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Applications of Differentiation Notes
Sec. 3.1 Extrema on an Interval
Theorem: The Extreme Value Theorem
If f is continuous on [a, b] then f has both an absolute maximum and minimum on [a, b].
This just says that if a function is continuous in going from the point (a, f(a)) to (b, f(b)), then
somewhere in the interval is a highest point and a lowest point. These points may occur at the
endpoints. Otherwise, they occur at critical numbers.
A number c is a critical number for the function f if f ′(c ) = 0 or if f ′(c ) D.N.E . If a function is
defined on an interval, then the endpoints of that interval are considered critical numbers too.
Theorem:
If f has a local maximum or local minimum at x = c, then c is a critical number.
When considering extrema on a closed interval, find the critical numbers. Then evaluate the
function at the critical numbers and endpoints and compare the function values to determine
which are maximum and minimum values.
Example: Find the extrema of f ( x ) = 3x 4 − 4 x 3 on [-1, 2].
f ′( x ) = 12x 3 −12x 2 = 0
12x 2 ( x −1) = 0
x = 0 or x = 1
The critical numbers are x = 0 and x = 1.
f (−1) = 7,
f (0) = 0,
f (1) = −1,
f (2) = 16
Comparing the function values above, the point (1, -1) is the abs. minimum and the point (2, 16)
is the abs. max.
Example: Find the extrema of f ( x ) = 2sin x − cos2x on [0, 2π].
f ′( x ) = 2cos x + 2sin2x
Using the double-angle identity, this is
f ′( x ) = 2cos x + 2(2sin x cos x )
= 2cos x + 4 sin x cos x
= 2cos x (1+ 2sin x )
Solving, we get
2cos x = 0
1+ 2sin x = 0
cos x = 0
sin x = −1/2
x = π 2 , 3π 2 x = 7π 6 ,11π 6
Now evaluate the function at the endpoints and critical points.
f (0) = −1,
f π 2 = 3,
f 7π 6 = − 3 2 ,
f 3π 2 = −1, f 11π 6 = − 3 2 ,
f (2π) = −1
The maximum value of the function is 3, which is at the point (π/2, 3). The minimum value of
the function is –3/2, which occurs twice, at (7π/6, -3/2) and (11π/6, -3/2).
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( )
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Sec. 3.2 The Mean Value Theorem
Theorem: The Mean Value Theorem
If f is continuous on [a, b] and differentiable on (a, b), then there exists some number c in the
interval (a, b) such that
f (b) − f ( a)
.
f ′(c ) =
b−a
f (b) − f ( a)
is the slope of the line joining the endpoints of the
b−a
curve, ( a, f ( a)) and (b, f (b)) . The expression f ′(c ) is the slope of the curve at x = c. Both are
slopes and they’re equal so think “parallel.” This theorem says that there is at least one point on
the curve where the slope of the tangent line is equal to the slope of the line between the
endpoints.
Discussion: The expression
This theorem isn’t used much on its own, but it is essential to the proof of many theorems,
including the Fundamental Theorem of Calculus. Here is an example.
Example: For f ( x ) = 5 −
4
, find all c in the interval (1, 4) for which the mean value theorem
x
applies.
To use the Mean Value Theorem, you’ll need the derivative and the slope of the line between the
endpoints of the interval. Set them equal and solve.
4
and the slope of the line between the endpoints of the curve is
x2
f ( 4 ) − f (1) 4 −1
m=
=
= 1. To find the values for which the Mean Value Theorem applies, solve
4 −1
4 −1
the equation f ′( x ) = 1:
4
= 1 ⇒ 4 = x 2 ⇒ x = ±2
2
x
f ′( x ) =
Since x = -2 isn’t in the interval (1, 4), the only value for which the Mean Value Theorem applies
is x = 2.
Sec. 3.3 Increasing and Decreasing Functions and the First
Derivative Test
Most of this we already know. If the derivative is positive, the function is increasing. If the
derivative is negative, the function is decreasing.
Theorem: The First Derivative Test for Local Extrema
Let c be a critical number for f.
• If the derivative changes sign from positive to negative at x = c, then f(c) is a local
maximum value.
• If the derivative changes sign from negative to positive at x = c, then f(c) is a local
minimum value.
In summary, find the critical points and do a sign chart. From this sign chart you can tell where
the function is increasing and decreasing, and you can also determine local extrema.
Example:
1
x − sin x on (0, 2π).
2
Find the derivative and its critical numbers.
1
f ′( x ) = − cos x
2
1
− cos x = 0
2
1
cos x =
2
π
5π
x = and x =
3
3
Then do a sign chart on the open interval (0, 2π).
This function is decreasing on (0, π/3) and (5π/3, 2π). It is increasing on (π/3, 5π/3). The local
maximum value occurs at x = 5π/3 and the local minimum is at x = π/3.
Determine the local extrema of f ( x ) =
Example:
Find the local extrema of f ( x ) = ( x 2 − 4 )
2/3
.
Again, find the critical numbers by finding where the derivative is 0 or undefined.
−1/ 3
2
f ′( x ) = ( x 2 − 4 ) (2x )
3
4x
=
1/ 3
2
3( x − 4 )
The derivative is 0 when 4x = 0, or x = 0. It is undefined when x2 – 4 = 0, or x = ±2.
The sign chart shows that the function is increasing on (-2, 0) and (2, ∞), and decreasing on (-∞,
-2) and (0, 2). The local maximum values occur at x = 0. The local minimum values are at x =
±2. Specifically, the points are 0, 3 16 and (2, 0) and (-2, 0). REMEMBER: to find the y-values,
(
)
substitute back into the ORIGINAL function.
Sec. 3.4 Concavity and the Second Derivative Test
Definition: A function is concave up if its derivative is increasing. A function is concave down
if its derivative is decreasing.
To determine the concavity of a function, find the values for which the second derivative is zero
or undefined. Then do a sign chart. Points where concavity changes are called inflection points.
Example:
Determine where f ( x ) = 6( x 2 + 3) is concave upward or downward.
−1
f ′( x ) = −6( x 2 + 3)
−2
(2x )
= −12x ( x 2 + 3)
f ′′( x ) = 24 x ( x 2 + 3)
−2
−3
(2x ) −12( x 2 + 3)
= 48x 2 ( x 2 + 3) −12( x 2 + 3)
−3
= 12( x 2 + 3)
=
−3
12( 3x 2 − 3)
(x
2
+ 3)
[4 x − ( x
2
2`
−2
−2
+ 3)
]
3
The second derivative is zero when x = ±1. It is never undefined since x2 + 3 ≥ 3 for all x.
To determine concavity, do a sign chart (shown above at the right). From the chart you can tell
that f is concave up on (-∞, -1) and (1, ∞) and concave down on (-1, 1.) The inflection points are
(-1, 3/2) and (1, 3/2). REMEMBER: to find inflection points, substitute back into the original
function.
Theorem:
If (c, f(c)) is an inflection point of the graph of f, then either f ′′(c ) = 0 or f ′′(c ) D.N.E.
The second derivative can also be used to find local extrema.
Theorem: The Second Derivative Test for Local Extrema
Let f be a function such that f ′(c ) = 0 and the second derivative exists on an interval containing
c.
• If f ′′(c ) > 0 then f(c) is a local minimum.
• If f ′′(c ) < 0 then f(c) is a local maximum.
• If f ′′(c ) = 0 the test fails. Use the first derivative test.
If you think about it, if a function is concave up at a critical point then that point must be a local
minimum value. Similarly, if a function is concave down at a critical point then that point must
be a local maximum value.
Example:
Use the second derivative test to determine the local extrema for f ( x ) = −3x 5 + 5x 3 .
f ′( x ) = −15x 4 + 15x 2
The first derivative is zero when x = -1, 0, or 1.
f ′′( x ) = −60x 3 + 30x = −30x (2x 2 −1)
f ′′(−1) = 30(1) = +30
f ′′(0) = 0
f ′′(1) = −30(1) = −30
The second derivative test indicates that f(-1) is a local minimum and that f(1) is a local
maximum. It provides no information about what happens at x = 0. Using the first derivative test
you can determine that there is no local extreme value at x = 0.