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### Course: AP®︎/College Calculus BC > Unit 5

Lesson 10: Connecting a function, its first derivative, and its second derivative- Calculus-based justification for function increasing
- Justification using first derivative
- Justification using first derivative
- Justification using first derivative
- Inflection points from graphs of function & derivatives
- Justification using second derivative: inflection point
- Justification using second derivative: maximum point
- Justification using second derivative
- Justification using second derivative
- Connecting f, f', and f'' graphically
- Connecting f, f', and f'' graphically (another example)
- Connecting f, f', and f'' graphically

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# Justification using first derivative

We can explain why a function is increasing, decreasing, or having a relative maximum using information about its first derivative. This is called calculus-based justification.

## Want to join the conversation?

- Is it possible to say that "g' crosses the x-axis from below it to above it at x= -3" for the AP exam and get full points? Wouldn't you have to be more mathematical about the justification by saying that "Since g' changes from negative to positive at x = -3, there is a relative minimum at x = -3"?(4 votes)
- Both statements are equivalent.

"Since g' changes from negative to positive at x = -3, there is a relative minimum at x = -3" another way of stating is "g' crosses the x-axis from below it to above it at x= -3"(2 votes)

- at3:40

How is the derivative negative before x=3 and positive after x=3?

I am looking at the orange graph of the derivative.(2 votes) - Is it correct to say that the value of the second derivative at some point is equal to the rate at which the slope of the original function is changing at that point?(1 vote)
- Yep! That is correct. And extending the same analogy, you can say that a third derivative at a point would be the rate of change of the rate of change of the slope of the function.(2 votes)

- One might want to "justify" the use of the 1st derivative given a function formula (though if we plot it, we obviously don't need its derivative to know where it's increasing/decreasing/constant). So this could be approached in both ways, right?(1 vote)
- If you want to know an extremum of a function that is defined over all real numbers, it is basically impossible to just plot and look by eye. You would want a shortcut, e.g. the derivative(s).(1 vote)

- @1.50 Sal said ". 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" I think Sal mean the function would be positive in that scenario , because I the derivative is decreasing but still positive , the function it self is decreasing(1 vote)
- Not quite. f' decreasing tells us two things:

1. f" is negative

2. The derivative is positive (in this case)

So, if the derivative is positive, the function is increasing. It needn't necessarily*be*positive. For example, graph x^2 - 4 and check the interval (0,2). Here, the function is increasing, the derivative is positive, but the function itself is negative.

Plus, to your last statement, if the derivative is positive, the function cannot decrease.(1 vote)

## Video transcript

- [Instructor] "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 the 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 three, we see that our function
is indeed decreasing. As x increases, the y value, the value of our function, decreases. So, a calculus-based justification, without even looking at the choices, well, 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. And we can see that for
x is greater than three, the derivative is less than zero. So, my justification,
I haven't even looked at these choices yet. I would say for x is greater than three, f prime of x is less than a 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 three." 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, than the function itself is decreasing. The slope of the tangent
line will be negative. 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
three, as x values increase, "the values of f of 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 three." 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 zero is
equal to negative three," so they're just pointing out this point. This isn't relative to the
interval that we care about, or this isn't even relevant
when x is greater than three, so we definitely wanna 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 six, and it looks like a relative
minimum point there. So, what's the best justification? So, once again, without
even looking at the choices, I would say a good justification is before we get to x equals negative three, before we get to x equals negative three, our derivative, and this is a
calculus-based justification, before we go to x equals negative three, our derivative is negative. And after x equals negative three, our derivative is positive. That would be my justification. Because 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 relative
minimum point right over there. So, let's see, "The point
where x equals negative three "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 zero comma three." At zero comma three, it actually does not. Oh, g prime, yes, g prime actually does
have a relative maximum at zero comma three, 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
three is equal to zero, so that tells us that the
slope of the tangent line of our function is going to
be zero right over there, but that by itself is not enough to say that we are at
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 go from being downward
sloping to upward sloping, which is an indication that we are at a relative minimum point.