
Applications
of differentiation  the graph of a function and its derivative 
The mean value theorem 
Generalization
of the mean value theorem

Cauchy's mean value theorem or generalized mean value theorem 






The
mean value theorem 
If a function f
is continuous on a closed interval [a,
b]
and differentiable between its endpoints, then
there is a point c
between a
and b
at which the slope of the tangent line to f
at c
equals the slope of the secant line through
the points (a,
f (a))
and (b,
f (b)), 

as
shows the right figure above. 
We
can also write the above formula as f
(b)

f (a) = f ' (c) (b 
a) 
and by substituting b
by x,
f
(x)
=
f (a) + f ' (c) (x 
a) 

Example:
Let
use the mean value theorem to prove that the abscissa of a point
of the parabola

f (x) = a_{2}x^{2} + a_{1}x
+ a_{0},
at
which the tangent is parallel to the secant line through
points (a,
f (a))
and 
(b,
f (b)),
is
the midpoint of the interval [a,
b]. 
Solution:
Since the derivative of the parabola
f
' (x) = 2a_{2}x + a_{1}
then f
' (c) = 2a_{2}c + a_{1},
and




Generalization
of the mean value theorem 
If
two functions,
f_{1
}(x)
and f_{2
}(x)
are continuous on a closed interval [a,
b]
and differentiable between its endpoints
have identical derivatives on that interval then they differ in a
constant. 
That is, if
we write j
(x)
= f_{1
}(x)

f_{2
}(x),
then its derivative
j
'
(x)
= f_{1}'
(x)

f_{2}'
(x)
= 0 at every
point of
the
interval, therefore j
(x)
= constant
or 
f_{1}(x)

f_{2}
(x)
= constant. 
This
theorem we use in the integral calculus. 

Example: Given
is the derivative f
' (x) =

x^{2}

8x

12 of a cubic function and
one of its roots x
=

1.
Find
the cubic and its source function and draw the graph of the cubic.

Solution:
Let factorize given quadratic to find its zeros,

f
' (x) =

x^{2}

8x

12 =

(x
+ 6)(x + 2),
x_{1}
= 
6 and x_{2}
= 
2 
and
calculate its translation in direction of the xaxis 
x_{0}
= 
a_{n1}/(n
a_{n}),
(x_{0})_{n
= 2} = 
(
8)/(2 · (
1)) = 
4, (x_{0})_{n
= 2} = 
4. 
Since,
a polynomial function and all its successive derivatives have the same
horizontal translation 

Then,
we calculate coefficient a_{1}
of the cubic 

where
a_{1} is the coefficient of the source cubic
f_{s}(x) = a_{3}x^{3} +
a_{1}x
or f_{s}(x) =
(1/3)x^{3} +
4x. 
Finally
we will find the coefficient a_{0}
by plugging given xintercept
(1,
0) into the cubic 

The cubic
function and its derivative are shown in the figure below. 

Let
find translation of the graph of the cubic in direction of the yaxis
by plugging x_{0}
= 
4
into its equation, 

Thus,
the graph of f
(x)
in the above figure is drawn translating the source cubic f_{s
}(x) =
(1/3)x^{3} +
4x
in direction
of the coordinate axes to the inflection point I(x_{0},
y_{0})
or I(
4, 3). 
To
check the source cubic expression mentioned above we should plug the coordinates
of translations, 
x_{0}
and y_{0}
into y
+ y_{0}
= a_{3}(x
+ x_{0})^{3} +
a_{2}(x + x_{0})^{2} +
a_{1}(x + x_{0}) +
a_{0}, 

Observe
that the problem can be solved almost straight considering we know how
find a function whose derivative
is given. The function we should find is called an antiderivative. 
Therefore,
to determine whether a function f
(x)
is
an antiderivative, we simply differentiate it and check if the result
is equal to given f
' (x). 
Let
denote f_{1}(x)
the cubic we solved and
differentiate it 

Now,
let's do reverse, integrate ( or antidifferentiate) f_{1}' (x), to return back the cubic 

Note
that the cubic functions, f_{1}(x)
and
f_{2}(x)
have the same derivatives and only differ in the constant term,
as was stated by the generalization of the mean value theorem above. 
The
graph of f_{2}(x),
the cubic missing the constant
term, passes through the origin. Therefore, we get its graph
by translating the graph of the cubic f_{1}(x), in the positive direction
of
the yaxis
by
25/3,
which is shown in the below
figure. 

Proof,
both
functions f_{1
}(x)
and
f_{2
}(x)
have
the same translation x_{0}
and the absolute value of the difference (
y_{0})_{1} 
(
y_{0})_{2}
of the translations in direction of the yaxis
equals
25/3.
That is, 

Thus,
the source function of f_{2} 

Therefore, 

Note,
that is why we should always add a constant C
to an antiderivative function as the substitution for the constant term lost by
differentiation. 
Let's mention here yet another straight method, that is
on some way related to what was just shown above, known as Maclaurin's
formula or theorem. 
An nth
degree polynomial P_{n
}(x) =
a_{n}x^{n} + a_{n}_{ }_{}_{
1}x^{n }^{}^{}
^{1} + ·
· · + a_{2}x^{2}^{}
+ a_{1}x + a_{0}, 
applying Maclaurin's
theorem,
can be written as 

Thus,
using Maclaurin's
formula we
can calculate coefficients of a polynomial
by evaluating all its successive derivatives
at the origin to the nth
order. 
Therefore,
we can find the coefficients of the cubic given in the
above example,
except the last term a_{0}. 
Since given
is, P_{3}' (x) =

x^{2}

8x

12
then P_{3}' (0) =
12 
P_{3}'' (x) =

2x

8,
P_{3}'' (0) =

8 
and
P_{3}''' (x) =

2,
P_{3}'''
(0) =

2 
then
by
plugging obtained values into above Maclaurin's
formula 


Cauchy's mean value theorem
or generalized mean value theorem 
If
f (x)
and g (x)
are differentiable in an interval (a,
b) and
continuous on [a,
b],
then 

or f
' (c) [g (b)

g
(a)]
=
g' (c) [f (b)

f
(a)],
a
< c < b. 
To
prove the above formula let 
j
(x)
= f (x) [g (b)

g
(a)]

g (x) [f (b)

f
(a)] 
then,
by calculating j
(a)
and j
(b)
we get 
j
(a)
=
j
(b)
=
f (a) g (b) 
g (a)
f (b) 
thus,
j
(x)
satisfies Rolle's theorem. 
Therefore,
there exists at least one point c
in the interval (a,
b) such
that 
j'
(c)
= f ' (c) [g (b)

g
(a)]

g' (c) [f (b)

f
(a)]
= 0,

what
proves the theorem. 









Calculus contents
D 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 