3d. Mean Value Theorem Handout.
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Topic 3 – Differentiation
d) Mean Value Theorem
Assigned Problems: pgs 100-101 (1-3, 5, 8, 10)
Recall:
• When we first looked at the definition of the derivative, we started by looking at the slope of secant
lines
•
We looked at the line connecting a and b, two values surrounding c, and we moved this up until it
was tangent to the curve at the point c
What conditions were important to us at that time?
The function needs to be continuous. For example, we would have problems if we had a function
that looked like:
Actually, it should even be continuous at a and b as well.
It should be differentiable on the interval (a,b), or we could run into a problem like:
•
•
where the derivative at c is not even defined.
So to be able to do this in a general sense, we need to specify that f is continuous on [a,b] and
differentiable on (a,b)
What would be the slope of this secant line if f(a) = f(b)?
C. Bellomo, revised 28-Mar-06
3d. Mean Value Theorem Handout.
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The Theorems:
• Rolle’s Theorem: Let f be a function that is continuous on [a,b] and differentiable on (a,b) with
f(a) = f(b). Then there is a c in (a,b) with f ′(c) = 0 .
Proof:
Assume (case 1) that f(x) is a constant function, i.e. f(x) = k
Then f ′( x) = 0 ∀x ∈ [a, b] , so c is any such value
A previous theorem states that if f :[a, b] → Re is continuous on [a, b] then f has an absolute
max and min on [a, b] (Thm 4.2)
By definition, an absolute max or min is a local max or min. So since f has an absolute max and
min on [a, b], f has a local max and min on [a, b]
A previous proposition states that if f has a local max or min at an interior point c of I, and f is
differentiable at c, then f ′(c) = 0 .
• Generalized Mean Value Theorem: Let f and g be continuous functions on [a, b] that are
differentiable on (a, b). Then there is a c in (a, b) such that [ f (b) − f (a )] g ′(c) = [ g (b) − g (a )] f ′(c)
Proof:
Define F :[a, b] → Re by F ( x) = [ f (b) − f (a )] g ( x ) − [ g (b) − g (a )] f ( x)
Then, F is continuous on [a, b] and differentiable on (a, b)
F (a ) = [ f (b) − f ( a )] g ( a) − [ g (b) − g ( a)] f ( a)
= f (b) g (a ) − f (a ) g (a ) − g (b) f (a ) + g (a ) f (a ) = f (b) g (a ) − g (b) f (a)
F (b) = [ f (b) − f (a)] g (b) − [ g (b) − g (a )] f (b) = − f (a ) g (b) + g (a ) f (b)
So F(a) = F(b)
By Rolles Theorem, F ′(c) = 0 for some c in (a, b)
Differentiating we find that F ′( x) = [ f (b) − f ( a)] g ′( x) − [ g (b) − g ( a)] f ′( x)
⇒ F ′(c) = [ f (b) − f ( a )] g ′(c) − [ g (b) − g ( a)] f ′(c) = 0
⇒ [ f (b) − f (a )] g ′( x) = [ g (b) − g (a )] f ′( x)
• Mean Value Theorem: Let f be a function that is continuous on [a,b] and differentiable on (a,b).
f (b ) − f ( a )
.
Then there is a c in (a,b) with f ′(c) =
b−a
Proof:
f satisfies the continuity and differentiability conditions of Rolle’s Theorem
f (b ) − f ( a )
The equation of the line through (a,b) and (f(a), f(b)) is y ( x) = f ( a ) +
( x − a)
b−a
Let h( x ) = f ( x ) − y ( x)
By construction, h(x) is continuous on [a,b] and differentiable on (a,b)
Therefore, there is a c in (a,b) with h′(c ) = 0 by Rolle’s Theorem
f (b) − f ( a )
f (b ) − f ( a )
.
Now, h′( x ) = f ′( x) − y ′( x) = f ′( x ) −
= 0 ⇒ f ′( x) =
b−a
b−a
C. Bellomo, revised 28-Mar-06
3d. Mean Value Theorem Handout.
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•
Corollary: If f is continuous on [a,b] and differentiable on (a,b) then
f ′( x) = 0 ∀x ∈ [a, b] ⇒ f ( x) = k on [a, b]
f ′( x) ≥ 0 ∀x ∈ [a, b] ⇒ f ( x) is monotone increasing on [ a, b]
f ′( x) > 0 ∀x ∈ [a, b] ⇒ f ( x) is strictly increasing on [a, b]
f ′( x) ≤ 0 ∀x ∈ [a, b] ⇒ f ( x) is monotone decreasing on [a, b]
f ′( x) < 0 ∀x ∈ [a, b] ⇒ f ( x) is strictly increasing on [ a, b]
Proof:
For x1 < x2 in [a, b] , the MVT implies that c in (a,b) with f ′(c) ( x2 − x1 ) = f ( x2 ) − f ( x1 )
If f ′( x) = 0 ∀x ∈ [ x1 , x2 ] ⇒ f ′(c) = 0 on [a, b] , which implies f ( x2 ) = f ( x1 )
If f ′( x) ≥ 0 ∀x ∈ [ x1 , x2 ] ⇒ f ′(c) ≥ 0 on [a, b] , which implies f ( x2 ) ≥ f ( x1 )
etc.
Examples:
• Example. Let f ( x) = x and g ( x) = x 2 on [0,1] , show it satisfies the generalized MVT and find c
f and g are continuous on [0,1] and differentiable on (0,1)
f(1) = g(1) = 1, and f(0) = g(0) = 0
f(1) – f(0) = g(1) – g(0) = 1
The generalized MVT implies that there is a c in (a, b) st [ f (1) − f (0)] g ′(c) = [ g (1) − g (0)] f ′(c)
1
Therefore, [1] ⋅ 2c = [1] ⋅1 ⇒ c =
2
•
Example. Verify f ( x) = x x + 6, [−6, 0] satisfies Rolles and find all such c’s.
f(x) is continuous on [–6,0] and is differentiable on (–6,0)
f (−6) = (−6) −6 + 6 = 0, f (0) = (0) 0 + 6 = 0 . So it satisfies Rolles.
x
3( x + 4)
+ x+6 =
f ′( x) =
2 x+6
2 x+6
f ′( x ) = 0 when x = −4
C. Bellomo, revised 28-Mar-06
3d. Mean Value Theorem Handout.
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•
Example. Let f ( x) = ( x − 1) −2 . Show that f(0)=f(2) but there is no number c in (0,2) so f ′(c) = 0 .
Why not?
First of all it is important to note that f has a vertical asymptote at x = 1, which is in [0,2].
Therefore, f is not continuous on [0,2], which is a requirement for Rolles
f (0) = (0 − 1) −2 = 1, f (2) = (2 − 1) −2 = 1
So f(0) = f(2)
−2
, which is never zero.
f ′( x) = −2( x − 1) −3 =
( x − 1)3
•
Example. Verify that f ( x) =
x
satisfies the MVT on [1,4]. Find c
x+2
f is continuous on [1,4] (there is an asymptote, but it is at x = -2)
( x + 2) − x(1)
2
f is differentiable on (1,4), and f ′( x) =
=
2
( x + 2)
( x + 2) 2
1
1
4
4 2
f (1) =
= , f (1) =
= =
1+ 2 3
4+2 6 3
f (b) − f (a ) 2 / 3 − 1/ 3 1/ 3 1
=
=
=
b−a
4 −1
3
9
2
1
= ⇒ ( x + 2) 2 = 18 ⇒ x = ± 18 − 2 = 2.24, − 6.24
2
( x + 2)
9
On the interval given, we find c = 2.24
C. Bellomo, revised 28-Mar-06