Main content

## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 6

Lesson 7: Justifying properties of functions using the second derivative- Inflection points from graphs of function & derivatives
- Justification using second derivative: maximum point
- Justification using second derivative: inflection point
- Justification using second derivative
- Justification using second derivative
- Worked example: Inflection points from first derivative
- Worked example: Inflection points from second derivative
- Inflection points from graphs of first & second derivatives
- Second derivative test
- Second derivative test

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Justification using second derivative: maximum point

Given that the derivative of a function is zero, we can justify whether the function has a relative maximum point by looking at the second derivative.

## Want to join the conversation?

- I don't think that this was an accurate justification because h''(-5) is also negative, does that mean that h(5) is the relative maximum? NO!!(10 votes)
- True, but h'(5) isn't zero.

The justification is not*just*that h''(x) is negative; it has to be h'(x) = 0 AND h''(x) is negative.

And we're told in the question that h'(-4) = 0.(27 votes)

- Why the graph of the function concaves downward when the second derivative is negative?(2 votes)
- If a function's second derivative is negative, then its slope is decreasing. This is equivalent to saying that a function is concave downward.

Remember: The first derivative gives the rate of change (slope) of the function, while the second derivative gives the rate of change of the first derivative. This means that the second derivative tells us how the slope is changing.(4 votes)

- How is it possible when f''(x) is negative that f'(x) is 0, shouldn't it be decreasing? am I missing something?2:00[h''(-4) is negative, h'(-4)=0](1 vote)
- f''(x)<0 means the function is concave down, and f'(x)=0 means it's neither increasing nor decreasing at that point.

Consider f(x)=-x^2 at x=0:

f'(x)=-2x, so f'(0)=0, and f''(x)=-2.(2 votes)

- Wouldn't you technically need both A and C to prove by the second derivative that x=-4 is a relative maximum point? If h''(-4) wasn't a relative minima and h''(-3) was, this wouldn't have the same result.(1 vote)
- No, because whether the h’’ is increasing, decreasing, or at an extrema is irrelevant here, we’re only looking for if it’s positive or negative (and we already know that h’(-4)=0.)

What this tells us is that if h’ is decreasing at -4, meaning it must have been positive before -4, and negative after that, therefore h(-4) is a relative maximum.(1 vote)

## Video 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.