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## Mean value theorem

# Mean value theorem example: polynomial

AP.CALC:

FUN‑1 (EU)

, FUN‑1.B (LO)

, FUN‑1.B.1 (EK)

## Video transcript

Let's say I have
some function f of x that is defined as being equal
to x squared minus 6x plus 8 for all x. And what I want to do is
show that for this function we can definitely find
a c in an interval where the derivative at the point c
is equal to the average rate of change over that interval. So let's give ourselves an
interval right over here. Let's say we care about the
interval between 2 and 5. And this function is definitely
continuous over this closed interval, and it's also
differentiable over it. And it just has to
be differentiable over the open interval, but
this is differentiable really for all x. And so let's show
that we can find a c that's inside the
open interval, that's a member of the open
interval, that's in the open interval such
that the derivative at c is equal to the average rate
of change over this interval. Or is equal to the slope of
the secant line between the two endpoints of the interval. So it's equal to f of 5
minus f of 2 over 5 minus 2. And so I encourage you to
pause the video now and try to find a c where
this is actually true. Well to do that, let's just
calculate what this has to be. Then let's just take the
derivative and set them equal and we should be able
to solve for our c. So let's see f of 5 minus f of
2, f of 5 is, let's see, f of 5 is equal to 25 minus 30 plus 8. So that's negative 5
plus 8 is equal to 3. f of 2 is equal to
2 squared minus 12. So it's 4 minus 12 plus 8. That's going to be a 0. So this is equal to 3/3,
which is equal to 1. f prime of c needs
to be equal to 1. And so what is the
derivative of this? Well let's see, f prime of
x is equal to 2x minus 6. And so we need to figure
out at what x value, especially it has to be
in this open interval, at what x value
is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7/2, which is
the same thing as 3 and 1/2. So it's definitely in this
interval right over here. So we've just found
our c is equal to 7/2. And let's just graph
this to really make sure that this makes sense. So this right over
here is our y-axis. And then this right
over here is our x-axis. Looks like all the
action is happening in the first and
fourth quadrants. So that is our x-axis. And let's see, let's say
this is 1, 2, 3, 4, and 5. So we already know that 2
is one of the zeroes here. So we know that our function
if we wanted to graph it, it intersects the
x-axis right over here. And you can factor this out
as x minus 2 times x minus 4. So the other place where
our function hits 0 is when x is equal
to 4 right over here. Our vertex is going to be right
in between at x is equal to 3. When x is equal to 3, let's
see, 9 minus 18 is negative 9 plus 8, so negative 1. So you have the point
3, negative 1 on it. And so that's enough
for us to graph. And we also know
at 5, we're at 3. So 1, 2, 3. So at 5, we are right over here. So over the interval
that we care about, our graph looks
something like this. So that's the interval
that we care about. And we're saying
that we were looking for a c whose slope is
the same as the slope of the secant line,
same as the slope of the line between
these two points. And if I were to just visually
look at it, I'd say well yeah, it looks like right around
there just based on my drawing, the slope of the tangent line
looks like it's parallel. It looks like it
has the same slope, looks like the tangent line is
parallel to the secant line. And that looks like it's
right at 3 and 1/2 or 7/2. So it makes sense. So this right over here is
our c. c is equal to 7/2.