MTH 3005 - Calculus I
Week 7: Derivative Theorems, Extrema, and Concavity
Adam Gilbert
Northeastern University
January 21, 2014
Objectives. After reviewing these notes the successful student will be prepared to
• State and use the definition of extema of a function on an open or closed interval
• Find extrema on a closed interval
• State and use Rolle’s Theorem
• State and use the Mean Value Theorem
• Determine intervals on which a function is increasing or decreasing
• Apply the first derivative test to find relative extrema of a function
• Determine the intervals on which a function is concave upward or concave downward
• Find any points of inflection of the graph of a function
• Apply the second derivative test to find relative extrema of a function
1
§3.1 Extrema on an Interval
Remark 1. Extrema are exactly what they sound like. Extreme points of functions. That
is, the highest of the high, and the lowest of the low. We are talking about maximums and
minimums here. The definitions follow.
Definition 1. Let f be a function defined on an interval I, containing the point x = c.
Then,
• f (c) is a maximum of f (x) on I if f (c) ≥ f (x) for all other x in the interval I.
• f (c) is a minimum of f (x) on I if f (c) ≤ f (x) for all other x in the interval I.
These minimums and maximums are called extrema.
Remark 2. Maximums of a function look like the tops of hills and minimums look like the
bottoms of valleys. It is possible for a function to have many hilltops and valley bottoms in
a closed interval. Each hilltop is called a local maximum and each valley bottom is called
a local minimum. The highest hilltop is called the global maximum and the lowest valley
bottom is called the global minimum.
Theorem 1 (The Extreme Value Theorem). Let f be a continuous function on a closed
interval [a, b]. Then f attains both a global maximum and global minimum on the interval.
Example 1. Given a continuous function, the EVT guarantees us the existence of maxima
and minima on a closed interval. Here are some of the possible situations for where the
maximum can occur:
y
y
maximum on interior
f
f
maximum at endpoint
a
b
x
a
2
b
x
Note 1. There is no extreme value theorem for open intervals. Maxima and minima cannot
be guaranteed if we only know that the function f is continuous on the interval (a, b). The
following graph illustrates this concept.
y
f
a
b
x
This function is continuous on (a, b), but attains neither a minimum nor a maximum on the
interval.
Remark 3. Now that we know about local and global minima and maxima, as well as about
when they are guaranteed to exist, we would really like to know how to find them.
Example 2. Take a moment and try to draw a few continuous functions over the interval
[1, 4]. Note where there minimums and maximums occur. Be as creative as you can - the
only restriction is that your functions must be continuous along the entire interval. Draw
your own before looking at my examples below. I bet that, at their maximums, your functions
look very much like mine.
y
y
y
f
f
1
4
x
Horizontal Tangent Line
1
4
Maximum at Endpoint
x
1
4
x
Sharp Corner
Remark 4. The same can be said for minima. Try it!
Theorem 2. If f (x) is a continuous function over an interval I, then any minima or maxima
occur at one of the following:
• A point c, where f 0 (c) = 0
• A point c, where f 0 (c) is undefined
3
• If the interval I is closed, at an endpoint of I
Definition 2. A point x = c is called a critical point if f 0 (c) = 0 or f 0 (c) is undefined.
Note 2. If I is an open interval, then if f (c) is a mimimum or a maximum of f , then c
must be a critical point.
Strategy 1 (Finding Minima and Maxima). Given a continuous function f (x), use the
following strategy for finding minima and maxima on a closed interval [a, b]:
1. Find f 0 (x)
2. Locate all critical points (solve f 0 (x) = 0 and find all points where f 0 (x) is undefined)
3. Evaluate f at each of the critical points
4. Evaluate f at each of the endpoints of the interval (find f (a) and f (b))
5. The least of these values is the minimum and the largest is the maximum
Example 3. Find the extrema of f (x) = 6x4 − 8x3 on [−1, 2].
Solution. We begin by finding f 0 (x)
f 0 (x) = 24x3 − 24x2
And now, find all critical points for f (x) by solving f 0 (x) = 0 and finding any values such
that f 0 (x) is undefined.
24x3 − 24x2 = 0
=⇒ 24 x3 − x2 = 0
=⇒ x3 − x2 = 0
=⇒ x2 (x − 1) = 0
=⇒ x = 0 or x = 1
Note that since f 0 (x) is a polynomial, it is never undefined! Therefore, we have found exactly
two critical points. Now we evaluate the function at each of the critical points as well as at
the endpoints of the interval!
Critical Point
−1
0
1
2
Function Value
f (−1) = 14
f (0) = 0
f (1) = −2
f (2) = 32
4
Recall that the largest value is the maximum, so f (x) has a maximum at x = 2. The smallest
value gives the minimum, so the function f (x) has a minimum at x = 1. The locations of
the extrema are as follows:
Minimum: (1, −2)
Maximum: (2, 32)
H
Example 4. Find the extrema of f (x) =
x4 2x3 3x2
−
−
+ 5 on the interval [−2, 2].
4
3
2
Solution. We find f 0 (x):
f 0 (x) = x3 − 2x2 − 3x
And, now we find the critical points:
x3 − 2x2 − 3x = 0
=⇒ x x2 − 2x − 3 = 0
=⇒ x (x − 3) (x + 1) = 0
=⇒ x = −1, 0, or 3
Warning 1. Notice that x = 3 is not in our interval, so we do not consider it moving
forward. Beware of extraneous critical points!
Similar the the previous problem, since f 0 (x) is a polynomial, there are no points which
make f 0 (x) undefined, so we have just two critical points: x = −1 and x = 0.
Now, we evaluate f (x) at each of the critical points and at the endpoints of the interval.
Critical Point
−2
−1
0
2
Function Value
f (−2) = 25
3
53
f (−1) = 12
f (0) = 5
f (2) = − 73
On the interval [−2, 2], the function f (x) a mimimum at the point (2, −7/3) and a maximum
at the point (−2, 25/3).
H
√
Example 5. Find the extrema of f (x) = x − x on the interval [0, 4].
5
Solution. We find f 0 (x)
1
f 0 (x) = 1 − √
2 x
Note that f 0 (x) is undefined at x = 0, which gives us a critical point. And now we find any
remaining critical points by solving f 0 (x) = 0. Let x 6= 0. Then,
1
1− √ =0
2 x
1
√ =1
2 x
√
=⇒ 2 x = 1
√
1
=⇒ x =
2
1
=⇒ x =
4
So we have found two critical points. One at x = 0 and another at x = 14 . Now, we examine
the function values at the critical points as well as at the endpoints of the interval we are
interested in:
=⇒
Critical Point
0
1
4
1
Function Value
f (0)
= 01
1
f 4 = −4
f (4) = 2
From the table, we have found the following about the extrema of f (x) on the interval [0, 1]
Minimum at: 14 , − 14
Maximum at: (4, 2)
H
Example 6. Find the extrema of f (x) = 3 cos(2x) on the interval [0, 2π].
Solution. Find f 0 (x)
f 0 (x) = −3 sin (2x) · 2
=⇒ f 0 (x) = −6 sin (2x)
Now, we wish to find critical points by solving f 0 (x) = 0 and finding any points such that
f 0 (x) is undefined:
−6 sin (2x) = 0
6
=⇒ sin (2x) = 0
=⇒ 2x = k · π (Recall that the Sine function is 0 at integer multiples of π)
π
=⇒ x = · k
2
Recall that the domain of the sine function is all real numbers, so there are no additional
critical points. Now, we need to find the critical points in our interval and put them into
our table:
Critical Point
0
π
2
π
3π
2
2π
Function Value
f (0)
=3
π
f 2 = −3
f (π)
=3
3π
f 2 = −3
f (2π) = 3
Note that here, there are multiple critical points which give the same function values. In
this situation we have multiple minima and maxima.
Minimums at: (π/2, −3) ; (3π/2, −3)
Maximums at: (0, 3) ; (π, 3) ; (2π, 3)
H
Remark 5 (Recap). In this discussion we have found the following:
• Continuous functions are guaranteed to achieve their extreme values on a closed interval
• Critical points for a function f (x) are values where f 0 (x) is either 0 or undefined
• All extrema occur at either a critical point or an endpoint of an interval
• There may be multiple minimums or maximums if the function values dictate that it is
the case
§3.2 Rolle’s Theorem and the Mean Value Theorem
Theorem 3 (Rolle’s Theorem). Let f (x) be a continuous function on the interval [a, b] such
that f (a) = f (b) and f (x) is differentiable on the open interval (a, b), then there exists at
least one c in (a, b) such that f 0 (c) = 0.
7
Remark 6. Rolle’s Theorem may seem intimidating to read, however, draw a picture of it.
If you have a smooth function, and you know that the function values at the endpoints of
the interval [a, b] are equal, then there must be at least one point inside the interval such
that the function has a horizontal tangent line. Think about it: if you begin from f (a) and
draw a smooth curve that increases, in order for the curve to connect to f (b), you must turn
back around, so there must be a point at which the line tangent to your curve is horizontal
(remember, your curve must be smooth, so no sharp corners are allowed). The same is true
if you begin from f (a) and draw a smooth curve that decreases.
Theorem 4 (Mean Value Theorem). Let f (x) be continuous on [a, b] and differentiable on
the open interval (a, b), then there is at least one c in (a, b) such that
f 0 (c) =
f (b) − f (a)
b−a
√
Example 7. Use Rolle’s Theorem to show that the function f (x) = x x − 4 has a critical
point on the interval [0, 4].
Solution.
Note that f (0) = 0 and f (4) = 0. Thus, we have a situation where
the
√
function values at the endpoints of our interval are equal. Also, note that f (x) = x 4 − x is
differentiable on the interval (0, 4) (we could find its derivative). By Rolle’s Theorem, there
must exist at least one value c in (0, 4) such that f 0 (c) = 0. That is, there exists at least one
critical point for f (x) on this interval!
H
Remark 7. To explain the Mean Value Theorem, think of two police cars sitting one mile
apart on the highway. The officers in the cars radio back and forth to one another. The first
officer takes particular notice of a car that is traveling 30mph as it passes the first officer.
The officer radios his companion down the road about the car. The second officer sees the car
go by 1 minute later, still traveling at 30mph. What has happened? We all know that the car
must have sped up in between the two police officers’ locations, because otherwise, it would
have taken the car a full two minutes before it passed the second police car. The mean value
theorem guarantees us that there was at least one point between the police officers where the
car was traveling at 60mph.
Example 8. Use the Mean Value Theorem to show that, for the function f (x) = x2 , there
exists a c in (0, 1) so that f 0 (c) = 1. Then find it.
Solution. Notice that f (0) = 0 and f (1) = 1. Then,
f (1) − f (0)
1−0
(the slope of the secant line)
=1
8
Now, by the MVT, since f (x) = x2 is differentiable on (0, 1), we are guaranteed the existence
of at least one c such that f 0 (c) is equal to the slope of the tangent line. To find c, we want
to solve the equation f 0 (x) = 1:
f 0 (x) = 1
=⇒ 2x = 1
1
=⇒ x =
2
2
So the slope of the function f (x) = x at the point where x =
the secant line through the points (0, f (0)) and (1, f (1)).
1
2
is the same as the slope of
H
Warning 2. Be mindful of the requirements for the use of these theorems. If a function
does not satisfy the hypotheses of a theorem, then you cannot apply the result of the theorem!
Remember, that these theorems require continuous and differentiable functions.
§3.3 Increasing, Decreasing, and the First Derivative
Test
Definition 3. A function is said to be increasing on an interval I if for every x1 and x2 in
I with x2 > x1 , we have f (x2 ) > f (x1 ).
Remark 8. The definition above is just a technical way of saying that a function is increasing
if as the values of x get larger, the function values also become larger.
Definition 4. A function is said to be decreasing on an interval I if for every x1 and x2 in
I with x2 > x1 , we have f (x2 ) < f (x1 ).
Remark 9. Similarly to the definition for increasing, the definition above is a technical way
to say that as the x values get larger, the function values become smaller.
Definition 5. A function f (x) is said to be constant on an interval I if for every x1 and x2
in I, we have f (x1 ) = f (x2 ).
Example 9. Using the graph of the function f (x) below, estimate the intervals over which
f (x) is increasing, decreasing, and constant (assume both axes are using a 1-unit scale).
y
f (x)
x
9
Solution. We can do this by looking at the graph.
Increasing: [−5, −2] ∪ [0, 1] ∪ [2, 5]
Decreasing: [−2, 0]
Constant: [1, 2]
H
As long as a function is differentiable its derivative will tell us whether a the function is
increasing, decreasing, or constant on an interval!
Theorem 5 (Increasing, Decreasing, and Constant). Let f (x) be a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then,
• f (x) is increasing on I if for every x in I, f 0 (x) > 0
• f (x) is decreasing on I if for every x in I, f 0 (x) < 0
• f (x) is constant on I if for every x in I, f 0 (x) = 0
Corollary 1 (Increasing, Decreasing, and Critical). Let f (x) be differentiable at x = c.
Then,
• If f 0 (c) > 0, then f (x) is increasing at x = c
• If f 0 (c) < 0, then f (x) is decreasing at x = c
• If f 0 (c) = 0, then x = c is a critical point of f (x)
Example 10. Let f (x) = x sin(x). Determine whether the function is increasing, decreasing,
or has a critical point at each of the following values of x:
a) x = 0
b) x = π
c) x =
5π
2
Solution. We can determine the behavior of the function at each of the points by evaluating
its derivative at each. Thus, we begin by computing the derivative:
f (x) = x sin(x)
=⇒ f 0 (x) = x cos(x) + sin(x)
10
Now, we can evaluate the derivative at each of the points in question:
f 0 (0) = 0 · cos(0) − sin(0)
=⇒ f 0 (0) = 0
=⇒ x = 0 is a critical point for f (x)
f 0 (π) = π · cos (π) + sin (π)
=⇒ f 0 (π) = −π
=⇒ f (x) is decreasing at x = π
5π
5π
5π
5π
0
=
· cos
+ sin
f
2
2
2
2
5π
=⇒ f 0
=1
2
=⇒ f (x) is increasing at x =
5π
2
H
The derivative can tell us a lot about the character of a function. We have seen that the
derivative tells us about the slope of a function. One of the most important uses of the
derivative is in identifying maxima and minima of functions. We have already seen that the
derivative can be used to identify critical points for a function.
After finding the critical points for a function in previous examples we were able to find
the global minimum and global maximum by comparing the function values at each of the
critical points and choosing the smallest and largest respectively. Sometimes, though, we are
interested in knowing about all of the local extrema rather than just their global versions.
Unfortunately we can’t identify local extrema using the method where we compare the
function values. So, what makes a minimum and what makes a maximum? Let’s take a
look!
y
y
f
local
minimum
local
maximum
f
p
x
p
11
x
The derivative can be used to discuss what is going on near local extrema.
y
y
f
f 0 (x) > 0
f 0 (x) < 0
local
minimum
f 0 (x) < 0
f 0 (x) > 0
local
maximum
f
p
x
p
x
The images above depict what is called the First Derivative Test for Minima and Maxima.
Theorem 6 (First Derivative Test). Let c be a critical point for the function f (x) which
is continuous on an open interval I which contains c, and is differentiable on the interval
(except possibly at c). Then f (c) can be classified as follows:
• If f 0 (x) changes from negative to positive at c, then f has a relative minimum at
(c, f (c))
• If f 0 (x) changes from positive to negative at c, then f has a relative maximum at
(c, f (c))
• If f 0 (x) is either positive on both sides of c or negative on both sides of c, then f has
neither a minimum nor a maximum at c.
Strategy 2 (Using the First Derivative Test). The following strategy can be employed when
locating and classifying relative extrema for a differentiable function f (x):
a. Find all critical points for f (x)
b. For each critical point (c), select a value (a) to the left of it and a value (b) to the right
of it and evaluate f 0 (a) and f 0 (b)
F Be sure to choose values that are relatively close to the critical point you are
examining. For example, consider a function with two critical points c1 and c2 ,
where c2 > c1 . Then, the point you choose to the right of c1 must be between c1
and c2 !
c. Use the first derivative test to classify each critical point.
Example 11. Find and classify all extrema for the function f (x) =
12
x4 2x3 3x2
+
−
+ 18.
4
3
2
Solution. We begin by finding the critical points for f (x)
f 0 (x) = x3 + 2x2 − 3x
Since the derivative is a polynomial there are no critical points making the derivative undefined, so we find all critical points by solving f 0 (x) = 0.
x3 + 2x2 − 3x = 0
=⇒ x x2 + 2x − 3 = 0
=⇒ x (x + 3) (x − 1) = 0
=⇒ x = −3, 0, and 1 are critical points
−3
0
x
1
Now we need to evaluate the derivative to the left and to the right of each critical point, so
we pick any value we like in each of the regions created between critical points:
−4
1/2
−2
−3
0
2
1
x
We evaluate the derivative at each of our chosen points:
f 0 (−4) = −4 (−4 + 3) (−4 − 1)
= −20
(< 0)
f 0 (−2) = −2 (−2 + 3) (−2 − 1)
= 6 (> 0)
1 1
1
0
f (1/2) =
+3
−1
2 2
2
1 7
−1
=
2 2
2
−7
=
(< 0)
8
f 0 (2) = 2 (2 + 3) (2 − 1)
= 10
(> 0)
Now that we have evaluated the derivative in each region we can use the first derivative test
to categorize each of our critical points:
13
1/2
−2
2
x
1
0
>
x)
f 0(
<
>
0
0
0
f 0(
x)
f 0(
x)
<
0
−3
f 0(
x)
−4
Since the derivative changes from positive to negative at x = 0, the function f (x) must have
a relative maximum here.
Since the derivative changes from negative to positive at −3, the function f (x) has a relative
minimum at x = −3. Similarly, the function has a relative minimum at x = 1.
H
§3.4 Concavity and the Second Derivative Test
Remark 10. The concavity of a function describes which way the function “bends”. A
function which is concave upwards has the shape of a vertical cross section of a cereal bowl
(or at least part of a cross section), while a function which is concave down looks more like
part of an umbrella.
Note 3. Using the ideas of increasing and decreasing in conjunction with this new idea of
concavity, we can classify the basic (local) “geography” of most of our functions. That is, if
you look at a small portion of the graph of a twice differentiable function it will fall into one
of the categories shown below (or it will be constant):
14
Increasing and
Concave Up
Decreasing and
Concave Up
Increasing and
Concave Down
Increasing and
Neutral Concavity
Decreasing and
Concave Down
Decreasing and
Neutral Concavity
Definition 6 (Concavity). Let f be differentiable on an open interval I. The graph of f is
concave upward on I when f 0 is increasing on the interval, and is concave down on I when
f 0 is decreasing on the interval.
Theorem 7 (Test for Concavity). Let f be a function whose second derivative exists on the
open interval I:
• If f 00 (x) > 0 for all x in I, then f (x) is concave up on I
• If f 00 (x) < 0 for all x in I, then f (x) is concave down on I
Example 12. Determine the concavity of f (x) = 5x2 − 4x + 8 on all of R.
Solution.
f (x) = 5x2 − 4x + 8
=⇒ f 0 (x) = 10x − 4
=⇒ f 00 (x) = 10
Since f 00 (x) > 0 for all values of x, the function f (x) is concave upwards.
H
Definition 7 (Inflection Point). Let f be a function whose second derivative exists on an
open interval I containing c. If the concavity of f changes from negative to positive (or
vice-versa) at x = c and the point (c, f (c)) exists, then that point is called an inflection
point.
15
Theorem 8. If the point (c, f (c)) is an inflection point, then f 00 (c) = 0 or f 00 (c) is undefined.
Example 13. Find all possible points of inflection for the function f (x) = 3x5 − 9x4 .
Solution.
f (x) = 3x5 − 9x4
=⇒ f 0 (x) = 15x4 − 36x3
=⇒ f 00 (x) = 60x3 − 108x2
Now, we want to find all values such that f 00 (x) = 0, so we set up and solve this equation.
f 00 (x) = 0
=⇒ 60x3 − 108x2 = 0
=⇒ 12x2 (5x − 9) = 0
9
are points of inflection
5
Note that should test points to ensure that the concavity does indeed change through these
values. However, in the interest of brevity, we will omit the verification, and you may
continue to use this practice in your work for this course.
H
=⇒ x = 0, and x =
The second derivative (f 00 (x)) is also useful in determining the character of the function
(f (x)) at its critical points. That is, we can use the second derivative to determine whether
critical points of the function f (x) are local minima or local maxima instead of using the
first derivative test. The test you choose to use is a matter of personal preference, there are
situations where the Second Derivative Test is inconclusive whereas the First Derivative Test
always yields a result. In the interest of completeness, the second derivative test is stated
below.
Theorem 9 (Second Derivative Test). Let f be a function such that f 0 (c) = 0 and the second
derivative of f exists on an open interval containing c. Then,
• If f 00 (c) > 0, then f has a relative minimum at c
• If f 00 (c) < 0, then f has a relative maximum at c
• If f 00 (c) = 0, then the test fails and you should consult the first derivative test
Example 14. Use the second derivative test to classify all relative extrema for the function
x2 − 1
. Also, find all of the points of inflection for the function.
f (x) =
2x − 1
16
Solution. Since we are asked to classify the relative extrema, we begin by finding all of the
critical points for f (x)
x2 − 1
f (x) =
2x − 1
=⇒ f 0 (x) =
(2x − 1) (2x) − (x2 − 1) (2)
(2x − 1)2
=⇒ f 0 (x) =
2x2 − 2x + 2
(2x − 1)2
Note that f 0 (x) is undefined when x = 1/2, so we have at least one critical point. We can
find the remaining critical points by solving f 0 (x) = 0:
f 0 (x) = 0
=⇒
2x2 − 2x + 2
=0
(2x − 1)2
=⇒ 2x2 − 2x + 2 = 0
=⇒ x2 − x + 1 = 0 which has no real solutions
Thus we have found a single critical. The location is at 1/2.
In order to classify the critical points we are asked to use the second derivative test, so we
must calculate f 00 (x).
2x2 − 2x + 2
=⇒ f 0 (x) =
(2x − 1)2
=⇒ f 00 (x) =
(2x − 1)2 (4x − 2) − (2x2 − 2x + 2) (2 (2x − 1) · (2))
2
(2x − 1)2
(2x − 1)2 (4x − 2) − 4 (2x2 − 2x + 2) (2x − 1)
=⇒ f (x) =
(2x − 1)4
00
=⇒ f 00 (x) =
(2x − 1) [(2x − 1) (4x − 2) − 4 (2x2 − 2x + 2)]
(2x − 1)4
(2x − 1) (4x − 2) − 4 (2x2 − 2x + 2)
=⇒ f (x) =
(2x − 1)3
00
=⇒ f 00 (x) =
(8x2 − 8x + 2) − (8x2 − 8x + 8)
(2x − 1)3
=⇒ f 00 (x) =
17
−6
(2x − 1)3
Now we can evaluate f 00 (x) for our critical point:
1
00
f
is undefined, so the test fails
2
H
Note 4. The second derivative test won’t always fail.
x4 4x3
f (x) =
+
− 16x2 + 9.
4
3
18
Try it on the function
Closing
Homework. For homework, please try the following problems from the textbook:
Section
§3.1
§3.2
§3.3
§3.4
Suggested Problems
1-43 (odd), 60, 61, 62
1-51 (odd), 59, 63, 69, 71
1-47 (odd), 57-62, 79, 80
3-41 (odd), 53, 56, 63, 64
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