# 3.10 Theorems about Differentiable Functions - Arkansas Tech ...

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8 Οκτ 2013 (πριν από 4 χρόνια και 7 μήνες)

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Arkansas Tech University
MATH 2914:Calculus I
Dr.Marcel B.Finan
Consider the following problem:Let f(x) be a continuous function on [a;b]
and dierentiable in (a;b):Is there a point on the graph of f where the
tangent line at that point is parallel to the line passing through the points
(a;f(a)) and (b;f(b))?The answer is yes according to the following theorem
Theorem 24.1 (Mean Value Theorem)
If f is continuous on [a;b] and dierentiable in (a;b) then there is a number
a < c < b such that the tangent line to the graph at (c;f(c)) is parallel to
the line passing through the points (a;f(a)) and (b;f(b)):That is,
f
0
(c) =
f(b) f(a)
b a
:
Example 24.1
Use the Mean Value Theorem to show the following:Suppose that f is
continuous on [a;b] and dierentiable on (a;b):If f
0
(x) = 0 on (a;b) then
f is a constant function.That is,f(x) = c for all x in [a;b] where c is a
constant.
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Solution.
We just need to show that any two values x
1
and x
2
in [a;b] have the same
y-value,that is,f(x
1
) = f(x
2
):By the MVT,there is x
1
< c < x
2
such that
f
0
(c) =
f(x
2
)f(x
1
)
x
2
x
1
:But f
0
(c) = 0 so that
f(x
2
)f(x
1
)
x
2
x
1
= 0:This implies that
f(x
2
) f(x
1
) = 0 or f(x
1
) = f(x
2
):
Theorem 24.2 (The Increasing Function Theorem)
Suppose that f(x) is continuous on [a;b] and dierentiable in (a;b):
(a) If f
0
(x) > 0 on [a;b] then f is increasing on [a,b](that is if x
1
< x
2
then
f(x
1
) < f(x
2
),where a  x
1
;x
2
 b).
(b) If f
0
(x)  0 on (a,b) then f is nondecreasing on [a,b](that is if x
1
< x
2
then f(x
1
)  f(x
2
),where a  x
1
;x
2
 b).
Proof.
Let x
1
and x
2
be two numbers in [a;b] such that x
1
< x
2
:By the Mean Value
Theorem,there is a number c inside the interval (x
1
;x
2
) such that
f(x
2
) f(x
1
) = f
0
(c)(x
2
x
1
)
(a) Since f
0
(x) > 0 in (a,b),the right hand side is positive.That is,f(x
2
) 
f(x
1
) > 0 or f(x
1
) < f(x
2
):
(b) Since f
0
(x)  0;f(x
2
) f(x
1
) = f
0
(c)(x
2
x
1
)  0 or f(x
1
)  f(x
2
):
Theorem 24.3 (The Racetrack Principle)
Suppose that f and g are continuous on [a;b] and dierentiable in (a;b):
Furthemore,suppose that f
0
(x)  g
0
(x) in (a;b):
(a) If f(a) = g(a) then f(x)  g(x) for all a  x  b:
(b) If f(b) = g(b) then f(x)  g(x) for all a  x  b:
Proof.
(a) Let h(x) = g(x)  f(x):Since f
0
(x)  g
0
(x) for all a < x < b we
have h
0
(x)  0 for all a < x < b.By part (b) of the previous theorem,
h(x) is nondecreasing.So,for any a  x  b we have h(a)  h(x):But
h(a) = f(a) g(a) = 0.Thus,h(x)  0 or f(x)  g(x):
(b) Now,for a  x  b we have h(x)  h(b):But h(b) = f(b) g(b) = 0:
Thus,h(x)  0 or f(x)  g(x):
Example 24.2
Use the Racetrack principle to show that e
x
 1 +x for all values of x:
2
Solution.
Let g(x) = x + 1 and f(x) = e
x
:We have g
0
(x) = 1  e
x
= f
0
(x) for all
x  0:Since g(0) = f(0) = 1;by the Racetrack principle (a) we can write
that 1 +x  e
x
for all x  0:On the other hand,f
0
(x)  g
0
(x) for all x  0:
Since f(0) = g(0) = 1,by the Racetrack principle (b) we can write e
x
 1+x
for all x  0:Combining the two results,we nd that e
x
 1+x for all values
of x
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