We want to consider the given function f
of x on the closed interval
from negative five to seven.
We're going to begin
by finding the average
or mean slope on this interval,
which would be the slope
of the line passing
through the points when
x equals negative five
and x equals seven.
Then by the Mean Value Theorem,
there exists a c in the open interval
from negative five to seven,
such that f prime of c,
which would be the slope
of the tangent line,
is equal to the mean or average slope.
So we want to find the
values of c in this interval.
Let's begin by looking
at this graphically.
The given function is graphed here in blue
on the closed interval from
negative five to seven.
So the slope of this red line passing
through the two end points
on the closed interval
from negative five to
seven would be the mean
or average slope.
Then our goal is to find
the x values on the function
where the slope of the tangent
lines would be the same
as the mean slope or the
slope of this red line.
So looking at it graphically, we can see
we should be able to find two values of c
where the slope of this tangent line
and the slope of this tangent
line would be the same
as the mean or average slope.
So to begin we'll find the mean slope,
where the slope is equal
to the change of y divided
by the change of x,
where the change at y would be the change
in the function values,
which would be f of seven
minus f of negative five,
and the change of x would be
seven minus negative five.
To save some time I've already
found these function values.
If we substitute seven into the function,
you would get 64,
so we'd have 64 minus f of negative five,
is negative 152,
divided by seven minus negative five,
would be seven plus five or 12.
This works out to 216 divided by 12,
so the mean or average
slope is equal to 18.
216 divided by 12 equals 18.
So now to find the values of c,
such that f prime of c
would be equal to 18,
we'll find the derivative function,
set it equal to 18 and solve for x.
If this is the given function,
then the derivative function f prime of x
would be equal to the derivative
of two x to the third,
that would be six x squared,
minus the derivative of six
x squared that'd be 12 x,
minus the derivative of 48 x that'd be 48,
plus the derivative of
eight that would be zero.
So we want to determine when
this derivative function
would be equal to the mean slope of 18.
So now I'll go ahead
and solve this equation
on the next slide.
Let's go and set this equal
to zero by subtracting 18
on both sides that would
give us six x squared
minus 12 x.
Negative 48 minus 18,
that'd be negative 66
or minus 66 equals zero.
We do have a common factor here of six.
Let's factor out 6,
leaving us with x squared minus two x
minus 11 equals zero.
But unfortunately this
is not going to factor.
So to find our solutions here,
we'll have to use the quadratic formula.
Where a is equal to one,
b is equal to negative two,
and c is equal to negative 11.
So we'd have x equals
negative negative two
plus or minus the square
root of b squared,
that's negative two squared,
minus four times a which is one,
times c which is negative 11,
divided by two times a or two times one.
So now we have x equals,
this would be positive two,
plus or minus the square root of,
this would be four plus 44 that'd be 48,
divided by two.
And we could simplify
the square root of 48.
The square root of 48 is
equal to the square root
of 16 times three.
The square root of 16 is equal to four,
so this simplifies to
four square root of three.
So we have two plus or minus
four square root of three
divided by two.
Simplify this.
I recommend breaking it up
into two separate fractions.
We have x equals two divided by two
plus or minus four square root of three,
divided by two.
If we do it this way,
we avoid making the mistake of
simplifying just these twos.
Because we cannot simplify
across addition or subtraction,
this simplifies nicely
to one plus or minus
two square root three.
And therefore we can say that c sub one,
guaranteed by the Mean Value Theorem,
would be one plus two square root of three
and c sub two would be equal to one minus
two square root of three.
So these are the values of c that we want,
but if we want to check this graphically,
we should also get decimal approximations
so we can more easily find
the locations on the graph.
And I've already done that.
One plus the square root of
three is approximately 4.4641
and one minus two square root of three is
approximately negative 2.4641.
Now on our homework though we
shouldn't round these values
unless we're told to.
We should use the exact values.
But if I go back to our graph,
notice here is where x
is approximately 4.4641,
where the slope of the
tangent line is the same
as the mean slope.
And here is where x equals
approximately negative 2.4641,
again where the slope of
the tangent line is the same
as the mean slope.
So these are the x values
or the c values guaranteed
by the Mean Value Theorem.
I hope you found this helpful.
