the derivative is less than zero.

So my justification, I haven't even looked at these choices yet, I would say for x greater than three, f prime of x is less than zero. That would be my justification, not even looking at these choices.

Now let's look at the choices. F prime is decreasing when x is greater than 3. Now this isn't right. What we care about is whether f prime is positive or negative. If f prime is negative, if it's less than zero, then the function itself is decreasing; the slope of the tangent line will be negative. You could—f prime could be positive while decreasing. For example, f prime could be doing something like this, and even though f prime would be decreasing in this situation, the actual value of the derivative would be positive, which means the function would be increasing in that scenario. So I would rule this one out.

For values of x larger than 3, as x values increase, the values of f(x) decrease. Now that is actually true. This is actually the definition that f is decreasing: as x values increase, the values of f of x decrease. But this is not a calculus based justification, so I am going to rule this one out as well.

F prime is negative when x is greater than 3. Well, that's exactly what I wrote up here. If f prime is negative, then that means that our slope of the tangent line of our original function f is going to be downward sloping, or that our function is decreasing. So this one is looking good.

And this one right over here says f prime of 0 is equal to negative 3. So they're just pointing out this point. This isn't even relative to the interval that we care about or this is even relative when x is greater than 3, so we definitely want to rule that one out.

Let's do one more of these. So here we're told the differentiable function g and its derivative g prime are graphed. So once again, g is in this bluish color and then g prime, its derivative is in this orange color. What is an appropriate calculus based justification for the fact that g has a relative minimum point at x is equal to negative three?

And we could see here when x is equal to negative three it looks like g is equal to negative 6, and it looks like a relative minimum point there. So what's the best justification? Once again, without even looking at the choices, I would say a good justification is before we get to x equals negative 3, before we get to x equals negative 3, our derivative—this is a calculus based justification—before we go to x equals negative 3, our derivative is negative, and after x equals negative 3, our derivative is positive. That would be my justification, because if our derivative is positive before that value, if our derivative is negative before that value, that means that we are downward sloping before that value. And if it's positive after that value, that means we're upward sloping after that, which is a good justification that we are at a relative minimum point right over there.

So let's see. The point where x equals negative 3 is the lowest point on the graph of g in its surrounding interval. That is true, but that's not a calculus based justification; you wouldn't even have to look at the derivative to make that statement. So let's rule that one out.

G prime has a relative maximum at (0, 3). At (0, 3), it actually does not—oh, g prime! Yes, g prime actually does have a relative maximum at (0, 3), but that doesn't tell us anything about whether we're at a relative minimum point at x equals negative three, so I would rule that out.

G prime of negative three is equal to zero. So g prime of negative 3 is equal to 0. So that tells us that the slope of the tangent line of our function is going to be 0 right over there. But that by itself is not enough to say that we are at a relative minimum point. For example, I could be at a point that does something like this where the slope of the tangent line is zero, and then it keeps increasing again, or it does something like this and it keeps decreasing. So even though you're at a point where the slope of your tangent line is zero, it doesn't mean you're at a relative minimum point, so I would rule that out.

G prime crosses the x-axis from below it to above it at x equals negative three. G prime crosses the x-axis from below it to above it? Yep! And that's the argument that I made—that we're going from below the x-axis. So g prime goes from being negative to positive, which means the slope of the tangent lines of our points as we approach x equals negative three goes from being downward sloping to upward sloping, which is an indication that we are at a relative minimum point.