Justifying a function's features using the second derivative: maximum point
At about 218s the narrator says h'' being concave up does not justify that the original function is 'concave up'; in context he means it does not justify concave down (a maximum). Interventions avoid this wording.
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Given that h'(-4) = 0, the video picks the calculus-based justification for h having a relative maximum at x = -4: the second derivative is negative there, so h is concave down. It then explains why the other choices fail: a change from increasing to decreasing is valid but not calculus-based, and facts about the shape of h'' (a relative minimum, concave up) say nothing about its sign.
You'll be able to
- use h'(c) = 0 and h''(c) < 0 to justify a relative maximum at x = c.
- explain why h''(c) > 0 with h'(c) = 0 gives a relative minimum instead.
- explain why features of the graph of h'' (its extrema or concavity) do not justify a maximum of h on their own.
Key words
- relative maximum
- A point where the function's value is higher than at all nearby points.
- second derivative
- The derivative of the first derivative; it tells you whether the slope is increasing or decreasing.
- concave downwards
- Curving like an upside-down U, which happens when the first derivative is decreasing.
- calculus-based justification
- A reason that uses derivatives (such as the sign of h' or h'') rather than just describing the graph.
Transcript
- [Instructor] We're told that given that h prime of negative four is equal to zero, what is an appropriate calculus-based justification for the fact that h has a relative maximum at x is equal to negative four? So right over here, we actually have the graph of our function h.
This is the graph y is equal to h of x. And we don't have graphed the first derivative, but we do have graphed the second derivative right here in this orange color, h prime prime. So they're telling us, given that h prime of negative four is equal to zero, so that's saying that given that the first derivative at x equals negative four is equal to zero, and you can see that the slope of the tangent line when x is equal to negative four does indeed equal zero.
So given that, what is a calculus-based, let me underline that, a calculus-based justification for the fact that h has a relative maximum at x equals negative four? So this first one says that the second derivative at x equals negative four is negative. Now, what does that tell us?
If the second derivative is negative, that means that the first derivative is decreasing, which is another way of saying that we are dealing with a situation where, at least at x equals negative four, we are concave downwards, downwards, which means that the general shape of our curve is going to look something like this around x equals negative four.
And if the slope at x equals negative four is zero, well, that tells us that, yes, we indeed are dealing with a relative maximum point. If the second derivative at that point was positive, then we would be concave upwards. And then if our derivative is zero there, we'd say, okay, that's a relative minimum point.
But this is indeed true. The second derivative is negative at x equals negative four, which means we are concave downwards, which means that we are a upside U, and that point where the derivative is zero is indeed a relative maximum. So let me, so that is the answer.
And we're done, but let's just rule out the other ones. H increase before x equals negative four. That is indeed true. Before x equals negative four, we are increasing, and h decreases after it. That is true, and that is one rationale for thinking that, hey, we must have a maximum point, assuming that our function is continuous at x equals negative four.
So this is true. It is a justification for a relative maximum, but it is not calculus-based. And so that's why we can rule this one out. The second derivative has a relative minimum at x equals negative four. Well, it does indeed seem to be true. There's a relative minimum there, but that's not a justification for why this is why h of negative four or why we have a relative maximum at x equals negative four.
For example, this, you could have a relative minimum in your second derivative, but your second derivative could still be positive there. So what if the second derivative was like that? That would still be a relative minimum. But if it was positive at that point, then you would be concave upwards, which would mean that at x equals negative four, your original function wouldn't have a maximum point, it would have a minimum point.
And so just a relative minimum isn't enough. In order to know that you are dealing with a relative maximum, you would have to know that the second derivative is negative there. Now, this fourth choice, h prime prime is concave up, it does indeed look like the second derivative is concave up, but that, by itself, does not justify that the original function is concave up.
For example, well, I could use this example right here. This is a potential second derivative that is concave upwards, but it is positive the entire time. And if your second derivative is positive the entire time, that means that your first derivative is increasing the entire time, which means that your original function is going to be concave upwards the entire time.
And so if you're concave upwards the entire time, then you would not have a relative maximum at x equals negative four. So we would rule that one out as well.
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