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Applications
of differentiation - the graph of a function and its derivative
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Definition
of increasing and decreasing of a function
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A
function f
is increasing on an interval I
if f (x1)
< f (x2)
for each x1
< x2 in I
. |
| A
function f
is decreasing on an interval I
if f (x1)
> f (x2)
for each x1
< x2 in I
. |
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Note
that we always examine the graph of a function moving from left to right (x1
< x2)
when determining intervals I
in which the function increases or decreases. |
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Increasing/decreasing
test
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| If
the first derivative of a function f
is positive for all x
in an interval, then f
is increasing on that interval. |
| If
the first derivative of a function f
is negative for all x
in an interval, then f
is decreasing on that interval. |
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Rolle's
theorem
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If a
function f
is continuous on a closed interval [a,
b]
and differentiable between
a
and b,
for which the function
has the same value, that is f (a)
= f (b),
then there exists point
c
inside the interval at which f
' (c) = 0,
as shows the left figure below. |
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The
mean value theorem
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| 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)), |
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as
shows the right figure above. |
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| We
can also write the above formula as f (b)
-
f (a) = f ' (c) (b -
a) |
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and by substituting b
by x,
f (x)
=
f (a) + f ' (c) (x -
a) |
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Example:
Let
use the mean value theorem to prove that the abscissa of a point
of the parabola
f (x) = a2x2 + a1x
+ a0,
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) = 2a2x + a1
then f
' (c) = 2a2c + a1,
and
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Generalization
of the mean value theorem
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| If
two functions,
f1(x)
and f2(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)
= f1(x)
-
f2(x),
then its derivative j
' (x)
= f1' (x)
-
f2'
(x)
= 0 at every
point of
the
interval, therefore j
(x)
= constant
or f1(x)
-
f2
(x)
= constant. |
| This
theorem we use in the integral calculus. |
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