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# 2011 Calculus AB free response #4d

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Part d, find the average rate of change of
f on the interval from x is greater than or equal, negative four is less than or
equal to x, which is less than or equal to
three. So really the interval that they've
depicted right over here. And then they say there's no point c
between in that interval for which f prime of c is equal to the average
rate of change. Explain why the statement does not
contradict the Mean Value theorem. Fascinating. Alright. Let's do this first part first. The average rate of change of f on the
interval. Sounds like a very fancy thing, but the
average rate of change on the interval is really just a slope of the line that
connects the end points of the interval. So these, this right over here, those are
the end points, and let's figure out the slope of
that line. So, going we, going from this point to
that point, our change in x, we do this in a color that you're
likely to see. Our change in x over here, our change in
x, change in x, is equal to 7. And you can get that by taking 3 minus
negative 4 or you could literally just count, 1, 2, 3,
4, 5, 6, 7. That's our change in x. And our change in y, when we, when we run
over 7, when we go 7 to the right, our change in y, our change
in y is equal to negative 2. We went from negative 1 to negative 3 is
equal to negative 2. So, slope, slope, which is change in y
over change in x, rise over run. Is equal to 7 over negative 2, or well
actually let me, the other way around. The change in y is negative 2 over 7. So negative two 7s and that's all it is. Now you could, you could kind of. Think of it in, in, kind of a more fancy sense, but you're gonna get the
exact same answer if you said, oh well look, you know, the
average, the average rate of change of f on the
interval. Well, the rate of change of f is f prime,
f prime of x. That is the rate of change of f at any
point x. And so if you want to find the average
value of this over the interval, you would
integrate it from our starting point, from negative 4 to 3
dx, and then you would divide it by your change in
x. So then you would divide it by, you would
have it over 1 over 7. Over 1 over 7. But then this, this part right over here
is just going to be the same thing as f of 3 minus f of 4. And then this over here you have a 7 in
the denominator. So this is really just your change in x. This is your change in x which is, which
was by definition or how we, how we actually set up this
average right over here. And this over here is really just your
change in y. So it really is just a slope between the
end points. So, we did the first part. Our average rate of change of f on the
interval is negative two 7s. And then let's think about the second
part. You see there is no point c in that
interval. For which f prime of c is equal to the
average rate of change. Explain why this statement does not
contradict the Mean Value Theorem. So the Mean Value Theorem, just as a
little bit of a review, as a little bit of review, it says that if we
have some type of an interval. If you have an interval. So, let me draw some axes right over here. If you have an interval. Let me draw an interval like this. And over that interval you have a
differentiable function. You have a differentiable function, so
maybe my function looks like this. It says at least, there is at least one
point c on that interval where the derivative at that point c is
equal to the average rate of change. So the way I've drawn it right over here,
the average rate of change of this function, I'll do it in
magenta, is this right over here. And the Mean Value Theorem says, is it, if
this is deferentiable, there's at least one
point c in this interval. Where I have the same slope. Where the tangent line has the same slope. Where the derivative is the same as the
slope, as that average slope. And you can see here, is probably right
over here you have one of those points and actually we
probably have multiple of them. We probably have another point right over
here that's like that, and then another point there
that's like that. And if you think about it, it's kind of. It's kind of intuitive that at some point, you know, here we're, we obviously have a
larger slope and over we have a smaller slope and since our, since it's differentialable,
our derivative is continuous. So at some point, at some point the slope has to get to exactly what the average
slope is. Now, let's think about our little conundrum with this question right over
here. Why is there not a point c for which f prime of c is equal to the average rate of
change? You can even verify that for yourself
because from negative 4 to 0 our slope is
positive. We have a positive slope here, and then
the slope just jumps down to negative 2. It just jumps down to negative 2 which is
a much more negative slope than this. So, it never goes to negative 2 7s, and
the reason is that. This is not a differentiable function at x
is equal to 0. It is not differentiable at x equals 0. Our slope jumps here and because it's not differentiable the Mean Value theorem
doesn't apply. Now you can imagine if this was
differentiable, if this did have a continuous derivative, then you
would find a point. So if this looks something like this
instead, I'll continue it over here. If it looks something like this instead where it was, it had a continuous
derivative then there would be a point where the
slope was the same as the average slope. Maybe it would have been right over there. So it's really because, it's really
because f is not differentiable over the entire
interval. It does not have a continuous derivative. The derivative jumps from a, from a
positive value approaching 0 here and it jumps just straight down to
negative 2 right over here. It doesn't go continuously through all the
values in between.